Cracking the AP Physics 1 Exam 2018
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ACKNOWLEDGMENTS
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Contents
Cover
Title Page
Copyright
Acknowledgments
Register Your Book Online!
Part I: Using This Book to Improve Your AP Score
Part II: Practice Test 1
Part III: About the AP Physics 2 Exam
Part IV: Test-Taking Strategies for the AP Physics 2 Exam
1 How to Approach Multiple-Choice Questions
2 How to Approach Free-Response Questions
Part V: Content Review for the AP Physics 2 Exam
3 Fluid Mechanics
4 Thermal Physics
5 Electronic Forces and Fields
6 Electronic Potential and Capacitance
7 Direct Current Circuits
8 Magnetic Forces and Fields
9 Electromagnetic Induction
10 Optics
11 Modern Physics
12 Solutions to Chapter Review Questions
Part VI: Practice Test 2
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Part I
Using This Book to Improve Your AP Score
• Preview: Your Knowledge, Your Expectations
• Your Guide to Using This Book
• How to Begin
PREVIEW: YOUR KNOWLEDGE, YOUR EXPECTATIONS
Your route to a high score on the AP Physics 2 Exam will depend on how you plan to use this book. Please respond to the following questions.
1. Rate your level of confidence about your knowledge of the content tested by the AP Physics 2 Exam.
A. Very confident—I know it all
B. I’m pretty confident, but there are topics for which I could use help
C. Not confident—I need quite a bit of support
D. I’m not sure
2. If you have a goal score in mind, highlight your goal score for the AP Physics 2 Exam.
5 4 3 2 1 I’m not sure yet
3. What do you expect to learn from this book? Highlight all that apply to you.
A. A general overview of the test and what to expect
B. Strategies for how to approach the test
C. The content tested by this exam
D. I’m not sure yet
YOUR GUIDE TO USING THIS BOOK
This book is organized to provide as much—or as little—support as you need, so you can use this book in whatever way will be most helpful for improving your score on the AP Physics 2 Exam.
• The remainder of Part I will provide guidance on how to use this book and help you determine your strengths and weaknesses.
• Part II of this book contains your first practice test, answers and explanations, and a scoring guide. (Bubble sheets can downloaded here.) This is where you should begin your test preparation in order to realistically determine:
your starting point right now
which question types you’re ready for and which you might need to practice
which content topics you are familiar with and which you will want to carefully review
Once you have nailed down your strengths and weaknesses with regard to this exam, you can focus your preparation and be efficient with your time.
• Part III of this book will:
provide information about the structure, scoring, and content of the AP Physics 2 Exam
help you to make a study plan
point you toward additional resources
• Part IV of this book will explore the following strategies:
how to attack multiple-choice questions
how to approach free-response questions
how to manage your time to maximize the number of points available to you
• Part V of this book covers the content you need for the AP Physics 2 Exam.
• Part VI of this book contains Practice Test 2, its answers and explanations, and a scoring guide. (Bubble sheets can be downloaded here.) If you skipped Practice Test 1, we recommend that you do both (with at least a day or two between them) so that you can compare your progress between the two. Additionally, this will help to identify any external issues: if you get a certain type of question wrong both times, you probably need to review it. If you only got it wrong once, you may have run out of time or been distracted by something. In either case, this will allow you to focus on the factors that caused the discrepancy in scores and to be as prepared as possible on the day of the test.
You may choose to use some parts of this book over others, or you may work through the entire book. This will depend on your needs and how much time you have. Let’s now look at how to make this determination.
Equation Sheet Sidebars
Throughout the content review chapters of this book you’ll see the words “Equation Sheet” pop up in the sidebars. These words will be seen next to physics equations that are featured on the equation sheet that the College Board gives you during your AP Physics 2 Exam. So these are equations that are extremely important for you to be familiar with.
HOW TO BEGIN
1. Take Practice Test 1
Before you can decide how to use this book, you need to take a practice test. Doing so will give you insight into your strengths and weaknesses, and the test will also help you make an effective study plan. If you’re feeling test-phobic, remind yourself that a practice test is a tool for diagnosing yourself—it’s not how well you do that matters but how you use information gleaned from your performance to guide your preparation.
So, before you read further, take AP Physics 2 Practice Test 1 starting at this page of this book. Be sure to do so in one sitting, following the instructions that appear before the test.
2. Check Your Answers
Using the answer key on this page, count how many multiple-choice questions you got right and how many you missed. Don’t worry about the explanations for now, and don’t worry about why you missed questions. We’ll get to that soon.
3. Reflect on the Test
After you take your first test, respond to the following questions:
• How much time did you spend on the multiple-choice questions?
• How much time did you spend on each free-response question?
• How many multiple-choice questions did you miss?
• Do you feel you had the knowledge to address the subject matter of the essays?
• Do you feel you wrote well-organized, thoughtful essays?
4. Read Part III of this Book and Complete the Self-Evaluation
Part III will provide information on how the test is structured and scored. It will also set out areas of content that are tested.
As you read Part III, re-evaluate your answers to the questions above. At the end of Part III, you will revisit the questions on the previous page and refine your answers to them. You will then be able to make a study plan, based on your needs and time available, that will allow you to use this book most effectively.
5. Engage with Parts IV and V as Needed
Notice the word engage. You’ll get more out of this book if you use it intentionally than if you read it passively, hoping for an improved score through osmosis.
Strategy chapters will help you think about your approach to the question types on this exam. Part IV will open with a reminder to think about how you approach questions now and then close with a reflection section asking you to think about how/whether you will change your approach in the future.
Content chapters are designed to provide a review of the content tested on the AP Physics 2 Exam, including the level of detail you need to know and how the content is tested. You will have the opportunity to assess your mastery of the content of each chapter through test-appropriate questions and a reflection section.
6. Take Practice Test 2 and Assess Your Performance
Once you feel you have developed the strategies you need and gained the knowledge you lacked, you should take Practice Test 2. You should do so in one sitting, following the instructions at the beginning of the test.
When you are done, check your answers to the multiple-choice sections. See if a teacher will read your essays and provide feedback.
Once you have taken the test, reflect on what areas you still need to work on, and revisit the chapters in this book that address those deficiencies. Through this type of reflection and engagement, you will continue to improve.
7. Keep Working
After you have revisited certain chapters in this book, continue the process of testing, reflecting, and engaging with the next practice test in this book. Consider what additional work you need to do and how you will change your strategic approach to different parts of the test. You can continue to explore areas that can stand to be improved and engage in those areas right up to the day of the test. As we will discuss in Part III, there are other resources available to you, including a wealth of information online at AP Students.
Part II
Practice Test 1
• Practice Test 1
• Practice Test 1: Answers and Explanations
Practice Test 1
Click here to download a PDF of Practice Test 1.
The Exam
AP® Physics 2 Exam
SECTION I: Multiple-Choice Questions
DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.
At a Glance
Total Time
90 minutes
Number of Questions
50
Percent of Total Grade
50%
Writing Instrument
Pen required
Instructions
Section I of this examination contains 50 multiple-choice questions. Fill in only the ovals for numbers 1 through 50 on your answer sheet.
CALCULATORS MAY BE USED IN BOTH SECTIONS OF THE EXAMINATION.
Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. After you have decided which of the suggested answers is best, completely fill in the corresponding oval on the answer sheet. Give only one answer to each question. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer.
Sample Question
Chicago is a
(A) state
(B) city
(C) country
(D) continent
Sample Answer
Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all the multiple-choice questions.
About Guessing
Many candidates wonder whether or not to guess the answers to questions about which they are not certain. Multiple choice scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers, and no points are awarded for unanswered questions. Because points are not deducted for incorrect answers, you are encouraged to answer all multiple-choice questions. On any questions you do not know the answer to, you should eliminate as many choices as you can, and then select the best answer among the remaining choices.
AP PHYSICS 2
SECTION I
Time—90 minutes
50 Questions
Note: To simplify calculations, you may use g = 10 m/s2 in all problems.
Directions: Each of the questions or incomplete statements below is followed by four suggested answers or completions. Select the one that is best in each case and mark it on your sheet.
Questions 1 and 2 refer to the following situation.
A proton is travelling along a straight line at a constant speed through a uniform electric field near the surface of the Earth.
1. Which of the following choices correctly describes the direction of the electric field and the relative magnitudes of the electric and gravitational fields?
2. Which of the following describes the equipotential lines for the electric and gravitational fields that the proton experiences?
(A) Equipotential lines are straight, horizontal lines for both fields.
(B) Equipotential lines are straight, vertical lines for both fields.
(C) Equipotential lines are straight, horizontal lines for the electric field and curve upwards for the gravitational field.
(D) Equipotential lines are straight, horizontal lines for the gravitational field and curve upwards for the electric field.
3. A pipe with a diameter of D splits into two smaller, identical pipes with diameter d. If the speed of the water in the small pipes is v, what is the speed of the water in the large pipe?
(A) dv/D
(B) 2dv/D
(C) d2v/D2
(D) 2d2v/D2
4. A fluid container is shaped as a rectangular prism. The areas of the three faces of the prism are different. Which face of the container should be placed on a horizontal table so that the fluid pressure against the face touching the table is at the the lowest value?
(A) The face with the smallest area
(B) The face with the median area
(C) The face with the largest area
(D) All three orientations will result in the same fluid pressure.
5. An ohmmeter is used to measure resistance. Measurements are made of the cross-sectional area, A, and length, L, of each resistor. What should be plotted so that the slope of the plot will yield the resistivity, ρ, of the resistors?
(A) Resistance on the y-axis versus ratio of L to A on the x-axis
(B) Resistance on the y-axis versus ratio of A to L on the x-axis
(C) Ratio of L to A on the y-axis versus resistance on the x-axis
(D) Ratio of A to L on the y-axis versus resistance on the x-axis
6. An ideal gas is taken from an initial set of conditions with pressure Pi and volume Vi, to a final set of conditions, with pressure Pf and volume Vf, through several different processes. At the end of the process, the gas is at both a higher pressure and a larger volume than when it started. Which process requires the least amount of work on the system?
(A) First, expansion at constant pressure Pi from Vi to Vf followed by increasing pressure from Pi to Pf at constant volume Vf.
(B) First, increasing pressure from Pi to Pf at constant Vi volume followed by expansion at constant pressure Pf from Vi to Vf
(C) A series of small increases in volume alternating with small increases in pressure, resulting in a nearly straight line on a PV graph from the beginning to the end of the process.
(D) Any set of steps will require the same amount of work because all gases have the same change in pressure and volume.
Questions 7 and 8 refer to the following diagram.
7. The image above shows a converging lens and an object represented as a bold vertical arrow. Which line correctly depicts the output path of the ray for the ray that is incident on the lens coming in through the focus?
(A) 1
(B) 2
(C) 3
(D) 4
8. The converging lens above has a focal length of 25 cm. The object is located at a distance of 65 cm from the lens. Where should a screen be placed so that the observer will see a focused image on the screen?
(A) 65 cm to the right side of the lens
(B) 40.6 cm to the right side of the lens
(C) At the focus of the lens
(D) The image in such an arrangement will be virtual and cannot be seen on a screen.
9. You are tasked with creating a real image using a concave mirror as your imaging system. Which of the following criteria is true about both the image and the object?
(A) A real image can be created only if the object is farther away from the lens than the radius of curvature of the lens.
(B) A real image can be created only with the object located between the center of the lens and the focal length of the lens.
(C) A real image can be created with the object located anywhere farther from the lens than the focal point.
(D) A real image cannot be created using only a concave lens.
10. An atom has its lowest four energy levels at −10 eV, −5 eV, −3.5 eV, −2 eV. Which of the following photons could not be absorbed by the atom?
(A) A 10 eV photon
(B) A 5 eV photon
(C) A 2.5 eV photon
(D) A 1.5 eV photon
11. A radioactive particle undergoes beta decay, emitting an electron from its nucleus. Which of the following explanations correctly explains why this process must also involve a neutron turning into a proton within the nucleus?
(A) The total charge of the system before and after is not the same if the proton is not created.
(B) The mass energy of the system is not conserved if the proton is not created.
(C) The momentum of the system could not be conserved without the generation of a proton.
(D) All nuclear decay processes involve the generation of two particles.
12. A thermodynamic process is conducted wherein an ideal gas is taken from state A to B to C to D to A. State A is at a pressure P and a volume 5V. State B is at pressure 4P and volume V. State C is at pressure 4P and volume 4V. state D is at pressure 2P and volume 10V. Which step in the process requires the largest change in internal energy of the system?
(A) State A to State B
(B) State B to State C
(C) State C to State D
(D) State D to State A
13. A circuit consists of a 50 V battery, a 100 Ω resistor and a 25 μF capacitor. Once the capacitor has become fully charged, how much energy is stored in the capacitor?
(A) 0.0013 J
(B) 0.031 J
(C) 0.062 J
(D) 0.125 J
14. As shown above, the +Q charge is fixed in position, and the +q charge is brought close to +Q and then released from rest. Which graph best shows the acceleration of the +q charge as a function of its distance r from +Q ?
(A)
(B)
(C)
(D)
15. A charged particle with mass m is moving at a speed v at one particular instant in time. The particle is later found at a position with an electrical potential of ∆V higher than its initial position, and a gravitational potential equal to its original position. Which mathematical routine could be used to determine its speed in the final position?
(A) Calculate the work from the electrical potential and use conservation of energy to find the final speed.
(B) Calculate the work from the gravitational potential and use conservation of energy to find the final speed.
(C) Calculate the impulse from the electric potential and use conservation of momentum to find the final speed.
(D) Calculate the impulse from the gravitational potential and use conservation of momentum to find the final speed.
Questions 16-18 refer to the following diagram.
16. Voltmeters are placed across the Resistor R1, the capacitor C, and the resistor R2. The switch S has been closed a long time. What is the rank of the value readings on the voltmeters?
(A) VR1 > (VC = VR2)
(B) (VR1 = VC = VR2)
(C) (VC = VR2) > VR1
(D) VR1 > VC > VR2
17. The circuit is reset and the capacitor is discharged. Then, the switch is closed again. At what time will the current through the resistor R2 be greatest?
(A) The current through the resistor will be constant.
(B) The current through the resistor will be greatest before closing the switch.
(C) The current through the resistor will be greatest immediate after closing the switch.
(D) The current through the resistor will be greatest a long time after closing the switch.
18. The capacitance of the capacitor is known initially. The capacitor is now altered to have a larger capacitance. Which of the following observations will occur with the new capacitor in the circuit a long time after the switch is closed as compared to what was observed a long time after the switch was closed with the original capacitor?
(A) The current flowing in R1 will be greater.
(B) The current flowing in R2 will be greater.
(C) The current flowing into the capacitor will be greater.
(D) There will be no observed change in the circuit.
19. An ideal gas is confined within a cube shaped container. In addition to the length of the side of the container, which of the following sets of measurements will allow a student to determine the pressure of the gas in the container?
I. The mass of gas in the container and the average speed of a gas molecule.
II. The impulse delivered to the gas by a wall in a measured time period.
III. The force of the gas against one of the walls.
(A) I only
(B) III only
(C) II or III
(D) I or II or III
20. Points P and Q lie between the plates of a fully charged parallel plate capacitor as shown above. The lower plate is negatively charged and the upper plate is positively charged. How do the magnitudes of the electric fields at points P and Q compare?
(A) The field is 0 N/C at both points.
(B) The field is the same at both points, but not 0 N/C.
(C) EP > EQ
(D) EQ > EP
21. A wire carries a constant current to the right. A positively charged particle is a distance d above the wire, and it is moving in the same direction as the current. The particle will experience a magnetic force in which direction?
(A) To the right
(B) To the top of the page
(C) To the left
(D) To the bottom of the page
22. A neutral conducting sphere is hung from a thin insulating string. A positively charged object is brought to point P. The two objects are not allowed to touch. What is true about the string when the positively charged object is present at point P?
(A) The tension is the same as before the object was present at point P.
(B) The tension is greater than when the object was not at point P and the string stretches to the left of its original orientation.
(C) The tension is greater than when the object was not at point P and the string stretches to the right of its original orientation.
(D) The tension is less than when the object was not at point P and the string stretches to the left of its original orientation.
23. An experiment is performed on a fixed volume of an ideal gas. The pressure, in pascals, of the gas is plotted on the vertical axis and the temperature of the gas, in degrees Kelvin, is plotted on the horizontal axis. During a second performance of the experiment at a greater volume, the pressure-temperature gas is expected to
(A) have a greater slope and the same intercept
(B) have a smaller slope and the same intercept
(C) have a greater slope and a greater intercept
(D) have a smaller slope and a greater intercept
24. Two gas samples contain different gases. The first gas sample contains more massive molecules than the second. The molecules in both gas samples have the same average speed. When the samples are brought into contact, what is the flow of energy?
(A) No energy will flow because the average speed of the particles in each gas is the same.
(B) No energy will flow because the gases are made up of different molecules.
(C) Energy will flow away from the first sample because its molecules are more massive.
(D) Energy will flow away from the first sample because its molecules have more kinetic energy.
25. Which statement correctly characterizes the work done by the gas during the ABCA cycle shown in the above P-V diagram?
(A) There is no work done by the gas because the system both starts and concludes in state A.
(B) There is no work done because the work done during the transition from A→B cancels out the work done in transition from C→D.
(C) The work done by the gas is positive because the work done during the transition from A→B is greater than the work done in transition from C→D.
(D) The work done by the gas is positive because the work done during the transition from B→C is greater than the work done in transition from D→A.
26. What is the equation for the electric field given by the above plot? The electric field crosses the axis at times 0 ps, 1.43 ps, 2.85 ps, and 4.28 ps.
(A) E = sin ((2.85 THz)t)
(B) E = sin ((2.85 THz)t)
(C) E = sin ((2.20 THz)t)
(D) E = cos ((0.45 THz)t)
27. A spherical balloon filled with helium is floating in air. If the balloon is inflated until its radius is doubled, how will the buoyant force on the balloon be affected?
(A) It will decrease by a factor of 4.
(B) It will increase by a factor of 4.
(C) It will increase by a factor of 8.
(D) It will not be affected.
28. Data is collected in an experiment preformed on an ideal gas. In the experiment, temperature (in K) is the independent variable and volume (in m3) is the dependent variable. If the data is graphed, which of the following is true about the slope and y-intercept of the graph?
(A) The slope will be directly proportional to the pressure of the gas and the intercept will be 0 m3.
(B) The slope will be inversely proportional to the pressure of the gas and the intercept will be 0 m3.
(C) The slope will be directly proportional to the pressure of the gas and the intercept will not be 0 m3.
(D) The slope will be inversely proportional to the pressure of the gas and the intercept will not be 0 m3.
29. Which of the following relationships, when plotted, will yield a curve which is inverse to the first power?
I. The electric potential versus distance from a positive point particle.
II. The volume versus pressure for an ideal gas.
III. The magnetic field from a current carrying wire versus distance from the wire.
(A) I only
(B) I and III
(C) II only
(D) I and II and III
30. A massive charged object is fixed in space with a separation of d meters. A quantity R is defined as the ratio of the gravitational potential energy to the electric potential energy. The separation is then increased slowly. What happens to R?
(A) R is constant for all separations.
(B) R increases as the separation increases.
(C) R decreases as the separation increases.
(D) R increases up to a distance D, then R decreases back to its original value.
31. An ideal gas is at a pressure P and a volume V. The gas is in a fixed volume, but is heated until the pressure doubles. What happens to the average speed of the molecules in the gas?
(A) The speed of the molecules on average remain unchanged.
(B) The speed of the molecules on average increases by a factor of .
(C) The speed of the molecules on average increases by a factor of 2.
(D) The speed of the molecules on average increases by a factor of 4.
32. Which of the following changes to a double-slit interference experiment would increase the widths of the fringes in the interference pattern that appears on the screen?
(A) Use light of a shorter wavelength.
(B) Move the screen closer to the slits.
(C) Move the slits closer together.
(D) Use light with a lower wave speed.
33. Tritium is an isotope of hydrogen consisting of one proton and two neutrons. The isotope has a mass of 5.008 × 10−27 kg. The mass of a proton is 1.673 × 10−27 kg and a neutron has a mass of 1.675 × 10−27 kg. What is the binding energy of tritium?
(A) 1.500 × 10−29 J
(B) 1.350 × 10−12 J
(C) 4.507 × 10−10 J
(D) 4.521 × 10−10 J
34. In an experiment designed to study the photoelectric effect, it is observed that low-intensity visible light of wavelength 550 nm produced no photoelectrons. Which of the following best describes what would occur if the intensity of this light were increased dramatically?
(A) Almost immediately, photoelectrons would be produced with a kinetic energy equal to the energy of the incident photons.
(B) Almost immediately, photoelectrons would be produced with a kinetic energy equal to the energy of the incident photons minus the work function of the metal.
(C) After several seconds, the electrons absorb sufficient energy from the incident light, photoelectrons would be produced with a kinetic energy equal to the energy of the incident photons minus the work function of the metal.
(D) Nothing would happen.
35. Radioactive carbon-14 undergoes beta-minus decay. The atomic number of carbon is 6. The number of nucleons in the products after the beta decay is
(A) 6
(B) 7
(C) 13
(D) 14
36. A pipe carries water with a density of ρ = 1000 kg/m3. One end of the pipe has a diameter of 0.02 m. At that location, a pressure gauge says the pressure is P = 1.25 × 105 Pa and the water is moving at a speed of 2.0 m/s. The other end of the pipe is 1.2 m higher than the end with the pressure gauge and a diameter of 0.01 m. What will a pressure gauge installed in the upper end of the pipe read?
(A) 7.90 × 104 Pa
(B) 8.30 × 104 Pa
(C) 1.07 × 105 Pa
(D) 1.43 × 105 Pa
37. A square loop of conducting wire with side s is moved at a constant rate v to the right into a region where there is a constant magnetic field directed out of the page. Which of the following graphs shows the flux through the loop as a function of distance?
(A)
(B)
(C)
(D)
38. A hollow conducting sphere is placed around point P so that the center is at P. What happens to the electric field strength at point P?
(A) It doubles.
(B) It halves.
(C) It remains unchanged.
(D) It becomes zero.
39. A machine shoots a proton, a neutron, or an electron into a magnetic field at various locations. The paths of two particles are shown above. Assume they are far enough apart so that they do not intersect and the magnetic field is going out of the page. What can you say about the paths that represent each particle?
(A) a is the proton and b is the electron.
(B) b is the proton and a is the electron.
(C) Either may be a neutron.
(D) You cannot make any conclusions without knowing the velocities.
40. In the figure above, a ray of light hits an object and travels parallel to the principal axis as shown by the dotted line. Which line shows the correct continuation of the ray after it hits the concave lens?
(A) a
(B) b
(C) c
(D) d
41. A tube with two T branches that has an open end is inserted in a liquid. However, the section of the tube above part B is hidden from view. The hidden section may be wider or narrower. Air is blown through the tube and the water levels rise as shown. You can conclude which of the following?
(A) The picture as drawn is impossible—A and B must be at equal heights.
(B) The tube is narrower and the air speed is greater above section B.
(C) The tube is narrower and the air speed is less above section B.
(D) The tube is wider and the air speed is greater above section B.
42. A beam of light goes from water to air. Depending on the actual angle that the light strikes the surface, which of the following rays are possible outcomes?
(A) A only
(B) B only
(C) A or D
(D) C or D
43. The instrument in an aircraft to measure airspeed is known as the pitot tube, shown in the figure above. The opening facing the incoming air (with the small aperture) is the part meant to capture the air at rest. The opening perpendicular to the flow of air (with the large aperture) is meant to capture air at speed. If h = 1 m and the fluid within the manometer is water, what is the airspeed? Take the density of air to be ρair = 1.2 kg/m3.
(A) 27 m/s
(B) 68 m/s
(C) 95 m/s
(D) 128 m/s
44. Two charged, massive particles are isolated from all influence except those between the particles. They have charge and mass such that the net force between them is 0 N. Which of the following is NOT true?
(A) The particles must have the same sign of charge.
(B) If the distance between the particles changes, the force will no longer be 0 N.
(C) The particles must have the same mass.
(D) The particle cannot have the same magnitude of charge.
45. The following three step process refers to a simple RC circuit with a battery and an initially open switch.
Step 1: The switch is closed, allowing the capacitor to charge.
Step 2: After the capacitor has charged, a slab of dielectric material is inserted between the plates of the capacitor and time passes.
Step 3: The switch is opened, and the dielectric is removed.
Which of the following describes the change in potential energy stored in the capacitor during each step?
(A) ∆U1 < 0; ∆U2 > 0; ∆U3 > 0
(B) ∆U1 > 0; ∆U2 > 0; ∆U3 > 0
(C) ∆U1 < 0; ∆U2 > 0; ∆U3 < 0
(D) ∆U1 > 0; ∆U2 < 0; ∆U3 < 0
Directions: For questions 46-50 below, two of the suggest answers will be correct. Select the two answers that are best in each case, and then fill in both of the corresponding circles on the answer sheet.
46. N resistors (N > 2) are connected in parallel with a battery of voltage V0. If one of the resistors is removed from the circuit, which of the following quantities will decrease? Select two answers.
(A) The voltage across any of the remaining resistors
(B) The current output by the battery
(C) The total power dissipated in the circuit
(D) The total resistance in the circuit
47. A metal conducting sphere of radius r in electrostatic equilibrium has a positive net charge of +5e. Which of the following is true? Select two answers.
(A) All of the charge will be located on the outside of the sphere.
(B) There will be a constant, non-zero electric field within the sphere.
(C) There will be a constant, non-zero electric potential within the sphere.
(D) When the sphere is connected to a ground, the +5e of charge on the sphere will flow into the ground to neutralize the sphere.
48. A circuit is created with a battery of negligible internal resistance and three identical resistors with resistance of 10 Ω. The resistors are originally arranged so that one is in series with the battery, and the other two are in parallel with one another. Which of the following changes to the circuit will result in an increase in the amount of current drawn from the battery? Select two answers.
(A) Rearranging the resistors so that all three are in parallel
(B) Replacing every resistor with a resistor of half the resistance
(C) Removing one of the two parallel branches entirely from the circuit
(D) Replacing the battery with a battery with half the voltage
49. An ideal gas is in state 1, with P1, V1, and T1. The final volume will be the same as V1 for which of the following processes? Select two answers.
(A) Triple P1 and decrease T1 by
(B) Triple T1 and decrease P1 by
(C) Quadruple P1 and T1
(D) Decrease P1 by and decrease T1 by
50. In a double slit experiment, students are attempting to increase the spacing of the fringes observed on the screen. Which modifications to the set up will result in increased fringe separation? Select two answers.
(A) Doubling the wavelength only
(B) Doubling the wavelength and doubling the slit separation
(C) Doubling the distance to the screen only
(D) Doubling the distance to the screen and doubling the slit separation
END OF SECTION I
AP PHYSICS 2
SECTION II
Time—90 minutes
4 Questions
Directions: Questions 1 and 2 are long free-response questions that require about 30 minutes to answer. Questions 3 and 4 are short answer questions that require about 15 minutes to answer. Show your work for each part in the space provided after that part.
1. An unplugged freezer is at room temperature. The door is closed and the freezer is plugged in.
(A) Your friend observes, “When you open the freezer, the cold air comes out of the freezer. That didn’t happen before the freezer was cooled down. It must therefore be the case that cold air is less dense than warm air, since the low density gas will expand more easily.”
(i) Is there any part of your friend’s statement you agree with? Why?
(ii) Is there any part of your friend’s statement you disagree with? Why?
(b) You set out to perform an experimental investigation of the relationship between gas density and temperature. The following equipment is available. Write down each item you will need to use:
(c) Write out a numbered procedure you will use to gather necessary data. Your description should be detailed enough that another student could reproduce your experiment.
(d) Your data analysis must include a graph. Explain what you would graph on the x-axis and the y-axis. Justify your decision and explain how your graph will help you understand the relationship between density of a gas and its temperature.
(e) What shape graph do you expect to see?
2. The figure below shows an electric circuit containing a source of emf, ε, a variable resistor (r) and a resistor of fixed resistance R. The resistor R is immersed in a sealed beaker containing a mass m of water, currently at temperature Ti. When the switch S is closed, current through the circuit causes the resistor in the water to dissipate heat, which is absorbed by the water. A stirrer at the bottom of the beaker simply ensures that the temperature is uniform throughout the water at any given moment. The apparatus is well-insulated (insulation not shown), and it may be assumed that no heat is lost to the walls or lid of the beaker or to the stirrer.
(a) Determine the current in the circuit once S is closed. Write your answer in terms of ε, r, and R.
(b) Determine the power dissipated by the resistor R in terms of ε, r, and R.
(c) Explain at the microscopic level why the water heats up when the switch is closed and how the stirrer helps ensure a constant temperature throughout.
(d) Assume the stirrer has a knob, which changes its speed. How can the temperature of the water be increased more rapidly by adjusting the rotation rate of the stirrer?
(e) As the temperature of the water increases, whether from the resistor or from the stirrer rate, explain the microscopic interactions responsible for the changing pressure in the container.
3. In an experiment conducted, two tests are run. In both trials you may ignore the effect of gravity. The following is a diagram of the tests.
Test 1: There are two large parallel plates separated by a distance d = 0.5 m with a potential difference of 0.12 V across them. There is a uniform magnetic field B pointing perpendicularly out of the paper of strength 0.002 T starting to the right of plate 2. An electron is released from rest at plate 1 as shown above. It passes through a hole in plate 2 and enters the magnetic field and only experiences forces due to the magnetic field.
Test 2: The same set-up is run with the following two exceptions.
1. The battery is switched so that plate 1 becomes positive and plate 2 becomes negative.
2. A proton is used instead of an electron.
(a) As the particles move from plate 1 towards plate 2 in each test, they experience unbalanced forces causing their speed to change. After plate 2, each particle still experiences unbalanced forces, but no longer change speed. Use the concept of work to explain how this occurs.
(b) Find the ratio of the speed of the proton as it emerges from the hole to the speed of the electron as it emerges from the hole.
(c) Make a sketch of each path each particle will follow after emerging from the hold in plate 2.
(d) A third test is conducted similar to test 2, except an alpha particle is used instead of a proton. Explain how the path of the alpha particle after emerging from the hole will differ from the proton path.
4. In a double-slit interference experiment, a parallel beam of monochromatic light is needed to illuminate two narrow parallel slits of width w that are a distance h apart, where h ≫ w, in an opaque card as shown in the figure above. The interference pattern is formed on a screen a distance D from the slits, where D ≫ h.
(a) Explain why it is important that the distance between S and the card is much greater than the distance h.
(b) Draw the first three wavefronts that emerge from each slit after the card.
(c) In the interference patterns on the screen, the distance from the central bright fringe to the third bright fringe on one side is measured to be y3. In terms of D, H, y3 and S, what is the wavelength of the light emitted from the source S?
(d) If the space between the slits and the screen was filled with a material having an index of refraction n > 1, would the distance between the bright fringes increase, decrease, or remain the same? Explain your reasoning.
STOP
END OF EXAM
Practice Test 1: Answers and Explanations
PRACTICE TEST 1 ANSWER KEY
1. D
2. A
3. D
4. C
5. A
6. A
7. B
8. B
9. C
10. C
11. A
12. D
13. B
14. A
15. A
16. C
17. D
18. D
19. D
20. B
21. D
22. C
23. B
24. D
25. C
26. C
27. C
28. B
29. D
30. A
31. B
32. C
33. B
34. D
35. D
36. B
37. A
38. D
39. A
40. A
41. B
42. D
43. D
44. A
45. B
46. B, C
47. A, C
48. A, B
49. C, D
50. A, C
SECTION I
1. D The forces must balance in order for the proton to travel along a straight line. If the forces balance, m(Eg) = q(Eel). Since m is less than q for a proton, the gravitation field is stronger. Gravity points down, so the electric force must be up, and for a positive particle, the field must also be up.
2. A Equipotential lines will be perpendicular to the force lines. Gravitational force lines point vertically downward. The equipotential lines for the electric field must also be horizontal lines for the particle to move in a straight line at a constant speed.
3. D The continuity equation, A1v1 = A2v2 applies here as A1v1 = 2A2v2 where position 1 is in the large pipe. Solving for v1 = 2v/π
4. C The fluid pressure is the force divided by the area. The force is the same regardless of the container orientation as the weight of the container is constant. To result in a minimum pressure, the largest face must be placed against the table.
5. A R = ρ
For any linear equation, y = mx+b, the slope is multiplied by the quantity which is graphed on the horizontal axis. In order to have the slope equal the resistivity, the horizontal axis must be the ration of length to area.
6. A All processes that end at higher volumes will require removing work to be done on the system. When pressure is constant, the work is found from W = P∆V, so to minimize the work required, we want the change in volume to occur at the lowest pressure possible.
7. B A ray which travels into a converging lens along a line that goes through the focal point will refract to travel along a line parallel to the optical axis upon leaving the lens.
8. B The screen must be placed at the image location in order for the image to appear in focus on the screen. We use the equation
9. C A concave mirror can only make a real image when the object is placed beyond the focus.
10. C Any photon that has energy that matches an energy difference in that atom can be absorbed. The 10 eV photon can be absorbed if there is an electron in the −10 eV state by ionizing the atom. A 5 eV photon can excite a photon from the −10 eV state to the −5 eV state or ionize an electron from the −5 eV state. The 1.5 eV photon can cause a transition from the −3.5 eV state to the −2 eV state.
11. A During nuclear decay processes, the total charge in the system before and after the decay must be the same.
12. D The internal energy is proportional to the temperature of the system, and the Ideal Gas Law explains that the product of pressure and volume is also proportional to the temperature. State A is at 5PV, State B is at 4PV, State C is at 16PV, and State D is at 20PV. The change from State D to State A requires the largest change in energy.
13. B The internal energy stored in a fully charged capacitor is UC = CV2. The voltage of the battery will be the voltage across the capacitor when the capacitor is fully charged.
UC = (25 × 10–6 F)(50 V)2 = 0.03125 J
14. A The acceleration of an object with charge +q and mass m is given by Graph (A) shows an inverse square law.
15. A There is a change in electrical potential and no change in gravitational potential. The work is done because of the change in the potential. Therefore, the Work–Energy Theorem, ∆Wnet = ∆KE can be used to find the change in kinetic energy. Finally, using ∆KE = mv2f ‒ mv2i and the known initial speed and mass, the final speed can be determined.
16. C A long time after the switch has been closed, the capacitor will be fully charged. There will be no current flowing across R1 so it will have a voltage drop of 0 V. The capacitor and R2 are both in parallel with the battery, and will have a voltage equal to the battery voltage.
17. D Before the switch is closed, current will flow through R2, but the resistance in the circuit will be R1 + R2. With a fully discharged capacitor, the instant the switch is closed, R2 will be shorted out of the circuit. As the capacitor charges, more current will flow across R2. The current through R2 will continue to increase until the capacitor is fully charged, at which time the current through R1 will stop. After the circuit is fully charged, the total resistance in the circuit will be R2. The current through R2 is greater after the capacitor is charged than before the switch is closed.
18. D The capacitance of a fully charged capacitor will have no effect on the circuit.
19. D One way to find the pressure is with the ideal gas law. The ideal gas law states that PV = nRT, so in addition to the volume, found from the length, one needs n and T to find the pressure, so I is correct. The impulse and time lets us calculate the force applied against the gas in that time. Newton’s Third Law states that this must be the same magnitude as the force applied to the wall. The force and area allow us to calculate the pressure. So II and III are also correct.
20. B Between the plates of a parallel plate capacitor, the field will be constant and non-zero.
21. D The magnetic field produced by the wire is found using the right hand rule. The current is to the right, so at a position above the wire the magnetic field points outward. We then use the right hand rule again to fine the force on the particle. The particle moves to the right in a magnetic field pointed outward. The force is towards the bottom.
22. C Negative charges will be drawn to the right side of the sphere, causing a force on the sphere to the right. The vertical tension will still have to balance with gravity, but there will also be a components of tension to the right.
23. B The ideal gas law is PV = nRT, or P = T. Comparing this to y = mx + b, the intercept should be 0 and the slope should be related to 1/V.
24. D Energy will flow from the hotter gas to the cooler gas. The speeds are the same, but the masses are different, so the more massive gas has more kinetic energy.
25. C Work is done when there is a change in volume. At constant pressure, the equation W = –P∆V tells us that when P is higher (as it is at path A→B) a greater amount of negative work is done on the gas than at lower P. Thus, the work done by the gas during the entire cycle is positive, and the constant volume paths have no influence on the amount of work.
26. C The field is a sine curve with an amplitude of 10 N/C. From the graph, the period of the wave is 2.85ps. The argument of sine needs to be 2πt/T.
27. C The buoyant force is found from F = ρairVsubg. Because the entire balloon is surrounded by air, the entire volume is Vsub. The volume of the spherical balloon is V = πr3. Doubling the radius increases the volume by a factor of 8.
28. B The ideal gas law equation is PV = nRT. Solving this for V (since volume is on the vertical axis of the graph described in the problem) yields V = T. Comparing this to the equation for a line, y = mx + b, the slope of the line will be nR/P, which is inversely proportional to P, and b = 0 is the intercept.
29. D The ideal gas law is P = (nRT)
The potential from a point particle is
The magnetic field from a wire is
All of these are inverse to the first power.
30. A The quantity R is defined as the ratio of the gravitational potential energy to the electric potential energy, so
R is constant.
31. B Doubling pressure at constant volume causes temperature to double. Doubling temperature doubles the average kinetic energy of the molecules. The speed squared is proportional to the kinetic energy, so the speed increases by a factor of .
32. C Relative to the central maximum, the locations of the bright fringes on the screen are given by the expression m, where λ is the wavelength, L is the distance to the screen, d is the slit separation, and m is any integer. The width of a fringe is therefore (m + 1) – = . The slit spacing will increase if there is a decrease in d.
33. B The binding energy is the mass defect, the difference in the mass of the constituent components and the mass of the nucleus, multiplied by c2. That mass defect is
mp + 2mn – mTr = 1.673 × 10–27 + 2 * 1.675 × 10–27 – 5.008 × 10–27 = 1.500 × 10–29 kg
E = ∆mc2 = 1.500 × 10–29 * (3 × 108)2
34. D If the photons of the incident light have insufficient energy to liberate electrons from the metal’s surface, then simply increasing the number of these weak photons (that is, increasing the intensity of the light) will do nothing. In order to produce photoelectrons, each photon of the incident light must have an energy at least as great as the work function of the metal.
35. D Beta-minus decay occurs when a neutron turns into a proton and a beta particle. The number of nucleons does not change in this process.
36. B Use the continuity equation to find the speeds.
Then use Bernoulli’s Equation
37. A Flux is the product of the area enclosed by the loop of wire and the field strength. The flux will increase linearly up to a maximum value when the wire is completely within the field, and will begin to decrease once the wire begins to leaves the region within the field.
38. D Within a conducting sphere, the electric field is always zero. Therefore, in the region of P, as well as everywhere else within the conducting sphere, there is zero electric field.
39. A Path a follows the right-hand rule so must be positively charged. Path b is opposite the right hand rule so must be negatively charged. Neutrons would follow a straight line path, so it is impossible for either a or b to be a neutron.
40. A The rules for ray-tracing diagrams state that a line parallel to the principle axis bends away from the focal point as shown.
41. B Because the fluid is higher in column B, the air speed must be greater above column B. This must be due to a narrower tube.
42. D Water has a higher index of refraction than air, so if light refracts through the water the light will bend away from the normal—so C is possible. Another possibility is D, total internal reflection when the angle in water is greater than the critical angle.
43. D First, we use Bernoulli’s Equation. We call the point with the air at rest B and the point with the speeding air A.
We have vB = 0 and yA = vB, so the equation reduces to
The pressure difference, PB = PA arises from the fluid depth ρfgh.
Solving for va and using the given values
44. A It is possible for the gravitation force to balance the electrical force, making (D) false. In order for this to occur, since gravity is always an attractive force, the electrical force must be repulse, and therefore the charges must have the same sign, making statement A true. Both the gravitational and electrical force are inverse square laws, so if the distance between the particles changes, both forces change by the same factor, making (B) false. And (C) is just false, both particles will experience the same gravitational force regardless of how the mass is distributed amongst them; the gravitation force depends on the product of the two masses.
45. B Step 1: Potential energy rises because the magnitude of charge on each plate increases, QC = CVC and UC = CV2. ∆U1 > 0, eliminate (A) and (C).
Step 2: Inserting a dielectric increases potential energy. Inserting a dielectric always increases capacitance. QC = CVC and UC = CV2; therefore, potential energy increases. ∆U2 > 0, so eliminate (D).
Step 3: Capacitor is still charged so there is still potential energy.
46. B, C The resistors are all in parallel, meaning their voltage is the same across each resistor and matches the voltage of the battery. Eliminate (A). The voltage provided by the battery does not increase or decrease, the battery still maintains the same voltage. Eliminate (D). The resistors are in parallel, so eliminating one of the resistors will increase the equivalent resistance (there is now one less path for current to go through so equivalent resistance goes up). Because I = V/R this increase in resistance would decrease the current. Power’s equation is P = IV. If our current decreases, our power decreases as well.
47. A, C Conductors will always have all of their excess charge located at the surface. The entire volume of a conductor must be an equipotential in electrostatics because otherwise the charges in the conductor would move as a result of the potential difference. Choice (B) is false because within a conductor in electrostatic equilibrium, there is an electric field of 0 N/C. Choice (D) is false because, while the sphere will neutralize when connected to a ground, it is not the + charges that flow out of the sphere, but – charges which will flow from the ground into the sphere.
48. A, B From Ohm’s Law, the current drawn out of the battery will increase when the total resistance of the circuit decreases, or when the total voltage supplied to the circuit increases, which makes (D) false. Resistors in parallel have a lower equivalent resistance than those in series, so (A) is true. To see that (B) decreases the overall resistance, we can think of a simpler version of the problem. Replacing a resistor with a wire will have the same effect of either increasing or decreasing the overall resistance, but with a larger effect. If any of the three resistors are replaced with a wire, the overall resistance decreases. Therefore, decreasing the resistance of any of the resistors will decrease the equivalent resistance, and replacing all of them most certainly will decrease the equivalent resistance.
49. C, D The ideal gas law states PV = nRT , so pressure is directly proportional to temperature. To keep volume constant, pressure and temperature must increase or decrease by the same amount.
50. A, C For the double slit experiment, mλ = dsin(θ) → mλ =dx\L.
The spacing of the fringes, x, increases for increasing λ and L and decreases by increasing d.
SECTION II
1. (a) i)I agree with “When you open the freezer, the cold air comes out of the freezer. That didn’t happen before the freezer was cooled down.” The reason is that energy flows, according to the Second Law of Thermodynamics, due to a difference in temperature. When the freezer wasn’t cooled down, there was no temperature difference.
ii) I disagree with “cold air is less dense than warm air, since the low-density gas will expand more easily.” A low-density gas will have a lot of space between molecules, so a higher density gas will expand more easily into the low-density gas.
(b) Using the chamber with the sliding lid allows for a changing volume at a constant amount of air, so that density can be the dependent variable in the experiment. The area of the lid is constant, so we can determine volume as the piston length changes using the meter stick. Using the large tubs of boiling water and ice water to heat and cool the chamber allows the independent variable to be the temperature, measured with the thermometer.
(c) 1. Place the thermometer in the chamber with the sliding piston.
2. Place the chamber in the ice water and let it cool.
3. Record the temperature with the thermometer and the piston length with the meter stick.
4. Heat the chamber by placing it in the hot water.
5. Record the temperature and the piston length.
6. Repeat steps 4 and 5 to get temperature and volume data.
(d) The density is n/V, so since we have data for ∆x and we know area, we need to plot versus temperature. The shape of the graph will tell us whether the increase in temperature causes the density to increase or decrease.
(e) Based on the ideal gas law
we expect that we will see an inverse relationship.
2. (a)The total resistance in the circuit is r + R. The current is then I = .
(b) Power is P = IV = .
(c) Temperature increases when the average kinetic energy of a molecule increases. When the switch is closed, the electrons move through the circuit. As the electrons move within the metal of the circuit, they collide with the positive nuclei of the atoms. The nuclei are set into motion, so the resistor heats up. The Second Law of Thermodynamics states that energy will flow from high energy to low energy parts of a system. The molecules in the water are moving with a lower energy than the hot resistor, so as the water molecules collide with the resistor, energy transfers into the water molecules. The now heated water molecules transfer energy to one another through further collisions, with the colder, slower molecules heating up and speeding up. The stirrer helps to ensure that the average energy is equal throughout the water by dispersing the fast moving water molecules so that they transfer energy throughout the container and not just locally near the resistor.
(d) Energy can be transferred to the water molecules by the motion of the stirrer as well as by the high temperature molecules in the resistor. If the stirrer spins at a rapid enough rate, an average water molecule will be sped up simply by its motion.
(e) The pressure in the container is a result of the collisions between the molecules within the container and its walls. As the molecules heat up and move more rapidly, they will collide more frequently with walls of the container and cause the pressure against the walls to increase.
3. (a) Between the plates, there is an electric field that points from one of the plates towards the other plate. As the particles change their position within a potential difference, work is done on the particles, causing them to accelerate. After leaving plate 2, the particles are subjected to a magnetic field. Because the magnetic field is always perpendicular to the direction of motion, the magnetic field does no work and the particles do not change their speed.
(b) Using the Work–Energy Theorem for each particle
vi is zero for each particle, and each particle has the same ∆V and q, so
The ratio for the speed of the proton to the electron is the square root reciprocal ratio of the masses
i.
(c) The alpha particle will have twice the charge and four times the mass of the proton. Because
the final speed coming out of the hole will decrease by . The path curves because of the magnetic force being the centripetal force.
The radius of the circle will be 2 times as large, but the curve will be in the same direction.
4. (a) The double slit experiment works by producing spherical wavefronts that interfere with one another. The input to the slits must be plane waves to see the interference. If the source if close to the slits, the wavefronts incident on the card will not be plane waves.
(b)
(c) In the double slit experiment mλ = dx/L. Solving for λ
d is the distance between the slits, x is the distance measured between fringes, L is the space from the slits to the screen and m is the “order” or the number of dark spots between fringes. Using the given variables
(d) When light enters a medium with a different index of refraction, its speed changes. The light wave will have a constant frequency, so this results from a change in its wavelength. When light goes from an index of n = 1 to an index of n > 1, the wave speed decreases, resulting in a decreased wavelength.
So x is directly proportional to l, and the distance between fringes will decrease.
Part III
About the AP Physics 2 Exam
• The Structure of the AP Physics 2 Exam
• How the AP Exams Are Used
• Other Resources
• Designing Your Study Plan
THE STRUCTURE OF THE AP PHYSICS 2 EXAM
The AP Physics 2 Exam consists of two sections: a multiple-choice section and a free-response section. Questions in the multiple-choice section are each followed by four possible responses. In the single-select subsection, only one of these possible answers is correct. In the multi-select subsection, two of these possible answers are correct. It is your job to choose the right answer. Each correct answer is worth one point.
The free-response section requires you to write out your solutions, showing your work. The total amount of time for this section is 90 minutes, so you have an average of 22.5 minutes per question. Unlike the multiple choice section, which is scored by computer, the free-response section is graded by high school and college teachers. They have guidelines for awarding partial credit, so you don’t need to correctly answer every part to get points. You are allowed to use a calculator on the entire AP Physics 2 Exam (programmable or graphing calculators are okay, but ones with a typewriter-style keyboard are not) and a table of equations is provided for your use. The two sections—multiple choice and free response—are weighted equally, so each is worth 50 percent of your grade.
Grades on the AP Physics 2 Exam are reported as a number: either 1, 2, 3, 4, or 5. The descriptions for each of these five numerical scores are as follows:
AP Exam Grade | Description |
5 | Extremely well qualified |
4 | Well qualified |
3 | Qualified |
2 | Possibly qualified |
1 | No recommendation |
Colleges are generally looking for a 4 or 5, but some may grant credit for a 3. How well do you have to do to earn such a grade? Each test is curved, and specific cut-offs for each grade vary a little from year to year, but here’s a rough idea of how many points you must earn—as a percentage of the maximum possible raw score—to achieve each of the grades 2 through 5:
AP Exam Grade | Percentage Needed |
5 | ≥ 75% |
4 | ≥ 60% |
3 | ≥ 45% |
2 | ≥ 35% |
AP Physics 2: Algebra-Based Course Content
You may be using this book as a supplementary text as you take an AP Physics 2 course at your high school or you may be using it on your own. The College Board is very detailed in what they require your AP teacher to cover in his or her AP Physics 2 course. They explain the course that you should know with the following summary.
Students explore principles of fluids, thermodynamics, electricity, magnetism, optics, and topics in modern physics. The course is based on seven Big Ideas, which encompass core scientific principles, theories, and processes that cut across traditional boundaries and provide a broad way of thinking about the physical world.
The following are Big Ideas:
• Objects and systems have properties such as mass and charge. Systems may have internal structure.
• Fields existing in space can be used to explain interactions.
• The interactions of an object with other objects can be described by forces.
• Interactions between systems can result in changes in those systems.
• Changes that occur as a result of interactions are constrained by conservation laws.
• Waves can transfer energy and momentum from one location to another without the permanent transfer of mass and serve as a mathematical model for the description of other phenomena.
• The mathematics of probability can be used to describe the behavior of complex systems and to interpret the behavior of quantum mechanical systems.
Questions will address the content of the Big Ideas in a variety of ways. These will include mathematical calculations similar to those found in many textbooks, proportional reasoning, verbal descriptions, and analysis of graphs including interpretations of slopes and x- or y-intercepts. Naturally, it’s important to be familiar with the topics—to understand the basics of the theory, to know the definitions of the fundamental quantities, and to recognize and be able to use the equations. Then, you must acquire practice at applying what you’ve learned to answering questions like you’ll see on the exam. This book is designed to review all of the content areas covered on the exam, illustrated by hundreds of examples. Also, each content review chapter is followed by practice multiple-choice and free-response questions, and perhaps even more important, answers and explanations are provided for every example and question in this book. You’ll learn as much—if not more—from actively reading the solutions as you will from reading the text and examples. Also, two full-length practice tests (with solutions) are provided in this book. The difficulty level of the examples and questions in this book is at or slightly above AP level, so if you have the time and motivation to attack these questions and learn from the solutions, you should feel confident that you can do your very best on the real thing.
The College Board also gives us a list of Physics 2 Principles that they describe as foundational physics principles. If you are taking an AP Physics 2 course, you will likely recognize these categories from your textbook sections. Be sure that you’re familiar with the following principles as you prepare for your exam:
• Thermodynamics: Laws of Thermodynamics, Ideal Gases, Kinetic Theory
• Fluid Statics and Dynamics
• Electrostatics: Electric Force, Electric Field, and Electric Potential
• DC Circuits and RC Circuits (steady-state only)
• Magnetism and Electromagnetic Induction
• Geometric and Physical Optics
• Quantum Physics, Atomic, and Nuclear Physics
HOW AP EXAMS ARE USED
Different colleges use AP Exams in different ways, so it is important that you go to a particular college’s web site to determine how it uses AP Exams. The three items below represent the main ways in which AP Exam scores can be used:
• College Credit. Some colleges will give you college credit if you score well on an AP Exam. These credits count toward your graduation requirements, meaning that you can take fewer courses while in college. Given the cost of college, this could be quite a benefit, indeed.
• Satisfy Requirements. Some colleges will allow you to “place out” of certain requirements if you do well on an AP Exam, even if they do not give you actual college credits. For example, you might not need to take an introductory-level course, or perhaps you might not need to take a class in a certain discipline at all.
• Admissions Plus. Even if your AP Exam will not result in college credit or even allow you to place out of certain courses, most colleges will respect your decision to push yourself by taking an AP Course or even an AP Exam outside of a course. A high score on an AP Exam shows mastery of more difficult content than is taught in many high school courses, and colleges may take that into account during the admissions process.
OTHER RESOURCES
There are many resources available to help you improve your score on the AP Physics 2 Exam, not the least of which are your teachers. If you are taking an AP class, you may be able to get extra attention from your teacher, such as obtaining feedback on your essays. If you are not in an AP course, reach out to a teacher who teaches AP Physics 2 and ask if the teacher will review your essays or otherwise help you with content.
Another wonderful resource is AP Students, the official site of the AP Exams. The scope of the information at this site is quite broad and includes:
• a Course Description, which includes details on what content is covered and sample questions
• sample questions from the AP Physics 2 Exam
The AP Students home page address is: http://apstudent.collegeboard.org.
Finally, The Princeton Review offers tutoring for the AP Physics 2 Exam. Our expert instructors can help you refine your strategic approach and add to your content knowledge. For more information, call 1-800-2REVIEW.
DESIGNING YOUR STUDY PLAN
As part of the Introduction, you identified some areas of potential improvement. Let’s now delve further into your performance on Test 1, with the goal of developing a study plan appropriate to your needs and time commitment.
Read the answers and explanations associated with the Multiple-Choice questions (starting at this page). After you have done so, respond to the following questions:
• Review the list of Big Ideas on this page. Next to each topic, indicate your rank of the topic as follows: “1” means “I need a lot of work on this,” “2” means “I need to beef up my knowledge,” and “3” means “I know this topic well.”
• How many days/weeks/months away is your exam?
• What time of day is your best, most focused study time?
• How much time per day/week/month will you devote to preparing for your exam?
• When will you do this preparation? (Be as specific as possible: Mondays and Wednesdays from 3:00 P.M. to 4:00 P.M., for example)
• Based on the answers above, will you focus on strategy (Part Two) or content (Part Three) or both?
• What are your overall goals in using this book?
Part IV
Test-Taking Strategies for the AP Physics 2 Exam
• Preview
1 How to Approach Multiple-Choice Questions
2 How to Approach Free-Response Questions
PREVIEW
Review your Practice Test 1 results and then respond to the following questions:
• How many multiple-choice questions did you miss even though you knew the answer?
• On how many multiple-choice questions did you guess blindly?
• How many multiple-choice questions did you miss after eliminating some answers and guessing based on the remaining answers?
• Did you find any of the free-response questions easier or harder than the others—and, if so, why?
HOW TO USE THE CHAPTERS IN THIS PART
For the following Strategy chapters, think about what you are doing now before you read the chapters. As you read and engage in the directed practice, be sure to appreciate the ways you can change your approach.
Chapter 1
How to Approach Multiple-Choice Questions
CRACKING THE MULTIPLE-CHOICE SECTION
All the multiple-choice questions will have a similar format: Each will be followed by four answer choices. At times, it may seem that there could be more than one possible correct answer. For the single choice problems, which represent 45 out of your 50 multiple choice problems, there is only one correct answer! For the last 5 problems of the multiple choice section, there will be two correct answers per question. Keep in mind that this section is identified before you start it. Answers resulting from common mistakes are often included in the four answer choices to trap you.
Use the Answer Sheet
For the multiple-choice section, you write the answers not in the test booklet but on a separate answer sheet (very similar to the ones we’ve supplied at the very end of this book). Four oval-shaped answer bubbles follow the question number, one for each possible answer. Don’t forget to fill in all your answers on the answer sheet. Don’t just mark them in the test booklet. Marks in the test booklet will not be graded. Also, make sure that your filled-in answers correspond to the correct question numbers! Check your answer sheet after every five answers to make sure you haven’t skipped any bubbles by mistake.
The Two-Pass System
The AP Physics 2 Exam covers a broad range of topics. There’s no way, even with our extensive review, that you will know everything about every topic in algebra-based physics. So, what should you do?
Adopt a two-pass system. The two-pass system entails going through the test and answering the easy questions first. Save the more time-consuming questions for later. (Don’t worry—you’ll have time to do them later!) First, read the question and decide if it is a “now” or “later” question. If you decide this is a “now” question, answer it in the test booklet. If it is a “later” question, come back to it. Once you have finished all the “now” questions on a double page, transfer the answers to your bubble sheet. Flip the page and repeat the process.
Once you’ve finished all the “now” questions, move on to the “later” questions. Start with the easier questions first. These are the ones that require calculations or that require you to eliminate the answer choices (in essence, the correct answer does not jump out at you immediately). Transfer your answers to your bubble sheet as soon as you answer these “later” questions.
Watch Out for Those Bubbles!
Because you’re skipping problems, you need to keep careful track of the bubbles on your answer sheet. One way to accomplish this is by answering all the questions on a page and then transferring your choices to the answer sheet. If you prefer to enter them one by one, make sure you double-check the number beside the ovals before filling them in. We’d hate to see you lose points because you forgot to skip a bubble!
Process of Elimination (POE)
On most tests, you need to know your material backward and forward in order to get the right answer. In other words, if you don’t know the answer beforehand, you probably won’t answer the question correctly. This is particularly true of fill-in-the-blank and essay questions. We’re taught to think that the only way to get a question right is by knowing the answer. However, that’s not the case on Section I of the AP Physics 2 Exam. You can get a perfect score on this portion of the test without knowing a single right answer—provided you know all the wrong answers!
What are we talking about? This is perhaps the most important technique to use on the multiple-choice section of the exam. Let’s take a look at an example.
41. A gas undergoes an expansion-compression cycle. If, plotted on a P–V diagram, the cycle is counterclockwise and the work is 300 J in magnitude, what was the heat transfer during this cycle?
(A) 300 J into the system
(B) 300 J out of the system
(C) 600 J into the system
(D) 600 J out of the system
Now, if this were a fill-in-the-blank-style question, you might be in a heap of trouble. But let’s take a look at what we’ve got. If we can eliminate a few of the answer choices, we can get closer to making an educated guess.
For a cycle, ∆U = 0, so we know the magnitude of the work and heat transfer have to be the same. This eliminates (C) and (D). Since the cycle is counterclockwise on a P–V diagram, the work has to be positive, so the heat transfer is negative. A negative heat transfer is heat leaving the system. Choice (B) is correct.
We think we’ve illustrated our point: Process of Elimination is the best way to approach the multiple-choice questions. Even when you don’t know the answer right off the bat, you’ll surely know that two or three of the answer choices are not correct. What then?
Aggressive Guessing
As mentioned earlier, you are scored only on the number of questions you get right, so we know guessing can’t hurt you. But can it help you? It sure can. Let’s say you guess on five questions; odds are you’ll get one right. So you’ve already increased your score by one point. Now, let’s add POE into the equation. If you can eliminate as many as two answer choices from each question, your chances of getting them right increase, and so does your overall score. Remember: Don’t leave any bubbles blank on test day!
General Advice
Answering 50 multiple-choice questions in 90 minutes can be challenging. Make sure to pace yourself accordingly and remember that you do not need to answer every question correctly to do well. Exploit the multiple-choice structure of this section. For 45 of your 50 multiple-choice questions, there are three wrong answers and only a single correct one (for the multi-correct section, the last 5 questions, there are two wrong and two correct), so even if you don’t know exactly which one is the right answer, you can eliminate some that you know for sure are wrong. Then you can make an educated guess from among the answers that are left and greatly increase your odds of getting that question correct.
Problems with graphs and diagrams are usually the fastest to solve and problems with an explanation for each answer usually take the longest to work through. Do not spend too much time on any one problem or you may not get to easier problems further into the test.
These practice exams are written to give you an idea of the format of the test, the difficulty of the questions, and to allow you to practice pacing yourself. Take them in the same circumstances as you will encounter during the real exam.
Reflect
Respond to the following questions:
• How long will you spend on multiple choice questions?
• How will you change your approach to multiple choice questions?
• What is your multiple choice guessing strategy?
• Will you seek further help, outside of this book (such as a teacher, tutor, or AP Students), on how to approach the questions that you will see on the AP Physics 2 Exam?
Chapter 2
How to Approach Free-Response Questions
FREE-RESPONSE SECTION
On the free-response section, be sure to show the graders what you’re thinking. Write clearly—that is very important—and show your steps. If you make a mistake in one part and carry an incorrect result to a later part of the question, you can still earn valuable points if your method is correct. But the graders cannot give you credit for work they can’t follow or can’t read. And, where appropriate, be sure to include units on your final answers.
The most important advice we can give you for the free-response section of the AP Physics 2 Exam is to read the questions carefully and answer according to exactly what the questions are asking you to do. Credit for the answers depends not only on the quality of the solutions but also on how they are explained. On the AP Physics 2 Exam, the words “justify,” “explain,” “calculate,” “what is,” “determine,” and “derive” have specific meanings, and the graders are looking for very precise approaches in your explanations in order to get maximum credit.
Questions that ask you to “justify” are looking for you both to show an understanding in words of the principles underlying physical phenomena and to perform the mathematical operations needed to arrive at the correct answer. The word “justify” as well as the word “explain” requires that you support your answers with text, equations, calculations, diagrams, or graphs. In some cases, the text or equations must elucidate physics fundamentals or laws, while in other cases they will serve to analyze the behavior of different values or different types of variables in the equation.
The word “calculate” requires you to show numerical or algebraic work to arrive at the final answer. In contrast, “what is” and “determine” questions signify that full credit may be given without showing mathematical work. Just remember, showing work that leads to the correct answer is always a good idea when possible, especially since showing work may still earn you partial credit even if the answer is not correct.
“Derive” questions are looking for a more specific approach, which entails beginning the solution with one or more fundamental equations and then arriving at the final answer through the proper use of mathematics, usually involving some algebra.
CRACKING THE FREE-RESPONSE SECTION
Section II is worth 50 percent of your grade on the AP Physics 2 Exam. This section is composed of four free-response questions. You’re given a total of 90 minutes for this section. We at The Princeton Review recommend that you spend 50 minutes on the first two questions (experimental design/translation) and then 40 minutes on the second two questions (the short response questions) because short response questions should require a bit less time. Pacing is a personal decision, though, so feel free to build your own timeline for the free-response section.
Clearly Explain and Justify Your Answers
Remember that your answers to the free-response questions are graded by readers and not by computers. Communication is a very important part of AP Physics 2. Compose your answers in precise sentences. Just getting the correct numerical answer is not enough. You should be able to explain your reasoning behind the technique that you selected and communicate your answer in the context of the problem. Even if the question does not explicitly say so, always explain and justify every step of your answer, including the final answer. Do not expect the graders to read between the lines. Explain everything as though somebody with no knowledge of physics is going to read it. Be sure to present your solution in a systematic manner using solid logic and appropriate language. And remember: Although you won’t earn points for neatness, the graders can’t give you a grade if they can’t read and understand your solution!
Use Only the Space You Need
Do not try to fill up the space provided for each question. The space given is usually more than enough. The people who design the tests realize that some students write in big letters and some students make mistakes and need extra space for corrections. So if you have a complete solution, don’t worry about the extra space. Writing more will not earn you extra credit. In fact, many students tend to go overboard and shoot themselves in the foot by making a mistake after they’ve already written the right answer.
Read the Whole Question!
Some questions might have several subparts. Try to answer them all, and don’t give up on the question if one part is giving you trouble. For example, if the answer to part (b) depends on the answer to part (a), but you think you got the answer to part (a) wrong, you should still go ahead and do part (b) using your answer to part (a) as required. Chances are that the grader will not mark you wrong twice, unless it is obvious from your answer that you should have discovered your mistake.
Use Common Sense
Always use your common sense in answering questions. For example, on one free-response question that asked students to compute the mean weight of newborn babies from given data, some students answered 70 pounds. It should have been immediately obvious that the answer was probably off by a decimal point. A 70-pound baby would be a giant! This is an important mistake that should be easy to fix. Some mistakes may not be so obvious from the answer. However, the grader will consider simple, easily recognizable errors to be very important.
Think Like a Grader
When answering questions, try to think about what kind of answer the grader is expecting. Look at past free-response questions and grading rubrics on the College Board website. These examples will give you some idea of how the answers should be phrased. The graders are told to keep in mind that there are two aspects to the scoring of free-response answers: showing statistical knowledge and communicating that knowledge. Again, responses should be written as clearly as possible in complete sentences. You don’t need to show all the steps of a calculation, but you must explain how you got your answer and why you chose the technique you used.
Think Before You Write
Abraham Lincoln once said that if he had eight hours to chop down a tree, he would spend six of them sharpening his axe. Be sure to spend some time thinking about what the question is, what answers are being asked for, what answers might make sense, and what your intuition is before starting to write. These questions aren’t meant to trick you, so all the information you need is given. If you think you don’t have the right information, you may have misunderstood the question. In some calculations, it is easy to get confused, so think about whether your answers make sense in terms of what the question is asking. If you have some idea of what the answer should look like before starting to write, then you will avoid getting sidetracked and wasting time on dead-ends.
Reflect
Respond to the following questions:
• How much time will you spend on the short free-response questions? What about the long free-response questions?
• What will you do before you begin writing your free response answers?
• Will you seek further help, outside of this book (such as a teacher, tutor, or AP Students), on how to approach the questions that you will see on the AP Physics 2 Exam?
Part V
Content Review for the AP Physics 2 Exam
3 Fluid Mechanics
4 Thermal Physics
5 Electronic Forces and Fields
6 Electronic Potential and Capacitance
7 Direct Current Circuits
8 Magnetic Forces and Fields
9 Electromagnetic Induction
10 Optics
11 Modern Physics
12 Solutions to Chapter Review Questions
Chapter 3
Fluid Mechanics
INTRODUCTION
In this chapter, we’ll discuss some of the fundamental concepts dealing with substances that can flow, which are known as fluids. The term fluid refers to both liquids and gases. While there are distinctions between liquids and gases, this chapter focuses on the similarities of all fluids.
DENSITY
Although the concept of mass is central to your study of mechanics (because of the all-important equation Fnet = ma), it is the substance’s density that turns out to be more useful in fluid mechanics.
By definition, the density of a substance is its mass per unit volume, and it’s typically denoted by the letter ρ (the Greek letter rho):
Note that this equation immediately implies that m is equal to ρV when ρ is a constant.
For example, if 10–3 m3 of oil has a mass of 0.8 kg, then the density of this oil is
PRESSURE
If we place an object in a fluid, the fluid exerts a contact force on the object. How that force is distributed over any small area of the object’s surface defines the pressure:
The subscript ⊥ (which means perpendicular) is meant to emphasize that the pressure is defined to be the magnitude of the force that acts perpendicular to the surface, divided by the area. Because force is measured in newtons (N) and area is expressed in square meters (m2), the SI unit for pressure is the newton per square meter. This unit is given its own name: the pascal, abbreviated Pa:
SI unit of pressure: 1 pascal = 1 Pa = 1
One pascal is a very tiny amount of pressure; for example, a nickel on a table exerts about 140 Pa of pressure, just due to its weight alone. For this reason, you’ll often see pressures expressed in kPa (kilopascals, where 1 kPa = 103 Pa) or even in MPa (megapascals, where 1 MPa = 106 Pa). Another common unit for pressure is the atmosphere (atm). At sea level, atmospheric pressure, Patm, is about 101,300 Pa; this is 1 atmosphere.
Example 1 A vertical column made of cement has a base area of 0.5 m2. If its height is 2 m, and the density of cement is 3000 kg/m3, how much pressure does this column exert on the ground?
Solution. The force the column exerts on the ground is equal to its weight, mg, so we’ll find the pressure it exerts by dividing this by the base area, A. The mass of the column is equal to ρV, which we calculate as follows:
m = ρV = ρAh = (3 × 103 )(0.5 m2) (2 m) = 3 × 103 kg
Therefore,
P = = 6 × 104 Pa = 60 kPa
Now, you may be used to the unit m/s2 for g. However, for this chapter, it is simpler to use the unit N/kg for g, as it is equivalent to m/s2. So don’t be confused when you see it later on.
HYDROSTATIC PRESSURE
Imagine that we have a tank with a lid on top, filled with some liquid. Suspended from this lid is a string, attached to a thin sheet of metal which hangs horizontally. The figures below show two views of this:
The weight of the liquid produces a force that pushes down on the metal sheet. If the sheet has length l and width w, and is at depth h below the surface of the liquid, then the weight of the liquid on top of the sheet is
Fg = mg = ρVg = ρ(lwh)g
where ρ is the liquid’s density. If we divide this weight by the area of the sheet (A = lw), we get the pressure due to the liquid:
Since the liquid is at rest, this is known as hydrostatic pressure.
Note that the hydrostatic pressure due to the liquid, Pliquid = ρgh, depends only on the density of the liquid and the depth below the surface; in fact, it’s proportional to both of these quantities. One important consequence of this is that the shape of the container doesn’t matter. For example, if all the containers in the figure below are filled with the same liquid, then the pressure is the same at every point along the horizontal dashed line (and within a container), simply because every point on this line is at the same depth, h, below the surface of the liquid.
If the liquid in the tank were open to the atmosphere, then the total (or absolute) pressure at depth h would be equal to the pressure pushing down on the surface—the atmospheric pressure, Patm—plus the pressure due to the liquid alone. In general, the pressure on the surface of the liquid is some pressure P0 and the total pressure is
total (absolute) pressure: Ptotal = P0 + Pliquid = P0 + ρgh
When the pressure at the top of the liquid is atmospheric pressure (as in the scenario above), then our equation for total pressure reduces to
Ptotal = Patm + ρgh
Because pressure is the magnitude of the force per area, pressure is a scalar. It has no direction. The direction of the force due to the pressure on any small surface is perpendicular to that surface. For example, in the figure below, the pressure at Point A is the same as the pressure at Point B, because they’re at the same depth.
But, as you can see, the direction of the force due to the pressure varies depending on the orientation of the surface—and even which side of the surface—the force is pushing on.
BUOYANCY
Let’s place a block in our tank of fluid. Because the pressure on each side of the block depends on its average depth, we see that there’s more pressure on the bottom of the block than there is on the top. Because the block is rectangular and the top and bottom have the same area, there’s a greater force pushing up on the block than there is pushing down on it. The forces due to the pressure on the other four sides cancel out (because they are at the same depth), so the net force on the block is upward.
This net upward force is called the buoyant force (or just buoyancy for short), denoted Fbuoy. We calculate the magnitude of the buoyant force using Archimedes’ principle; in words, Archimedes’ principle says
The strength of the buoyant force on the object is equal to the weight of the fluid displaced by the object.
When an object is partially or completely submerged in a fluid, the volume of the object submerged, which we call Vsub, is the volume of the fluid displaced. By multiplying this volume by the density of the fluid, we get the mass of the fluid displaced; then, multiplying this mass by g gives us the weight of the fluid displaced. So, here’s Archimedes’ principle as a mathematical equation:
Buoyant force: Fbuoy = ρfluidVsub g
When an object floats, its submerged volume is just enough to make the buoyant force it feels balance its weight. So, if an object’s density is ρobject and its (total) volume is V, its weight will be mg = ρVg. The buoyant force it feels is ρfluidVsub g. Setting these equal to each other, we find that
So, if ρobject < ρfluid, then the object will float; and the fraction of its volume submerged is the same as the ratio of its density to the fluid’s density. For example, if the object’s density is 2/3 the density of the fluid, then the object will float, and 2/3 of the object will be submerged.
If an object is denser than the fluid, it will sink. In this case, even if the entire object is submerged (in an attempt to maximize Vsub and the buoyant force), its weight is still greater than the buoyant force, and down it goes. And if an object just happens to have the same density as the fluid, it will be happy hovering (in static equilibrium) anywhere underneath the fluid.
Let’s summarize, and go through the steps of an object floating and then an object as it sinks and hits the bottom of a tank. In the diagram on the next page we have an object that is floating in a liquid.
Below is a diagram of an object as it sinks to the bottom of the tank.
Because the buoyant force is always directed upwards, and the force of gravity is defined to point downwards, these forces are always directed in opposite directions. The result of the buoyant force opposing gravity is that any object in a fluid appears to weigh less than when it is not in the fluid, and as the fluid density increases, the apparent weight decreases. The apparent weight of an object is commonly referred to as the normal force of that object.
Example 2 An object with a mass of 150 kg and a volume of 0.75 m3 is floating in ethyl alcohol, whose density is 800 kg/m3. What fraction of the object’s volume is above the surface of the fluid?
Solution. The density of the object is
The ratio of the object’s density to the fluid’s density is
This means that 1/4 of the object’s volume is below the surface of the fluid; therefore, the fraction above the surface is 1 – (1/4) = 3/4. Make sure you know exactly what the question is asking. Typically you use this equation to find the volume submerged, but this question (Example 2) asks for the volume not submerged.
Example 3 A brick, of density 2000 kg/m3 and volume 1.5 × 10–3 m3, is dropped into a swimming pool full of water.
(a) Explain briefly why the brick will sink.
(b) When the brick is lying on the bottom of the pool, what is the apparent weight of the brick?
Solution.
(a) The brick has a greater density than the surrounding fluid (water), so it will sink.
(b) When the brick is lying on the bottom surface of the pool, it is totally submerged, so Vsub = V; this means the buoyant force on the brick is
The weight of the brick is
When the brick is lying on the bottom of the pool, the net force it feels is zero.
Therefore, we must have Fbuoy + FN = Fg, so FN = Fg – Fbuoy = 30 N – 15 N = 15 N. So the apparent weight of the brick, that is, the normal force it feels, is 15 N.
Example 4 A helium balloon has a volume of 0.03 m3. What is the net force on the balloon if it’s surrounded by air? (Note: The density of helium is 0.2 kg/m3, and the density of air is 1.2 kg/m3.)
Solution. The balloon will feel a buoyant force upward, and the force of gravity—the balloon’s weight—downward. Because the balloon is completely surrounded by air, Vsub = V, and the buoyant force is
The weight of the balloon is
Because Fbuoy > w, the net force on the balloon is upward and has magnitude
Fnet = Fbuoy – Fg = 0.36 N – 0.06 N = 0.3 N
This is why a helium balloon floats away if you let go of its string.
FLOW RATE AND THE CONTINUITY EQUATION
Consider a pipe through which fluid is flowing. The flow rate, f, is the volume of fluid that passes a particular point per unit time. In SI units, flow rate is expressed in m3/s. To find the flow rate, all we need to do is multiply the size of the pipe at a particular location by the average speed of the flow at that point. In this context, the size of the pipe means the area of a cross-section of the pipe that is perpendicular to the direction of the flow.
Be careful not to confuse flow rate with flow speed; flow rate tells us how much fluid flows per unit time; flow speed tells us how fast it’s moving. The flow rate is measured in cubic meters per second, while the flow speed is measured in meters per second.
If the pipe is carrying a fluid that is incompressible—that is, its density cannot be changed—then the flow rate must be the same everywhere along the pipe. A liquid is typically an incompressible fluid, whereas a gas is compressible. Rewriting this using f = Av, we get the Continuity Equation:
A1v1 = A2v2
Because the product Av is a constant, the flow speed will increase where the pipe narrows, (decrease in area) and decrease where the pipe widens (increase in area). In fact, we can say that the flow speed is inversely proportional to the cross-sectional area—or to the square of the radius (in the case of a circular cross-section)—of the pipe.
Example 5 A circular pipe of non-uniform diameter carries water. At one point in the pipe, the radius is 2 cm and the flow speed is 6 m/s.
(a) What is the flow rate?
(b) What is the flow speed at a point where the pipe constricts to a radius of 1 cm?
Solution.
(a) At any point, the flow rate, f, is equal to the cross-sectional area of the pipe multiplied by the flow speed
f = Av = πr2v = π(2 × 10‒2 m)2 (6 m/s) ≈ 75 × 10‒4 m3/s = 7.5 × 10‒3 m3 /s
(b) By the Continuity Equation, we know that the flow rate must be the same at all points, so the flow rate must be the same as in part (a).
f = 7.510‒3 m3/s = Av = πr2 = π(110‒2 m)2 (v)
v ≈ 24 m/s
A second approach to part (b) is to use proportional reasoning. From the Continuity Equation, we know that v, the flow speed, is inversely proportional to A, the cross-sectional area of the pipe. If the pipe’s radius decreases by a factor of 2 (from 2 cm to 1 cm), then A decreases by a factor of 4, because A is proportional to r2.
If A decreases by a factor of 4, then v will increase by a factor of 4. Therefore, the flow speed at a point where the pipe’s radius is 1 cm will be 4 • (6 m/s) = 24 m/s.
BERNOULLI’S EQUATION
The most important equation in fluid mechanics is Bernoulli’s Equation, which is the statement of conservation of energy for ideal fluid flow. First, let’s describe the conditions that make fluid flow ideal.
• The fluid is incompressible.
This works very well for liquids and also applies to gases if the pressure changes are small.
• The fluid’s viscosity is negligible.
Viscosity is the force of cohesion between molecules in a fluid; think of viscosity as internal friction for fluids. For example, maple syrup is sticky and has a greater viscosity than water: there’s more resistance to a flow of maple syrup than to a flow of water. While Bernoulli’s Equation would give good results when applied to a flow of water, it would not give good results if it were applied to a flow of maple syrup.
• The flow is streamline.
In a tube carrying a flowing fluid, a streamline is just what it sounds like: it’s a line in the stream. If we were to inject a drop of dye into a clear glass pipe carrying, say, water, we’d see a streak of dye in the pipe, indicating a streamline.
When the flow is streamline, the fluid moves smoothly through the tube. (The opposite of streamline flow is turbulent flow, which is characterized by rapidly swirling whirlpools; such chaotic flow is unpredictable.)
If the three conditions described above hold, and the flow rate, f, is steady (meaning it doesn’t change with time), Bernoulli’s Equation can be applied to any pair of points along a streamline within the flow. Let ρ be the density of the fluid that’s flowing. Label the points we want to compare as Point 1 and Point 2. Choose a horizontal reference level, and let y1 and y2 be the heights of these points above this level. If the pressures at Points 1 and 2 are P1 and P2, and if the flow speeds at these points are v1 and v2, then Bernoulli’s Equation says
An alternative, but equivalent, way of stating Bernoulli’s Equation is to say that the quantity
P + ρ gy + ρv2
is constant along a streamline. We mentioned earlier that Bernoulli’s Equation is a statement of conversation of energy. You should notice the similarity between ρgy and mgh (gravitational potential energy) as well as between ρv2 and mv2 (kinetic energy).
Torricelli’s Theorem
Imagine that we punch a small hole in the side of an open tank of liquid. We can use Bernoulli’s Equation to figure out how fast the liquid will flow out of the hole.
Let the bottom of the tank be our horizontal reference level, and choose Point 1 to be at the surface of the liquid and Point 2 to be at the hole where the water shoots out. First, the pressure at Point 1 is atmospheric pressure since the tank is open; and the emerging stream at Point 2 is open to the air, so it’s at atmospheric pressure too. Therefore, P1 = P2 , and these terms cancel out of Bernoulli’s Equation. Next, since the area at Point 1 is so much greater than at Point 2, we can assume that v1, the speed at which the water level in the tank drops, is much lower than v2 , the speed at which the water shoots out of the hole. (Remember that by the Continuity Equation, A1v1 = A2v2 ; since A1 ≫ A2 , we’ll have v1 ≪ v2.) Because v1 ≪ v2, we can say that the water level at the top of the tank is approximately constant, and the speed of the water at the top is zero. So, Bernoulli’s Equation becomes
ρ gy1 = ρ gy2 + ρv
Solving for v2, we get
This is called Torricelli’s Theorem.
Example 6 In the figure below, a pump forces water at a constant flow rate through a pipe whose cross-sectional area, A, gradually decreases: at the exit point, A has decreased to 1/3 its value at the beginning of the pipe. If y = 60 cm and the flow speed of the water just after it leaves the pump (Point 1 in the figure) is 1 m/s, what is the gauge pressure at Point 1?
Solution. We’ll apply Bernoulli’s Equation to Point 1 and the exit point, Point 2. We’ll choose the level of Point 1 as the horizontal reference level; this makes y1 = 0. Now, because the cross-sectional area of the pipe decreases by a factor of 3 between Points 1 and 2, the flow speed must increase by a factor of 3; that is, v2 = 3v1. Since the pressure at Point 2 is patm , Bernoulli’s Equation becomes
P1 + ρv = Patm + ρ gy2 + ρv
Now, P1 is the total pressure at point 1. Recall that gauge pressure is Ptot – Patm, so Pgauge = P1 – Patm.
Therefore,
Example 7 The side of an above-ground pool is punctured, and water gushes out through the hole. If the total depth of the pool is 2.5 m, and the puncture is 1 m above ground level, what is the efflux speed of the water?
Solution. Torricelli’s Theorem is v = , where h is the distance from the surface of the pool down to the hole. If the puncture is 1 m above ground level, then it’s 2.5 – 1 = 1.5 m below the surface of the water (because the pool is 2.5 m deep). Therefore, the efflux speed will be
The Bernoulli Effect
Consider the two points labeled in the pipe shown below:
Since the heights y1 and y2 are equal in this case, the terms in Bernoulli’s Equation that involve the heights will cancel, leaving us with
P1 + ρv = P2 + ρv
We already know from the Continuity Equation (f = Av) that the speed increases as the cross-sectional area of the pipe decreases; that is, since A2 < A1, we know that v2 > v1, so the equation above tells us that P2 < P1. This shows that
At comparable heights, the pressure is lower where the flow speed is greater.
This is known as the Bernoulli (or Venturi) Effect, and is illustrated in the figure below.
The height of the liquid column above Point 2 is less than the height of the liquid column above Point 1, because the pressure at Point 2 is lower than at Point 1, due to the fact that the flow speed at Point 2 is greater than at Point 1.
The Bernoulli Effect also accounts for many everyday phenomena. It’s what allows airplanes to fly, curve balls to curve, and tennis balls hit with top spin to drop quickly. You may have seen sky divers or motorcycle riders wearing a jacket that seems to puff out as they move rapidly through the air. The essentially stagnant air trapped inside the jacket is at a much higher pressure than the air whizzing by outside, and as a result, the jacket expands outward. The drastic drop in air pressure that accompanies the high winds in a hurricane or tornado is yet another example. In fact, if high winds streak across the roof of a house whose windows are closed, the outside air pressure can be reduced so much that the air pressure inside the house (where the air speed is essentially zero) can be great enough to blow the roof off.
Chapter 3 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 3 Review Questions.
Section I: Multiple Choice
1. A large tank is filled with water to a depth of 6 m. If Point X is 1 m from the bottom and Point Y is 2 m from the bottom, how does ρX, the hydrostatic pressure due to the water at Point X, compare to ρY, the hydrostatic pressure due to the water at Point Y ?
(A) ρX = 2ρY
(B) 2ρX = ρY
(C) 5ρX = 4ρY
(D) 4ρX = 5ρY
2. In the figure above, a box of dimensions x, y, and z rests on the bottom of a tank filled to depth D with a liquid of density ρ. If the tank is open to the atmosphere, what is the force on the (shaded) top of the box?
(A) xy(Patm + ρgD)
(B) xyz[Patm – ρg(z ‒ D)]
(C) xy[Patm + ρg(D ‒ z)]
(D) xyz[Patm + ρgD]
3. The figure above shows a closed container partially filled with liquid. Point Y is at a depth of 1 m, and Point Z is at a depth of 3 m. If the pressure at Point Y is 13,000 Pa, and the pressure at Point Z is 29,000 Pa, what is the pressure at the surface, Point X ?
(A) 3,000 Pa
(B) 4,000 Pa
(C) 5,000 Pa
(D) 7,000 Pa
4. A plastic cube 0.5 m on each side and with a mass of 100 kg floats in water. What fraction of the cube’s volume is above the surface of the water?
(A)
(B)
(C)
(D)
5. A block of Styrofoam, with a density of ρS and volume V, is pushed completely beneath the surface of a liquid whose density is ρL, and released from rest. Given that ρL > ρS, which of the following expressions gives the magnitude of the block’s initial upward acceleration?
(A) (ρL ‒ ρS)g
(B)
(C)
(D)
6. In the figure above, a ball with a density of 400 kg/m3 and volume 5 × 10–3 m3 is attached to a string, the other end of which is fastened to the bottom of the tank. If the tank is filled with water, what is the tension in the string?
(A) 20 N
(B) 30 N
(C) 40 N
(D) 50 N
7. An object with a density of 2000 kg/m3 weighs 100 N less when it’s weighed while completely submerged in water than when it’s weighed in air. What is the actual weight of this object?
(A) 200 N
(B) 300 N
(C) 400 N
(D) 600 N
8. In the pipe shown above, which carries water, the flow speed at Point X is 6 m/s. What is the flow speed at Point Y ?
(A) m/s
(B) 2 m/s
(C) 18 m/s
(D) 54 m/s
9. The figure above shows a portion of a conduit for water, one with rectangular cross sections. If the flow speed at the top is v, what is the flow speed at the bottom?
(A) 4v
(B) 8v
(C) 12v
(D) 16v
10. A pump is used to send water through a hose, the diameter of which is 10 times that of the nozzle through which the water exits. If the nozzle is 1 m higher than the pump, and the water flows through the hose at 0.4 m/s, what is the difference in pressure between the pump and the atmosphere?
(A) 108 kPa
(B) 260 kPa
(C) 400 kPa
(D) 810 kPa
Section II: Free Response
1. The figure below shows a tank open to the atmosphere and filled to depth D with a liquid of density ρL. Suspended from a string is a block of density ρB (which is greater than ρL), whose dimensions are x, y, and z (meters). The top of the block is at depth h meters below the surface of the liquid.
In each of the following, write your answer in simplest form in terms of ρL, ρB, x, y, z, h, D, and g.
(a) Find the force due to the pressure on the top surface of the block and on the bottom surface. Sketch these forces in the diagram below:
(b) What are the average forces due to the pressure on the other four sides of the block? Sketch these forces in the diagram above.
(c) What is the total force on the block due to the pressure?
(d) Find an expression for the buoyant force on the block. How does your answer here compare to your answer to part (c)?
(e) What is the tension in the string?
2. The figure below shows a large, cylindrical tank of water, open to the atmosphere, filled with water to depth D. The radius of the tank is R. At a depth h below the surface, a small circular hole of radius r is punctured in the side of the tank, and the point where the emerging stream strikes the level ground is labeled X.
In parts (a) through (c), assume that the speed with which the water level in the tank drops is negligible.
(a) At what speed does the water emerge from the hole?
(b) How far is point X from the edge of the tank?
(c) Assume that a second small hole is punctured in the side of the tank, a distance of h/2 directly above the hole shown in the figure. If the stream of water emerging from this second hole also lands at Point X, find h in terms of D.
(d) For this part, do not assume that the speed with which the water level in the tank drops is negligible, and derive an expression for the speed of efflux from the hole punctured at depth h below the surface of the water. Write your answer in terms of r, R, h, and g.
3. The figure below shows a pipe fitted with a Venturi U-tube. Fluid of density ρF flows at a constant flow rate and with negligible viscosity through the pipe, which constricts from a cross-sectional area A1 at Point 1 to a smaller cross-sectional area A2 at Point 2. The upper portion of both sides of the Venturi U-tube contain the same fluid that’s flowing through the pipe, while the lower portion is filled with a fluid of density ρV (which is greater than ρF). At Point 1 in the pipe, the pressure is P1 and the flow speed is v1 ; at Point 2 in the pipe, the pressure is P2 and the flow speed is v2. All the fluid within the Venturi U-tube is stationary.
(a) What is ρX, the hydrostatic pressure at Point X? Write your answer in terms of P1, ρF, h1, and g.
(b) What is ρY, the hydrostatic pressure at Point Y? Write your answer in terms of P2, ρF, ρV , h2, d, and g.
(c) Write down the result of Bernoulli’s Equation applied to Points 1 and 2 in the pipe, and solve for P1 – P2.
(d) Since ρX = ρY, set the expressions you derived in parts (a) and (b) equal to each other, and use this equation to find P1 – P2.
(e) Derive an expression for the flow speed, v2 , and the flow rate, f, in terms of A1, A2, d, ρF, ρV, and g. Show that v2 and f are proportional to .
Summary
Density is given by ρ = . Pressure is given by P = . Hydrostatic pressure can be found using P = Po + ρ gh where ρ gh is the pressure at a given depth below the surface of the fluid and P0 is the pressure right above the surface of the fluid.
The buoyant force is an upward force any object immersed in a fluid experiences due to the displaced fluid. The buoyant force is given by Fbuoy = ρVg, where ρ is the density of the fluid and V is the volume of the fluid displaced.
The Continuity Equation says that the flow rate through a pipe (area times velocity) is constant so that A1v1 = A2v2. This expresses the idea that a larger cross sectional area of pipe will experience fluids traveling at a lower velocity.
Bernoulli’s Equation is a statement of conservation of energy. It states P + ρgy + ρv2 = constant
Chapter 4
Thermal Physics
INTRODUCTION
This chapter looks at heat and temperature, concepts that seem familiar from our everyday experience. Technically, heat is defined as thermal energy transmitted from one body to another. While an object can contain thermal energy (due to the random motion of its molecules), an object doesn’t contain heat; heat is energy in transit. Temperature, on the other hand, is a measure of an object’s internal thermal energy.
HEAT TRANSFER
There are three principal modes by which energy can be transferred:
Conduction
An iron skillet is sitting on a hot stove, and you accidentally touch the handle. You notice right away that there’s been a transfer of thermal energy to your hand. The process by which this happens is known as conduction. The highly agitated atoms in the handle of the hot skillet bump into the atoms of your hand, making them vibrate more rapidly, thus heating up your hand.
Convection
As the air around a candle flame warms, it expands, becomes less dense than the surrounding cooler air, and thus rises due to buoyancy. As a result, heat is transferred away from the flame by the large scale (from the atoms’ point of view anyway) motion of a fluid (in the case, air). This is convection.
Radiation
Sunlight on your face warms your skin. Radiant energy from the sun’s fusion reactions is transferred across millions of kilometers of essentially empty space via electromagnetic waves. Absorption of the energy carried by these light waves defines heat transfer by radiation.
Heat conducts from one point to another only if there is a temperature difference between the two objects. The rate at which heat is transferred is given by
THE KINETIC THEORY OF GASES
Unlike the condensed phases of matter—solid and liquid—the atoms or molecules that make up a gas do not move around relatively fixed positions. Rather, the molecules of a gas move freely and rapidly, in a chaotic swarm. A confined gas exerts a force on the walls of its container, because the molecules are zipping around inside the container, striking the walls and rebounding. The magnitude of the force per unit area is just the pressure, P, as defined in the previous chapter.
As we’ll see, the faster the gas molecules are moving, the more pressure they exert.
THE IDEAL GAS LAW
Three physical properties—pressure (P), volume (V), and temperature (T)— describe a gas. At low densities, all gases approach ideal behavior; this means that these three variables are related by the equation
PV = nRT
where n is the number of moles of gas and R is a constant (8.31 J/mol∙K) called the universal gas constant. This equation is known as the Ideal Gas Law. It can also be written as PV = NkBT , where kB is Boltzmann’s constant (kB = 1.38 × 10–23 J/K). The ideal gas law tells us how these variables are related to one another for a gas with no intermolecular forces. This is a bad assumption for a charged gas, where the individual charges would interact electrically (a strong interaction), but it is a good assumption for a neutral gas, where the atoms would only interact gravitationally (a very weak interaction).
An important consequence of this equation is that, for a fixed volume of gas, an increase in P gives a proportional increase in T. The pressure increases when the gas molecules strike the walls of their container with more force, which occurs if they move more rapidly. Using Newton’s Second Law (rate of change of momentum = force) we can find that the pressure exerted by N molecules of gas in a container of volume V is related to the average kinetic energy of the molecules by the equation PV = NKavg. Comparing this to the Ideal Gas Law we see that NKavg = nRT. We can rewrite this equation in the form NAKavg = RT, since, by definition, N = nNA. The ratio R/NA is a fundamental constant of nature called Boltzmann’s constant (kB = 1.38 × 10–23 J/K), so our equation becomes
Kavg = kBT
This tells us that the average translational kinetic energy of the gas molecules is directly proportional to the absolute temperature of the sample. Remember, this means you must use kelvins as your temperature unit.
Since the average kinetic energy of the gas molecules is Kavg = m(v2)avg, the equation above becomes m(v2)avg = kBT, so
The quantity on the left-hand side of this equation, the square root of the average of the square of v, is called the root-mean-square speed, vrms , so
vrms =
It’s important to realize that the molecules in the container have a wide range of speeds; some are much slower and others are much faster than vrms. The root-mean-square speed is important because it gives us a type of average speed that’s easy to calculate from the temperature of the gas.
Example 1 In order for the rms speed of the molecules in a given sample of gas to double, what must happen to the temperature?
Solution. Temperature is a measure of the average kinetic energy. The velocity is determined from the following equation:
vrms =
Since vrms is proportional to the square root of T, the temperature must quadruple, again, assuming the temperature is given in kelvins.
Example 2 A cylindrical container of radius 15 cm and height 30 cm contains 0.6 mole of gas at 433 K. How much force does the confined gas exert on the lid of the container?
Solution. The volume of the cylinder is πr2h, where r is the radius and h is the height. Since we know V and T, we can use the Ideal Gas Law to find P. Because pressure is force per unit area, we can find the force on the lid by multiplying the gas pressure times the area of the lid.
P = = 1.018 × 105 Pa
So, since the area of the lid is πr2, the force exerted by the confined gas on the lid is:
F = PA = (1.018 × 105 Pa)∙π(0.15 m)2 = 7,200 N
This is about 1,600 pounds of force, which seems like a lot. Why doesn’t this pressure pop the lid off? Because, while the bottom of the lid is feeling a pressure (due to the confined gas) of 1.018 × 105 Pa that exerts a force upward, the top of the lid feels a pressure of 1.013 × 105 Pa (due to the atmosphere) that exerts a force downward. The net force on the lid is
Fnet = (ΔP)A = (0.005 × 105 Pa)∙π(0.15 m)2 = 35 N
which is only about 8 pounds.
THE LAWS OF THERMODYNAMICS
We’ve learned about two ways in which energy may be transferred between a system and its environment. One is work, which takes place when a force acts over a distance. The other is heat, which takes place when energy is transferred due to a difference in temperature. The study of the energy transfers involving work and heat, and the resulting changes in internal energy, temperature, volume, and pressure is called thermodynamics.
The Zeroth Law of Thermodynamics
When two objects are brought into contact, heat will flow from the warmer object to the cooler one until they reach thermal equilibrium. This property of temperature is expressed by the Zeroth Law of Thermodynamics:
If Objects 1 and 2 are each in thermal equilibrium with Object 3, then Objects 1 and 2 are in thermal equilibrium with each other.
The First Law of Thermodynamics
Simply put, the First Law of Thermodynamics is a statement of the Conservation of Energy that includes heat.
Energy (in the form of heat) is neither created nor destroyed in any thermodynamic system.
Consider the following example, which is the prototype that’s studied extensively in thermodynamics.
An insulated container filled with an ideal gas rests on a heat reservoir (that is, something that can act as a heat source or a heat sink). The container is fitted with a snug, but frictionless, weighted piston that can be raised or lowered. The confined gas is the system, and the piston and heat reservoir are the surroundings.
The state of the gas is given once its pressure, volume, and temperature are known, and the equation that connects these state variables is the Ideal Gas Law, PV = nRT. We’ll imagine performing different experiments with the gas, such as heating it or allowing it to cool or increasing or decreasing the weight on the piston, and study the energy transfers (work and heat) and the changes in the state variables. If each process is carried out such that, at each moment, the system and its surroundings are in thermal equilibrium, we can plot the pressure (P) versus the volume (V) on a diagram. By following the path of this P–V diagram, we can study how the system is affected as it moves from one state to another.
Work is done on or by the system when the piston is moved and the volume of the gas changes. For example, imagine that the weight pushes the piston downward a distance d, causing a decrease in volume. Assume that the pressure stays constant at P. (Heat must be removed via the reservoir to accomplish this.) We can calculate the work done on the gas during this compression as W = –Fd, but since F = PA, we have W = –PAd, and because Ad = ∆V, we have
W = –P∆V
Textbooks differ about the circumstances under which work in thermodynamics is defined to be positive or negative. The negative signs we have included in the equations above are consistent with those used in the AP Physics 2 Exam (appearing in the equation sheet given for Section II). For the exam, work in thermodynamics is considered to be positive when the work is being done on the system. This means that the volume of the system is decreasing, ∆V is negative, and, in agreement with intuition, energy is being added to the system. In other words, when work is done in compressing a system, ∆V is negative and the work done, W = –P∆V, is positive. This also means, conversely, that when the system is doing work on the surroundings (volume increasing, ∆V positive), the work is negative, in agreement with intuition that energy is leaving the system.
The equation W = –P∆V assumes that the pressure P does not change during the process. If P does change, then the work is equal to the area under the curve in the P–V diagram; moving left to right as the volume increases gives a negative area (and negative work), while moving right to left as the volume decreases gives a positive area (and positive work). Because of this difference when moving left to right or right to left, processes in P–V diagrams are drawn with arrows to indicate which direction the process follows. In the diagram below, the process is taken from an initial state a, with a particular P and V value to a final state b, where P and V have changed.
Example 3 What’s the value of W for the process ab following path 1 and for the same process following path 2 (from a to d to b), shown in the P–V diagram below?
Solution.
Path 1. Since, in path 1, P remains constant, the work done is just –P ∆ V:
W = –P∆V = – (1.5 × 105 Pa)[(30 × 10–3 m3) – (10 × 10–3 m3)] = –3,000 J
Path 2. If the gas is brought from state a to state b, along path 2, then work is done only along the part from a to d. From d to b, the volume of the gas does not change, so no work can be performed. The area under the graph from a to d is
As this example shows, the amount of work done by a gas depends not only on the initial and final states of the system, but also on the path between the two. In general, different paths give different values for W.
Experiments have shown that the value of Q + W is not path dependent; it depends only on the initial and the final state of the system, so it describes a change in a fundamental property. This property is called the system’s internal energy, denoted U, and the change in the system’s internal energy, ∆U, is equal to Q + W. This is true regardless of the process that brought the system from its initial to final state. This statement is known as
The First Law of Thermodynamics
∆U = Q + W
∆U depends only on temperature change
(If volume is constant, W = 0 and U = Q.)
This statement of the First Law is consistent with the interpretation of work (W = –PΔV) explained above. The First Law identifies W and Q as separate physical mechanisms for adding to or removing energy from the system, and the signs of both W and Q are defined consistently: Both are positive when they are adding energy to the system and negative when they are removing energy from the system.
Example 4 A 0.5 mol sample of an ideal gas is brought from state a to state b when 7500 J of heat is added along the path shown in the following P–V diagram:
What are the values of each of the following?
(a) the temperature at a
(b) the temperature at b
(c) the work done by the gas during process ab
(d) the change in the internal energy of the gas
Solution.
(a, b) Both of these questions can be answered using the Ideal Gas Law, T = PV/(nR):
(c) Since the pressure remains constant during the process, we can use the equation W = –P∆V. Because ∆V = (30 – 10) × 10–3 m3 = 20 × 10–3 m3, we find that
W = –P∆V = (1.5 × 105 Pa)(20 × 10–3 m3) = –3,000 J
The expanding gas did negative work against its surroundings, pushing the piston upward. Important note: If the pressure remains constant (which is designated by a horizontal line in the P–V diagram), the process is called isobaric.
(d) By the First Law of Thermodynamics,
∆U = Q + W = 7,500 – 3,000 J = 4,500 J
Example 5 A 0.5 mol sample of an ideal monatomic gas is brought from state a to state b along the path shown in the following P–V diagram:
What are the values of each of the following?
(a) the work done by the gas during process ab
(b) the change in the internal energy of the gas
(c) the heat added to the gas during process ab
Solution. Note that the initial and final states of the gas are the same as in the preceding example, but the path is different.
(a) Let’s break the path into 3 pieces:
Over paths 1 and 3, the volume does not change, so no work is done. Work is done only over path 2:
W = –P∆V = –(0.5 × 105 Pa)(20 × 10–3 m3) = –1,000 J
Once again, the expanding gas does negative work against its surroundings, pushing the piston upward.
(b) Because the initial and final states of the gas are the same here as they were in the preceding example, the change in internal energy, ∆U, must be the same. Therefore, ∆U = 4,500 J.
(c) By the First Law of Thermodynamics, ∆U = Q + W, so
Q = ∆U ‒W = 4,500 J ‒ (‒1,000 J) = 5,500 J
Example 6 An isochoric process is one that takes place with no change in volume. What can you say about the change in the internal energy of a gas if it undergoes an isochoric change of state?
Solution. An isochoric process is illustrated by a vertical line in a P–V diagram and, since no change in volume occurs, W = 0. By the First Law of Thermodynamics, ∆U = Q + W = Q. Therefore, the change in internal energy is entirely due to (and equal to) the heat transferred. If heat is transferred into the system (positive Q), then ∆U is positive; if heat is transferred out of the system (negative Q), then ∆U is negative.
Example 7 A 0.5 mol sample of an ideal gas is brought from state a back to state a along the path shown in the following P–V diagram:
What are the values of each of the following?
(a) the change in the internal energy of the gas
(b) the work done on the gas during the process
(c) the heat added to the gas during the process
Solution. A process such as this, which begins and ends at the same state, is said to be cyclical.
(a) Because the final state is the same as the initial state, the internal energy of the system cannot have changed, so ∆U = 0.
(b) The total work involved in the process is equal to the work done from c to d plus the work done from b to a,
because only along these paths does the volume change. Along these portions, we find that
So the total work done is W = +2,000 J. The fact that W is positive means that, overall, work was done on the gas by the surroundings. Notice that for a cyclical process, the total work done is equal to the area enclosed by the loop, with clockwise travel taken as negative and counterclockwise travel taken as positive.
(c) The First Law of Thermodynamics states that ∆U = Q + W. Since ∆U = 0, it must be true that Q = –W (which will always be the case for a cyclical process), so Q = –2,000 J.
Example 8 A 0.5 mol sample of an ideal gas is brought from state a to state d along an isotherm, then isobarically to state c and isochorically back to state a, as shown in the following P–V diagram:
A process that takes place with no variation in temperature is said to be isothermal. Given that the work done during the isothermal part of the cycle is –1,650 J, how much heat is transferred during the isothermal process, from a to d?
Solution. Be careful that you don’t confuse isothermal with adiabatic. A process is isothermal if the temperature remains constant; a process is adiabatic if Q = 0. You might ask, How could a process be isothermal without also being adiabatic at the same time? Remember that the temperature is determined by the internal energy of the gas, which is affected by changes in Q, W, or both. Therefore, it’s possible for U to remain unchanged even if Q is not 0 (because there can be an equal but opposite W to cancel it out). In fact, this is the key to this problem. Since T doesn’t change from a to d, neither can the internal energy, which depends entirely on T. Because ∆Uad = 0, it must be true that Qad = –Wad. Since Wad equals –1,650 J, Qad must be +1,650 J. The gas absorbs heat from the reservoir and uses all this energy to do negative work as it expands, pushing the piston upward.
The Second Law of Thermodynamics
Entropy
Consider a box containing two pure gases separated by a partition. What would happen if the partition were removed? The gases would mix, and the positions of the gas molecules would be random.
A closed system that shows a high degree of order tends to evolve in such a way that its degree of order decreases. In other words, disorder (or, as it’s technically called, entropy) increases. The term “disorder” often makes the concept of entropy sound negative. Rather, entropy is better described as increasing molecular freedom. This is the reason why broken glass does not put itself back together and the reason why chemical reactions take place. If we started with the box on the right, containing the mixture of the gases, it would be virtually impossible that at any later time all the molecules of Gas 1 would happen to move to the left side of the box and, at the same time, the molecules of Gas 2 would spontaneously move to the right side of the box. If we were to watch a movie of this process, and saw the mixed-up molecules suddenly separate and move to opposite sides of the box, we’d assume that the film was running backwards. In a way, the Second Law of Thermodynamics defines the direction of time. Time flows in such a way that ordered systems become disordered. Disordered states do not spontaneously become ordered without any other changes taking place. The following is the essence of one form of the Second Law of Thermodynamics:
The total amount of disorder—the total entropy—of a system plus its surroundings will never decrease.
It is possible for the entropy of a system to decrease, but it will always be at the expense of a greater increase in entropy in the surroundings. For example, when water freezes, its entropy decreases. The molecules making up an ice crystal have a more structured order than the random collection of water molecules in the liquid phase, so the entropy of the water decreases when it freezes. But when water freezes, it releases heat energy into its environment, which creates disorder in the surroundings. If we were to figure out the total change in entropy of the water plus its surroundings, we would find that although the entropy of the water itself decreased, it was more than compensated by a greater amount of entropy increase in the surroundings. So, the total entropy of the system and its surroundings increased, in agreement with the Second Law of Thermodynamics.
There are several equivalent statements of the second law. In addition to the entropy form says that heat always flows from an object at higher temperature to an object at lower temperature, never the other way around. Heat always flows from hot to cold, never cold to hot. Another form, which will be considered more fully in a moment, says that a heat engine can never operate at 100% efficiency, or equivalently, that it is impossible to convert heat completely into work.
Heat Engines
The form of the Second Law of Thermodynamics that we’ll study for the AP Physics 2 Exam deals with heat engines.
Converting work to heat is easy—rubbing your hands together in order to warm them up shows that work can be completely converted to heat. What we’ll look at it is the reverse process: How efficiently can heat be converted into work? A device that uses heat to produce useful work is called a heat engine. The internal-combustion engine in a car is an example. In particular, we’re only interested in engines that take their working substance (a mixture of air and fuel in this case) through a cyclic process, so that the cycle can be repeated. The basic components of any cyclic heat engine are simple: Energy in the form of heat comes into the engine from a high-temperature source, some of this energy is converted into useful work, the remainder is ejected as exhaust heat into a low-temperature sink, and the system returns to its original state to run through the cycle again.
All engines function to convert heat into work by exchanging energy from a reservoir at higher temperature to reservoirs of lower temperature. A good example of this is a refrigerator. In a fridge, a liquid moves through tubes to remove heat from the internal chamber (such as the freezer) to the outer air. Thus, cold is not actually a concept, but the sensation of low heat content. This is why the back of a refrigerator feels hot.
Since we’re looking at cyclic engines only, the system returns to its original state at the end of each cycle, so ∆U must be 0. Therefore, by the First Law of Thermodynamics, Qnet = –W. That is, the net heat absorbed by the system is equal to the work performed by the system. The heat absorbed from the high- temperature source is denoted QH (H for hot), and the heat that is discharged into the low-temperature reservoir is denoted QC (C for cold). Because heat coming in is positive and heat going out is negative, QH is positive and QC is negative, and the net heat absorbed is QH + QC. Instead of writing Qnet in this way, it’s customary to write it as QH – |QC|, to show explicitly that Qnet is less than QH.
This is one of the forms of
The Second Law of Thermodynamics
For any cyclic heat engine, some exhaust heat is always produced. It’s impossible to completely convert heat into useful work.
Example 9 A heat engine draws 800 J of heat from its high- temperature source and discards 450 J of exhaust heat into its cold-temperature reservoir during each cycle. How much work does this engine perform per cycle?
Solution. The absolute value of the work output per cycle is equal to the difference between the heat energy drawn in and the heat energy discarded:
|W| = QH ‒|QC| = 800 J ‒ 450 J = 350 J
Chapter 4 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 4 Review Questions.
Section I: Multiple Choice
1. A container holds a mixture of two gases, CO2 and H2, in thermal equilibrium. Let KC and KH denote the average kinetic energy of a CO2 molecule and an H2 molecule, respectively. Given that a molecule of CO2 has 22 times the mass of a molecule of H2, the ratio KC/ KH is equal to
(A) 1/22
(B) 1
(C)
(D) 22
2. If the temperature and volume of a sample of an ideal gas are both doubled, then the pressure
(A) decreases by a factor of 2
(B) increases by a factor of 2
(C) increases by a factor of 4
(D) remains unchanged
3. In three separate experiments, a gas is transformed from state Pi, Vi to state Pf, Vf along the paths (1, 2, and 3) illustrated in the figure below:
The work done on the gas is
(A) greatest for path 1
(B) least for path 2
(C) greatest for path 2
(D) the same for all three paths
4. An ideal gas is compressed isothermally from 20 m3 to 10 m3. During this process, 5 J of work is done to compress the gas. What is the change of internal energy for this gas?
(A) –10 J
(B) –5 J
(C) 0 J
(D) 5 J
5. An ideal gas is confined to a container whose volume is fixed. If the container holds n moles of gas, by what factor will the pressure increase if the absolute temperature is increased by a factor of 2 ?
(A) 2/(nR)
(B) 2
(C) 2nR
(D) 2/n
6. Two large glass containers of equal volume each hold 1 mole of gas. Container 1 is filled with hydrogen gas (2 g/mol), and Container 2 holds helium (4 g/mol). If the pressure of the gas in Container 1 equals the pressure of the gas in Container 2, which of the following is true?
(A) The temperature of the gas in Container 1 is lower than the temperature of the gas in Container 2.
(B) The temperature of the gas in Container 1 is greater than the temperature of the gas in Container 2.
(C) The rms speed of the gas molecules in Container 1 is lower than the rms speed of the gas molecules in Container 2.
(D) The rms speed of the gas molecules in Container 1 is greater than the rms speed of the gas molecules in Container 2.
7. Through a series of thermodynamic processes, the internal energy of a sample of confined gas is increased by 560 J. If the net amount of work done on the sample by its surroundings is 320 J, how much heat was transferred between the gas and its environment?
(A) 240 J absorbed
(B) 240 J dissipated
(C) 880 J absorbed
(D) 880 J dissipated
8. What’s the total work performed on the gas as it’s transformed from state a to state c, along the path indicated?
(A) 1,500 J
(B) 3,000 J
(C) 4,500 J
(D) 9,500 J
9. In one of the steps of the Carnot cycle, the gas undergoes an isothermal expansion. Which of the following statements is true concerning this step?
(A) No heat is exchanged between the gas and its surroundings, because the process is isothermal.
(B) The temperature decreases because the gas expands.
(C) The internal energy of the gas remains constant.
(D) The internal energy of the gas decreases due to the expansion.
10. A cup of hot coffee is sealed inside a perfectly thermally insulating container. A long time is allowed to pass. Which of the following correctly explains the final thermal configuration within the box?
(A) The coffee has not changed temperature because the box is perfectly insulating.
(B) The coffee has gotten warmer and the air in the box has gotten cooler because of an exchange of thermal energy between the air and the coffee.
(C) The coffee has gotten cooler and the air in the box has gotten warmer because of an exchange of thermal energy between the air and the coffee.
(D) The coffee has gotten cooler but the air in the box has not changed its temperature. The energy from the coffee has caused an increase in entropy within the box.
Section II: Free Response
1. When a system is taken from state a to state b along the path acb shown in the figure below, 70 J of heat flows into the system, and the system does 30 J of work.
(a) When the system is returned from state b to state a along the curved path shown, 60 J of heat flows out of the system. Does the system perform work on its surroundings or do the surroundings perform work on the system? How much work is done?
(b) If the system does 10 J of work in transforming from state a to state b along path adb, does the system absorb or does it emit heat? How much heat is transferred?
(c) If Ua = 0 J and Ud = 30 J, determine the heat absorbed in the processes db and ad.
(d) For the process adbca, identify each of the following quantities as positive, negative, or zero:
W =________________
Q =________________
U =________________
2. A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown in the P–V diagram below:
Path ab is an isotherm, and it can be shown that the work done by the gas as it changes isothermally from state a to state b is given by the equation
Wab = −nRT × ln
The molar heat capacities for the gas are CV = 20.8 J/mol ∙ K and CP = 29.1 J/mol ∙K.
(a) What’s the temperature of
(i) state a?
(ii) state b?
(iii) state c?
(b) Determine the change in the internal energy of the gas for
(i) step ab
(ii) step bc
(iii) step ca
(c) How much work, Wab, is done by the gas during step ab?
(d) What is the total work done over cycle abca?
(e) (i) Is heat absorbed or discarded during step ab?
(ii) If so, how much?
(f) What is the maximum possible efficiency (without violating the Second Law of Thermodynamics) for a cyclical heat engine that operates between the temperatures of states a and c?
Summary
In an isolated system, Qgained + Qlost = 0.
For phase changes there is no temperature change and the equation that relates the amount of heat needed to change a certain mass to its new state is given by Q = mL.
The rate at which heat is transferred is given by H = or H = , where k is the thermal conductivity (a property of the material), A is the cross-sectional area, ∆T is the temperature difference between the two sides and L is the thickness or distance between the two ends of the material.
For gases there are a few common equations to understand:
• Pressure is defined as the fore per unit area (P = ).
• The ideal gas law is expressed as either PV = nRT or PV = NkBT.
• The average kinetic energy of the gas molecules is given by Kavg = >kbT.
• The average speed of a molecule of the gas is given by vrms = or vrms = .
The work done on a gas is given by W = –P∆V or can be found by the area under a pressure-vs.-volume graph.
The First Law of Thermodynamics is ∆U = Q+W where ∆U depends only on the temperature change.
W positive (or negative) means energy is being added to (or subtracted from) the system by means of work done on the system by the surroundings (or on the surroundings by the system).
Q positive (or negative) means energy is being added to (or subtracted from) the system by means of a flow of heat from the higher temperature surroundings (or system) to the lower temperature system (or surroundings).
Note: Some textbooks define work in thermodynamics in a different way: Work is considered to be positive when work is done on the surroundings. This is consistent with the idea that the overall objective of a heat engine is to produce external (positive) work. Under this definition, the First Law must be written as U = Q – W (or U + W = Q), and W must then be interpreted differently from Q. That is, while Q is still positive when heat is being added to the system, W is now positive when work is being done by the system on the surroundings (thus decreasing the internal energy of the system).
Chapter 5
Electronic Forces and Fields
INTRODUCTION
Long before the knowledge of electricity many people were aware of the mysterious force. Descriptions of electricity was reported by Egyptians, Greeks, and Romans. Electricity would remain a mystery until 1600, when English scientist William Gilbert made a careful study of electricity and magnetism. Further work was continued by other notable scientists including Ben Franklin (his famous metal key and kite experiment in a storm). Real progress came in the nineteenth century, with electricity and magnetism, by Hans Christian Orsted, Andre-Marie Ampere, Michael Faraday, and Georg Ohm. James Clerk Maxwell would definitively link them together into what we call today electromagnetism.
ELECTRIC CHARGE
The basic components of atoms are protons, neutrons, and electrons. Protons and neutrons form the nucleus (and are referred to collectively as nucleons), while the electrons keep their distance, swarming around the nucleus. Most of an atom consists of empty space. In fact, if a nucleus were the size of the period at the end of this sentence, then the electrons would be 5 meters away. So what holds such an apparently tenuous structure together? One of the most powerful forces in nature: the electromagnetic force. Protons and electrons have a quality called electric charge that gives them an attractive force. Electric charge comes in two varieties; positive and negative. A positive particle always attracts a negative particle, and particles of the same charge always repel each other. Protons are positively charged, and electrons are negatively charged.
Protons and electrons are intrinsically charged, but bulk matter is not. This is because the amount of charge on a proton exactly balances the charge on an electron, which is quite remarkable in light of the fact that protons and electrons are very different particles. Since most atoms contain an equal number of protons and electrons, their overall electric charge is 0, because the negative charges cancel out the positive charges. Therefore, in order for matter to be charged, an imbalance between the numbers of protons and electrons must exist. This can be accomplished by either the removal or addition of electrons (that is, by the ionization of some of the object’s atoms). If you remove electrons, then the object becomes positively charged, while if you add electrons, then it becomes negatively charged. Furthermore, charge is conserved. For example, if you rub a glass rod with a piece of silk, then the silk will acquire a negative charge and the glass will be left with an equal positive charge. Net charge cannot be created or destroyed. (Charge can be created or destroyed—it happens all the time—but net charge cannot.)
The magnitude of charge on an electron (and therefore on a proton) is denoted e. This stands for elementary charge, because it’s the basic unit of electric charge. The charge of an ionized atom must be a whole number times e, because charge can be added or subtracted only in lumps of size e. For this reason we say that charge is quantized. To remind us of the quantized nature of electric charge, the charge of a particle (or object) is denoted by the letter q. In the SI system of units, charge is expressed in coulombs (abbreviated C). One coulomb is a tremendous amount of charge, about 1018 electrons. The value of e is about 1.6 × 10–19 C.
COULOMB’S LAW
The electric force between two charged particles obeys a law that is very similar to that describing the gravitational force between two masses: they are both inverse-square laws. The electric force between two particles with charges of q1 and q2, separated by a distance r, is given by the equation
FE =
This is Coulomb’s Law. We interpret a negative FE as an attraction between the charges and a positive FE as a repulsion. The value of the proportionality constant, k, depends on the material between the charged particles. In empty space (vacuum)—or air, for all practical purposes—it is called Coulomb’s constant and has the approximate value k0 = 9 × 109 N∙m2/C2. k0 is sometimes written in terms of a fundamental constant known as the permittivity of free space, denoted ε0:
k0 =
Example 1 Consider the proton and electron in hydrogen. The proton has a mass of 1.6 × 10–27 kg and a charge of +e. The electron has a mass of 9.1 × 10–31 kg and a charge of –e. In hydrogen, they are separated by a Bohr radius, about 0.5 × 10–10 m. What is the electric force between the proton and electron in hydrogen? What about the gravitational force between the proton and electron?
Solution. The electric force between the proton and the electron is given by Coulomb’s Law:
The fact that FE is negative means that the force is one of attraction, which we naturally expect, since one charge is positive and the other is negative. The force between the proton and electron is along the line that joins the charges, as we’ve illustrated below. The two forces shown form an action/reaction pair.
Now, the gravitational force between the two charges is given by Newton’s law of gravitation:
Now, compare the orders of magnitude of the electric force to the gravitational force. The electric force has an order of magnitude of 10–8 N, but the gravitational force has an order of magnitude of 10–47 N! This means that the electric force is something like 1039 times larger!
Addition of Forces
Consider three point charges: q1, q2, and q3. The total electric force acting on, say, q2 is simply the sum of F1-on-2, the electric force on q2 due to q1, and F3-on-2, the electric force on q2 due to q3:
Example 2 Consider four equal, positive point charges that are situated at the vertices of a square. Find the net electric force on a negative point charge placed at the square’s center.
Solution. Refer to the diagram below. The attractive forces due to the two charges on each diagonal cancel out: F1 + F3 = 0, and F2 + F4 = 0, because the distances between the negative charge and the positive charges are all the same and the positive charges are all equivalent. Therefore, by symmetry, the net force on the center charge is zero.
Example 3 If the two positive charges on the bottom side of the square in the previous example were removed, what would be the net electric force on the negative charge? Assume that each side of the square is 4.0 cm, each positive charge is 1.5 µC, and the negative charge is –6.2 nC. (µ is the symbol for “micro-,” which equals 10–6.)
Solution. If we break down F1 and F2 into horizontal and vertical components, then by symmetry the two horizontal components will cancel each other out, and the two vertical components will add:
Since the diagram on the left shows the components of F1 and F2 making right triangles with legs each of length 2 cm, it must be that F1y = F1 sin 45° and F2y = F2 sin 45°. Also, the magnitude of F1 equals that of F2. So the net electric force on the negative charge is F1y + F2y = 2F sin 45°, where F is the strength of the force between the negative charge and each of the positive charges.
If s is the length of each side of the square, then the distance r between each positive charge and the negative charge is r = s and
The direction of the net force is straight upward, toward the center of the line that joins the two positive charges.
Example 4 Two pith balls of mass m are each given a charge of +q. They are hung side-by-side from two threads each of length L, and move apart as a result of their electrical repulsion. Find the equilibrium separation distance x in terms of m, q, and L. (Use the fact that if θ is small, then tan θ ≈ sin θ.)
Solution. Three forces act on each ball: weight, tension, and electrical repulsion:
When the balls are in equilibrium, the net force each feels is zero. Therefore, the vertical component of FT must cancel out Fw and the horizontal component of FT must cancel out FE:
FT cos θ = Fw and FT sin θ = FE
Dividing the second equation by the first, we get tan θ = FE/Fw. Therefore,
Now, to approximate: If θ is small, the tan θ ≈ sin θ and, from the diagram sin θ = x/L. Therefore, the equation above becomes
THE ELECTRIC FIELD
The presence of a massive body such as the Earth causes objects to experience a gravitational force directed toward the Earth’s center. For objects located outside the Earth, this force varies inversely with the square of the distance and directly with the mass of the gravitational source. A vector diagram of the gravitational field surrounding the Earth looks like this:
We can think of the space surrounding the Earth as permeated by a gravitational field created by the Earth. Any mass placed in this field then experiences a gravitational force due to an interaction with this field.
The same process is used to describe the electric force. Rather than having two charges reach out across empty space to each other to produce a force, we will instead interpret the interaction in the following way: The presence of a charge creates an electric field in the space that surrounds it. Another charge placed in the field created by the first will experience a force due to the field.
Consider a point charge Q in a fixed position and assume that it’s positive. Now imagine moving a tiny positive test charge q around to various locations near Q. At each location, measure the force that the test charge experiences, and call it Fon q. Divide this force by the test charge q; the resulting vector is the electric field vector, E, at that location:
E =
The reason for dividing by the test charge is simple. If we were to use a different test charge with, say, twice the charge of the first one, then each of the forces F we’d measure would be twice as much as before. But when we divided this new, stronger force by the new, greater test charge, the factors of 2 would cancel, leaving the same ratio as before. So this ratio tells us the intrinsic strength of the field due to the source charge, independent of whatever test charge we may use to measure it.
What would the electric field of a positive charge Q look like? Since the test charge used to measure the field is positive, every electric field vector would point radially away from the source charge. If the source charge is positive, the electric field vectors point away from it; if the source charge is negative, then the field vectors point toward it. And, since the force decreases as we get farther away from the charge (as 1/r2), so does the electric field. This is why the electric field vectors farther from the source charge are shorter than those that are closer.
Since the force on any test charge q, due to some source Q, has a strength of qQ/4πε0r2, when we divide this by q, we get the expression for the strength of the electric field created by a point-charge source of magnitude Q:
E =
This is the electric field in the space surrounding for a point charge.
To make it easier to sketch an electric field, lines are drawn through the vectors such that the electric field vector is always tangent to the line everywhere it’s drawn.
Now, your first thought might be that obliterating the individual field vectors deprives us of information, since the length of the field vectors told us how strong the field was. Well, although the individual field vectors are gone, the strength of the field can be figured out by looking at the density of the field lines. Where the field lines are closer together, the field is stronger.
Electric fields obey the same addition properties as the electric force. If we had two source charges, their fields would overlap and effectively add; a third charge wandering by would feel the effect of all three fields. At each position in space, add the electric field vector due to one of the charges to the electric field vector due to the other charge: Etotal = E1 + E2. In the diagram below, E1 is the electric field vector at a particular location due to the charge +Q, and E2 is the electric field vector at that same location due to the other charge, –Q. Adding these vectors gives the overall field vector Etotal at that location.
If this is done at enough locations, the electric field lines can be sketched.
Note that, like electric field vectors, electric field lines always point away from positive source charges and toward negative ones. Two equal but opposite charges, like the ones shown in the diagram above, form a pair called an electric dipole.
If a positive charge +q were placed in the electric field above, it would experience a force that is tangent to, and in the same direction as, the field line passing through +q’s location. After all, electric fields are sketched from the point of view of what a positive test charge would do. On the other hand, if a negative charge –q were placed in the electric field, it would experience a force that is tangent to, but in the direction opposite from, the field line passing through –q’s location.
Finally, notice that electric field lines never cross.
Example 5 A charge q = +3.0 nC is placed at a location at which the electric field strength is 400 N/C. Find the force felt by the charge q.
Solution. From the definition of the electric field, we have the following equation:
Fon q = qE
Therefore, in this case, Fon q = qE = (3 × 10–9 C)(400 N/C) = 1.2 × 10–6 N.
Example 6 A dipole is formed by two point charges, each of magnitude 4.0 nC, separated by a distance of 6.0 cm. What is the strength of the electric field at the point midway between them?
Solution. Let the two source charges be denoted +Q and –Q. At Point P, the electric field vector due to +Q would point directly away from +Q, and the electric field vector due to –Q would point directly toward –Q. Therefore, these two vectors point in the same direction (from +Q to –Q), so their magnitudes would add.
Using the equation for the electric field strength due to a single point charge, we find that
Example 7 If a charge q = –5.0 pC were placed at the midway point described in the previous example, describe the force it would feel. (“p” is the abbreviation for “pico-,” which means 10–12.)
Solution. Since the field E at this location is known, the force felt by q is easy to calculate:
Fon q = qE = (–5.0 × 10−12 C)(8.0 × 104 N/C to the right) = 4.0 × 10−7 N to the left
Example 8 What can you say about the electric force that a charge would feel if it were placed at a location at which the electric field was zero?
Solution. Remember that Fon q = qE. So if E = 0, then Fon q = 0. (Zero field means zero force.)
The Uniform Electric Field
An important subset of problems deal with uniform electric fields. One method of creating this uniform field is to have two large conducting sheets, each storing some charge Q, some distance d apart. Near the edges of each sheet, the field may not be uniform, but near the middle, for all practical purposes the field is uniform. Having a uniform field means having a constant force (and therefore a constant acceleration), so you can use kinematic equations just as if you had a uniform gravitational field (as you do near the surface of the Earth).
Example 9 Positive charge is distributed uniformly over a large, horizontal plate, which then acts as the source of a vertical electric field. An object of mass 5 g is placed at a distance of 2 cm above the plate. If the strength of the electric field at this location is 106 N/C, how much charge would the object need to have in order for the electrical repulsion to balance the gravitational pull?
Solution. Clearly, since the plate is positively charged, the object would also have to carry a positive charge so that the electric force would be repulsive.
Let q be the charge on the object. Then, in order for FE to balance mg, we must have
Example 10 A proton, neutron, and electron are in a uniform electric field of 20 N/C that is caused by two large charged plates that are 30 cm apart. The particles are far enough apart that they don’t interact with each other. They are released from rest equidistant from each plate.
(a) What is the magnitude of the net force acting on each particle?
(b) What is the magnitude of the acceleration of each particle?
(c) How much work will be done on the particle when it collides with one of the charged plates?
(d) What is the speed of each particle when it strikes the plate?
(e) How long does it take to reach the plate?
Solution.
(a) Since E = , F = qE. Plugging in the values, we get
proton: F = (1.6 × 10–19 C)(20 N/C) = 3.2 × 10–18 N
electron: F = (1.6 × 10–19 C)(20 N/C) = 3.2 × 10–18 N
neutron: F = (0 C)(20 N/C) = 0 N
Note: Because the proton and electron have the same magnitude, they will experience the same force. If you’re asked for the direction, the proton travels in the same direction as the electric field and the electron travels in the opposite direction as the electric field.
(b) Since F = ma, a = . Plugging in the values, we get
proton: a = = 1.9 × 109 m/s2
electron: a = = 3.5 × 1012 m/s12
neutron: a = = 0 m/s2
Notice that although the charges have the same magnitude of force, the electron experiences an acceleration almost 2000 times greater due to its mass being almost 2000 times smaller than the proton’s mass.
(c) Since W = Fd, we get W = qEd. Plugging in the values, recalling that the charges start midway between the plates, we get
proton: W = (1.6 × 10–19 C)(20 N/C) (0.15 m) = 4.8 × 10–19 N
electron: W = (1.6 × 10–19 C)(20 N/C) (0.15 m) = 4.8 × 10–19 N
neutron: W = (0 C) (20 N/C)(0.15 m) = 0 N
(d) Recall one of the big five kinematics equations:
vf2 = v2 + 2a(x − x0) → vf =
If the particles are midway between the 30 cm plates, they will travel 0.15 m.
proton: vf = = 24,000 m/s
electron: vf = = 1.0 × 106 m/s
neutron: vf = = 0 m/s
Notice that, even though the force is the same and the same work is done on both charges, there is a significant difference in final velocities due to the large mass difference. An alternative solution to this would be using W = ∆KE. You would have obtained the same answers.
(e) Recall one of the big five kinematics equations:
vf = vi + at → t = → t =
proton: t = → 1.3 × 10–5 s
electron: t = → 2.9 × 10–7 s
neutron: The neutron never accelerates, so it will never hit the plate
CONDUCTORS AND INSULATORS
Materials can be classified into broad categories based on their ability to permit the flow of charge. If electrons were placed on a metal sphere, they would quickly spread out and cover the outside of the sphere uniformly. These electrons would be free to flow through the metal and redistribute themselves, moving to get as far away from each other as they could. Materials that permit the flow of excess charge are called conductors; they conduct electricity. Metals are the best examples of conductors. Aqueous solutions that contain dissolved electrolytes (such as salt water) are also conductors. Metals conduct electricity because they bind all but their outermost electron very tightly. That outermost electron is free to move about the metal. This creates a sort of sea of mobile (or conduction) electrons.
Insulators, on the other hand, closely guard their electrons—and even extra ones that might be added. Electrons are not free to roam throughout the atomic lattice. Examples of insulators are glass, wood, rubber, and plastic. If excess charge is placed on an insulator, it stays put.
Example 11 A solid sphere of copper is given a negative charge. Discuss the electric field inside and outside the sphere.
Solution. The electric field inside a conductor is zero. Therefore, the excess electrons that are deposited on the sphere move quickly to the outer surface (copper is a great conductor). Any excess charge on a conductor resides entirely on the outer surface.
Once these excess electrons establish a uniform distribution on the outer surface of the sphere, there will be no net electric field within the sphere. Why not? Since there is no additional excess charge inside the conductor, there are no excess charges to serve as a source or sink of an electric field line cutting down into the sphere, because field lines begin or end on excess charges. There can be no electrostatic field within the body of a conductor.
In fact, you can shield yourself from electric fields simply by surrounding yourself with metal. Charges may move around on the outer surface of your cage, but within the cage, the electric field will be zero. For points outside the sphere, it can be shown that the sphere behaves as if all its excess charge were concentrated at its center. (Remember that this is just like the gravitational field due to a uniform spherical mass.) Also, the electric field is always perpendicular to the surface, no matter what shape the surface may be. See the diagram below.
Now let’s take our previous sphere and put it in a different situation: Start with a neutral metal sphere and bring a charge (a positive charge) Q nearby without touching the original metal sphere. What will happen? The positive charge will attract free electrons in the metal, leaving the far side of the sphere positively charged. Since the negative charge is closer to Q than the positive charge, there will be a net attraction between Q and the sphere. So, even though the sphere as a whole is electrically neutral, the separation of charge induced by the presence of Q will create a force of electrical attraction between them.
Charging by Induction
This process may be used to redistribute charges among a pair of neutrally charged spheres, so that in the end both spheres are charged. Imagine two neutrally charged spheres that are each set on an insulating stand. The spheres are arranged so that they are in contact with one another.
A charge is brought near the side of one of the spheres resulting in the same attraction as occurred with the single sphere. However, because we have two conducting spheres, the excess negative charges are located on the right sphere and the positive excess charges are on the left sphere.
With the external charge still nearby, the two constructing spheres are separated. Because the conductors are no longer in contact, the negative charge has been trapped on the right sphere and the positive charge has been trapped on the left sphere.
Note that the net charge is unchanged between the two spheres—the distribution of the charges has simply been changed.
Now what if the sphere was made of glass (an insulator)? Although there aren’t free electrons that can move to the near side of the sphere, the atoms that make up the sphere will become polarized. That is, their electrons will feel a tug toward Q, and so will spend more time on the side of the atom closer to Q than on the side opposite Q, causing the atoms to develop a partial negative charge pointing toward Q (and a partial positive charge pointing away from Q). The effect isn’t as dramatic as the mass movement of free electrons in the case of a metal sphere, but the polarization is still enough to cause an electrical attraction between the sphere and Q. For example, if you comb your hair, the comb will pick up extra electrons, making it negatively charged. If you place this electric field source near little bits of paper the paper will become polarized and will then be attracted to the comb.
The same phenomenon, in which the presence of a charge tends to cause polarization in a nearby collection of charges, is responsible for a kind of intermolecular force. Dipole-induced forces are caused by a shifting of the electron cloud of a neutral molecule toward positively charged ions or away from a negatively charged ion; in either case, the resulting force between the ion and the atom is attractive.
Chapter 5 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 5 Review Questions.
Section I: Multiple Choice
1. If the distance between two positive point charges is tripled, then the strength of the electrostatic repulsion between them will decrease by a factor of
(A) 3
(B) 6
(C) 8
(D) 9
2. Two 1 kg spheres each carry a charge of magnitude 1 C. How does FE, the strength of the electric force between the spheres, compare to FG, the strength of their gravitational attraction?
(A) FE < FG
(B) FE = FG
(C) FE > FG
(D) If the charges on the spheres are of the same sign, then FE > FG ; but if the charges on the spheres are of the opposite sign, then FE < FG.
3. The figure below shows three point charges, all positive. If the net electric force on the center charge is zero, what is the value of y/x ?
(A)
(B)
(C)
(D)
4.
The figure above shows two point charges, +Q and –Q. If the negative charge were absent, the electric field at Point P due to +Q would have strength E. With –Q in place, what is the strength of the total electric field at P, which lies at the midpoint of the line segment joining the charges?
(A) 0
(B)
(C) E
(D) 2E
5. A sphere of charge +Q is fixed in position. A smaller sphere of charge +q is placed near the larger sphere and released from rest. The small sphere will move away from the large sphere with
(A) decreasing velocity and decreasing acceleration
(B) decreasing velocity and increasing acceleration
(C) increasing velocity and decreasing acceleration
(D) increasing velocity and increasing acceleration
6. An object of charge +q feels an electric force FE when placed at a particular location in an electric field, E. Therefore, if an object of charge –2q were placed at the same location where the first charge was, it would feel an electric force of
(A)
(B) −2FE
(C) −2qFE
(D)
7. A charge of −3Q is transferred to a solid metal sphere of radius r. How will this excess charge be distributed?
(A) −Q at the center, and −2Q on the outer surface
(B) −3Q at the center
(C) −3Q on the outer surface
(D) −Q at the center, −Q in a ring of radius r, and −Q on the outer surface
Section II: Free Response
1. In the figure shown, all four charges (+Q, +Q, −q, and −q) are situated at the corners of a square. The net electric force on each charge +Q is zero.
(a) Express the magnitude of q in terms of Q.
(b) Is the net electric force on each charge −q also equal to zero? Justify your answer.
(c) Determine the electric field at the center of the square.
2. Two charges, +Q and +2Q, are fixed in place along the y-axis of an x-y coordinate system as shown in the figure below. Charge 1 is at the point (0, a), and Charge 2 is at the point (0, −2a).
(a) Find the electric force (magnitude and direction) felt by Charge 1 due to Charge 2.
(b) Find the electric field (magnitude and direction) at the origin created by both Charges 1 and 2.
(c) Is there a point on the x-axis where the total electric field is zero? If so, where? If not, explain briefly.
(d) Is there a point on the y-axis where the total electric field is zero? If so, where? If not, explain briefly.
(e) If a small negative charge, −q, of mass m were placed at the origin, determine its initial acceleration (magnitude and direction).
Summary
Coulomb’s Law describes the force acting on two point charges and is given by
Fe =
where is a constant equal to 9.0 × 109 N·m2/C2. You can use all the strategies you used in the Newton’s Laws chapter to solve these types of problems.
The electric field is given by E =
The electric field from a point charge is E =
Both the electric force and field are vector quantities and therefore all the rules for vector addition apply.
Chapter 6
Electronic Potential and Capacitance
INTRODUCTION
When an object moves in a gravitational field, it usually experiences a change in kinetic energy and in gravitational potential energy due to the work done on the object by gravity. Similarly, when a charge moves in an electric field, it generally experiences a change in kinetic energy and in electrical potential energy due to the work done on it by the electric field. By exploring the idea of electric potential, we can simplify our calculations of work and energy changes within electric fields.
ELECTRICAL POTENTIAL ENERGY
When a charge moves in an electric field, then unless its displacement is always perpendicular to the field, the electric force does work on the charge. If WE is the work done by the electric force, then the change in the charge’s electrical potential energy is defined by
∆UE = −WE
Notice that this is the same equation that defined the change in the gravitational potential energy of an object of mass m undergoing a displacement in a gravitational field (∆UG = −WG).
Example 1 A positive charge +q moves from position A to position B in a uniform electric field E:
What is its change in electrical potential energy?
Solution. Since the field is uniform, the electric force that the charge feels, FE = qE, is constant. Since q is positive, FE points in the same direction as E, and, as the figure shows, they point in the same direction as the displacement, r. This makes the work done by the electric field equal to WE = FEr = qEr, so the change in the electrical potential energy is
∆UE = −qEr
Note that the change in potential energy is negative, which means that potential energy has decreased; this always happens when the field does positive work. It’s just like dropping a rock to the ground: Gravity does positive work, and the rock loses gravitational potential energy.
Example 2 Do the previous problem, but consider the case of a negative charge, −q.
Solution. In this case, an outside agent must be pushing the charge to make it move, because the electric force naturally pushes negative charges against field lines. Therefore, we expect that the work done by the electric field is negative. The electric force, FE = (−q)E, points in the direction opposite to the displacement, so the work it does is WE = −FEr = −qEr = −qEr. Thus, the change in electrical potential energy is positive: ∆UE = −WE = −(−qEr) = qEr. Because the change in potential energy is positive, the potential energy increases; this always happens when the field does negative work. It’s like lifting a rock off the ground: Gravity does negative work, and the rock gains gravitational potential energy.
Example 3 A positive charge +q moves from position A to position B in a uniform electric field E:
What is its change in electrical potential energy?
Solution. The electric force felt by the charge q is FE = qE and this force is parallel to E, because q is positive. In this case, because FE is not parallel to r (as it was in Example 1), we will use the more general definition of work for a constant force:
WE = FE · r = FE r cos θ = qEr cos θ
But r cos θ = d, so
WE = qEd and ∆UE = − WE = −qEd
Because the electric force is a conservative force, which means that the work done does not depend on the path that connects the positions A and B, the work calculated above could have been figured out by considering the path from A to B composed of the segments r1 and r2:
Along r1, the electric force does no work since this displacement is perpendicular to the force. Thus, the work done by the electric field as q moves from A to B is simply equal to the work it does along r2. And since the length of r2 is d = r cos θ, we have WE = FEd = qEd, just as before.
Electric Potential Energy from a Point Charge
Example 4 A positive charge q1 = +2 × 10−6 C is held stationary, while a negative charge, q2 = −1 × 10−8 C, is released from rest at a distance of 10 cm from q1. Find the kinetic energy of charge q2 when it’s 1 cm from q1.
Solution. The gain in kinetic energy is equal to the loss in potential energy; you know this from Conservation of Energy. Previously, we looked at constant electric fields and were able to use the equation for work from a constant force. However, when the field (and therefore the force) changes, we need another equation. Electrical potential energy required to move along the field lines surrounding a point charge is given by
Therefore, if q1 is fixed and q2 moves from rA to rB, the change in potential energy is
Since q2 lost 0.016 J of potential energy, the gain in kinetic energy is 0.016 J. Since q2 started from rest (with no kinetic energy), this is the kinetic energy of q2 when it’s 1 cm from q1.
Example 5 Two positive charges, q1 and q2, are held in the positions shown below. How much work would be required to bring (from infinity) a third positive charge, q3, and place it so that the three charges form the corners of an equilateral triangle of side length s?
Solution. An external agent would need to do positive work, equal in magnitude to the negative work done by the electric force on q3 as it is brought into place, so let’s first compute this quantity. Let’s first compute the work done by the electric force as q3 is brought in. Since q3 is fighting against both q1’s and q2’s electric field, the total work done on q3 by the electric force, WE, is equal to the work done on q3 by q1 (W1−3 ) plus the work done on q3 by q2 (W2−3 ). Using the equation WE = −∆UE and the one we gave above for ∆UE, we have
Therefore, the work that an external agent must do to bring q3 into position is
ELECTRIC POTENTIAL
Let WE be the work done by the electric field on a charge q as it undergoes a displacement. If another charge, say 2q, were to undergo the same displacement, the electric force would be twice as great on this second charge, and the work done by the electric field would be twice as much, 2WE. Since the work would be twice as much in the second case, the change in electrical potential energy would be twice as great as well, but the ratio of the change in potential energy to the charge would be the same: UE/q = UE/2q. This ratio says something about the field and the displacement but not the charge that made the move. The change in electric potential ∆V, is defined as this ratio:
There is a similar concept with gravitational fields. The work done by gravity (near the Earth) on an object as it changes its position is ∆W = mgh. If twice the mass were moved, the work would have to be ∆W = (2m)gh. The ration of the work to the mass moved is technically the “gravitational potential,” but near the Earth, this quantity is simply the gravitational field strength, g, multiplied by the change in the height of the object. Electric potential is the electrical equivalent of a change in height.
Electric potential is electrical potential energy per unit charge; the units of electric potential are joules per coulomb. One joule per coulomb is called one volt (abbreviated V); so 1 J/C = 1 V.
Electric Potential from a Point Charge
Consider the electric field created by a point source charge Q. If a charge q moves from a distance rA to a distance rB from Q, then the change in the potential energy is
The difference in electric potential between positions A and B in the field created by Q is
If we designate VA → 0 as rA → ∞ (an assumption that’s stated on the AP Physics 2 Exam), then the electric potential at a distance r from Q is
Note that the potential depends on the strength of the source charge making the field and the distance from the source charge.
Example 6 Let Q = 2 × 10–9 C. What is the potential at a Point P that is 2 cm from Q?
Solution. Relative to V = 0 at infinity, we have
This means that the work done by the electric field on a charge of q coulombs brought to a point 2 cm from Q would be –900q joules.
Note that, like potential energy, electric potential is a scalar. In the preceding example, we didn’t have to specify the direction of the vector from the position of Q to the Point P, because it didn’t matter. Imagine a sphere with a surface of 2 cm from Q; at any point on that sphere, the potential will be 900 V. These spheres around Q are called equipotential surfaces, and they’re surfaces of constant potential. The equipotentials are always perpendicular to the electric field lines.
Example 7 How much work is done by the electric field as a charge moves along an equipotential surface?
Solution. If the charge always remains on a single equipotential, then, by definition, the potential, V, never changes. Therefore, ∆V = 0, so ∆UE = 0. Since WE = –∆UE, the work done by the electric field is zero.
Addition of Electric Potential
The formula V = kQ/r tells us how to find the potential due to a single point source charge, Q. Potential is scalar (we will not be concerned with orientation just the sign of charge). When we add up individual potentials, we’re simply adding numbers; we’re not adding vectors.
Example 8 If charges q1 = 4 × 10–9 C and q2 = –6 × 10–9 C are stationary, calculate the potential at Point A in the figure below:
Solution. Potentials add like ordinary numbers. Therefore, the potential at A is just the sum of the potentials at A due to q1 and q2. Note that the distance from q1 to A is 5 cm.
Example 9 How much work would it take to move a charge q = +1 × 10–2 C from Point A to Point B (the point midway between q1 and q2)? Note: We are using the data from Example 8.
Solution. ∆UE = q∆V, so if we calculate the potential difference between Points A and B and multiply by q, we will have found the change in the electrical potential energy: ΔUA → B = qΔVA → B. Then, since the work by the electric field is –∆U, the work required by an external agent is ∆U. In this case, the potential at Point B is
In the preceding example, we calculated the potential at Point A: VA = –1080 V, so ∆VA → B = VB – VA = (–900 V) – (–1080 V) = +180 V. This means that the change in electrical potential energy as q moves from A to B is
∆VA → B = qΔVA → B = (+1 × 10–2 C)(+180 V) = 1.8 J
This is the work required by an external agent to move q from A to B.
THE ELECTRIC POTENTIAL OF A UNIFORM FIELD
Example 10 Consider a very large, flat plate that contains a uniform surface charge density σ. At points that are not too far from the plate, the electric field is uniform and given by the equation
What is the potential at a point which is a distance d from the sheet (close to the plates), relative to the potential of the sheet itself?
Solution. Let A be a point on the plate and let B be a point a distance d from the sheet. Then
Since the field is constant, the force that a charge q would feel is also constant, and is equal to
Therefore,
so applying the definition gives us
This says that the potential decreases linearly as we move away from the plate.
Example 11 Two large flat plates—one carrying a charge of +Q, the other –Q—are separated by a distance d. The electric field between the plates, E, is uniform. Determine the potential difference between the plates.
Solution. Imagine a positive charge q moving from the positive plate to the negative plate:
Since the work done by the electric field is
WE,+→– = FEd = qEd
the potential difference between the plates is
This tells us that the potential of the positive plate is greater than the potential of the negative plate, by the amount Ed. This equation can also be written as
Therefore, if the potential difference and the distance between the plates are known, then the magnitude of the electric field can be determined quickly. The magnitude is simply
CAPACITANCE
Consider two conductors, separated by some distance, that carry equal but opposite charges, +Q and –Q. Such a pair of conductors comprise a system called a capacitor. Work must be done to create this separation of charge, and, as a result, potential energy is stored. Capacitors are basically storage devices for electrical potential energy.
The conductors may have any shape, but the most common conductors are parallel metal plates or sheets. These types of capacitors are called parallel-plate capacitors. We’ll assume that the distance d between the plates is small compared to the dimensions of the plates since, in this case, the electric field between the plates is uniform. The electric field due to one such plate, if its surface charge density is σ = Q/A, is given by the equation E = σ/(2ε0), with E pointing away from the sheet if σ is positive and toward the plate if σ is negative.
Therefore, with two plates, one with surface charge density +σ and the other –σ, the electric fields combine to give a field that’s zero outside the plates and that has the magnitude
in between.
In Example 11, we learned that the magnitude of the potential difference, ∆V, between the plates satisfies the relationship ∆V = Ed, so combining this with the previous equation, we get
The ratio of Q to ∆V, for any capacitor, is called its capacitance (C),
C = Q/∆V
so for a parallel-plate capacitor, we get
The capacitance measures the capacity for holding charge. The greater the capacitance, the more charge can be stored on the plates at a given potential difference. The capacitance of any capacitor depends only on the size, shape, and separation of the conductors. From the definition, C = Q/∆V, the units of C are coulombs per volt. One coulomb per volt is renamed one farad (abbreviated F): 1 C/V = 1 F.
Example 12 A 10-nanofarad parallel-plate capacitor holds a charge of magnitude 50 µC on each plate.
(a) What is the potential difference between the plates?
(b) If the plates are separated by a distance of 0.2 mm, what is the area of each plate?
Solution.
(a) From the definition, C = Q/∆V, we find that
ΔV = = = 5000 V
(b) From the equation C = ε0A/d, we can calculate the area, A, of each plate:
A = = = 0.23 m2
ELECTRIC FIELD AND CAPACITORS
When we studied electric fields in the preceding chapter, we noticed that the electric field created by one or more point source charges varied, depending on the location. For example, as we move further away from the source charge, the electric field gets weaker. Even if we stay at the same distance from, say, a single source charge, the direction of the field changes as we move around. Therefore, we could never obtain an electric field that was constant in both magnitude and direction throughout some region of space from point-source charges. However, the electric field that is created between the plates of a charged parallel-plate capacitor is constant in both magnitude and direction throughout the region between the plates; in other words, a charged parallel-plate capacitor can create a uniform electric field. The electric field, E, always points from the positive plate toward the negative plate, and its magnitude remains the same at every point between the plates, whether we choose a point close to the positive plate, closer to the negative plate, or between them.
Because E is so straightforward (it’s the same everywhere between the plates), the equation for calculating it is equally straightforward, V = Ed. The strength of E depends upon the voltage between the plates, ΔV.
The equation F = qE showed us that the units of E are N/C. The formula ΔV = Ed tells us that the units of E are also V/m. You’ll see both newtons- per-coulomb and volts-per-meter used as units for the electric field; it turns out these units are exactly the same.
Example 13 The charge on a parallel-plate capacitor is 4 × 10–6 C. If the distance between the plates is 2 mm and the capacitance is 1 μF, what’s the strength of the electric field between the plates?
Solution. Since C = Q/ΔV, we have ΔV = Q/C = (4 × 10–6 C)/(10–6 F) = 4 V. Now, using the equation ΔV = Ed,
E = ΔV/d = 4ΔV/(2 × 10–3 m) = 2000 V/m
Example 14 The plates of a parallel-plate capacitor are separated by a distance of 2 mm. The device’s capacitance is 1 μF. How much charge needs to be transferred from one plate to the other in order to create a uniform electric field whose strength is 104 V/m?
Solution. Because Q = CΔV and ΔV = Ed, we find that
Q = CEd = (1 × 10–6 F)(1 × 104 V/m)(2 × 10–3 m) = 2 × 10–5 C = 20 μC
Example 15 A proton (whose mass is m) is placed on top of the positively charged plate of a parallel-plate capacitor, as shown below.
The charge on the capacitor is Q, and the capacitance is C. If the electric field in the region between the plates has magnitude E, give an expression that shows the time required for the proton to move up to the other plate.
Solution. Once we find the acceleration of the proton, we can use Big Five #3, with v0 = 0, to find the time it will take for the proton to move the distance y = d. The acceleration of the proton is F/m, where F = qE is the force the proton feels; this gives a = qE/m. (We’re ignoring the gravitational force on the proton because it is so much weaker than the electric force.) Now, since E = ΔV/d and ΔV = Q/C, the expression becomes a = eQ/mdC for a. Note that q = +e for a proton. If we consider y = d, then Big Five #3 gives us
Solving for t,
THE ENERGY STORED IN A CAPACITOR
To figure out the electrical potential energy stored in a capacitor, imagine taking a small amount of negative charge off the positive plate and transferring it to the negative plate. This requires that positive work is done by an external agent, and this is the reason that the capacitor stores energy. If the final charge on the capacitor is Q, then we transferred an amount of charge equal to Q, fighting against the prevailing voltage at each stage. If the final voltage is ∆V, then the average voltage during the charging process is ∆V; so, because ∆UE is equal to charge times voltage, we can write ΔUE = Q · ΔV = QΔV. At the beginning of the charging process, when there was no charge on the capacitor, we had Ui = 0, so ∆UE = Uf – Ui = Uf – 0 = Uf; therefore, we have
UE = QΔV
This is the electrical potential energy stored in a capacitor. Because of the definition C = Q/ΔV, the equation for the stored potential energy can be written as:
or
CAPACITORS AND DIELECTRICS
One method of keeping the plates of a capacitor apart, which is necessary to maintain charge separation and store potential energy, is to insert an insulator (called a dielectric) between the plates.
A dielectric always increases the capacitance of a capacitor.
Let’s see why this is true. Imagine charging a capacitor to a potential difference of ΔV with charge +Q on one plate and –Q on the other. Now disconnect the capacitor from the charging source and insert a dielectric. What happens? Although the dielectric is not a conductor, the electric field that existed between the plates causes the molecules within the dielectric material to polarize; there is more electron density on the side of the molecule near the positive plate.
The effect of this is to form a layer of negative charge along the top surface of the dielectric and a layer of positive charge along the bottom surface; this separation of charge induces its own electric field (Ei), within the dielectric, which opposes the original electric field, E, within the capacitor.
So the overall electric field has been reduced from its previous value: Etotal = E + Ei, and Etotal= E – Ei. Let’s say that the electric field has been reduced by a factor of κ (the Greek letter kappa) from its original value as follows
Ewith dielectric = Ewithout dielectric − Ei =
Since ΔV = Ed for a parallel-plate capacitor, we see that ΔV must have decreased by a factor of κ. But C = , so if ΔV decreases by a factor of κ, then C increases by a factor of κ:
Cwith dielectric = κCwithout dielectric
The value of κ, called the dielectric constant, varies from material to material, but it’s always greater than 1. In general, the capacitance of parallel-plate capacitors is:
C = κε0 A/d
Chapter 6 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 6 Review Questions.
Section I: Multiple Choice
1. If the electric field does negative work on a negative charge as the charge undergoes a displacement from Position A to Position B within an electric field, then the electrical potential energy
(A) is negative
(B) is positive
(C) increased
(D) decreased
2.
The work required to assemble the system shown above, bringing each charge in from an infinite distance, is equal to
(A)
(B)
(C)
(D)
3. Negative charges are accelerated by electric fields toward points
(A) at lower electric potential
(B) at higher electric potential
(C) where the electric field is weaker
(D) where the electric field is stronger
4. A charge q experiences a displacement within an electric field from Position A to Position B. The change in the electrical potential energy is ∆UE, and the work done by the electric field during this displacement is WE. Then
(A) VA – VB = qWE
(B) VB – VA = qWE
(C) VA – VB = ∆UE/q
(D) VB – VA = ∆UE/q
5.
Which points in this uniform electric field (between the plates of the capacitor) shown above lie on the same equipotential?
(A) 1 and 3 only
(B) 2 and 4 only
(C) 3 and 4 only
(D) 1, 2, 3, and 4 all lie on the same equipotential since the electric field is uniform.
6.
How much work would the electric field (created by the stationary charge Q) perform as a charge q is moved from Point A to B along the curved path shown? VA = 200 V, VB = 100 V, q = –0.05 C, length of line segment AB = 10 cm, length of curved path = 20 cm.
(A) –10 J
(B) –5 J
(C) +5 J
(D) +10 J
Section II: Free Response
1. In the figure shown, all four charges are situated at the corners of a square with sides s.
(a) What is the total electrical potential energy of this array of fixed charges?
(b) What is the electric field at the center of the square?
(c) What is the electric potential at the center of the square?
(d) Sketch (on the diagram) the portion of the equipotential surface that lies in the plane of the figure and passes through the center of the square.
(e) How much work would the electric field perform on a charge q as it moved from the midpoint of the right side of the square to the midpoint of the top of the square?
2. The figure below shows a parallel-plate capacitor. Each rectangular plate has length L and width w, and the plates are separated by a distance d.
(a) Determine the capacitance.
An electron (mass m, charge –e) is shot horizontally into the empty space between the plates, midway between them, with an initial velocity of magnitude v0. The electron just barely misses hitting the end of the top plate as it exits. (Ignore gravity.)
(b) In the diagram, sketch the electric field vector at the position of the electron when it has traveled a horizontal distance of L/2.
(c) In the diagram, sketch the electric force vector on the electron at the same position as in part (b).
(d) Determine the strength of the electric field between the plates. Write your answer in terms of L, d, m, e, and v0.
(e) Determine the charge on the top plate.
(f) How much potential energy is stored in the capacitor?
3. A solid conducting sphere of radius a carries an excess charge of Q.
(a) Determine the electric field magnitude, E(r), as a function of r, the distance from the sphere’s center.
(b) Determine the potential, V(r), as a function of r. Take the zero of potential at r = ∞.
(c) On the diagrams below, sketch E(r) and V(r). (Cover at least the range 0 < r < 2a.
Summary
The electric potential (commonly referred to as the voltage) is defined as V = .
The electrical potential energy is defined by
ΔUE = −WE or UE = qV or UE = .
The work done moving a charge through an electric field is given by W = qEd.
A capacitor is basically a storage device for electrical potential energy. Capacitance is given by C = . For parallel plate capacitors, we have C = where ε0 = 8.85 × 10–12 C2/Nm2, A is the area of the plate and d is the distance between the plates.
The electrical energy stored in a capacitor is given by either
UC = QV or UC = CV2.
Chapter 7
Direct Current Circuits
INTRODUCTION
Today, electricity is well-known and appears all around us, from the lighting in our houses to the computing in our personal computers. All of these are powered by complex circuits. In this chapter, we will study the basics of simple direct current circuits.
ELECTRIC CURRENT
Picture a piece of metal wire. Within the metal, electrons are zooming around at speeds of about a million m/s in random directions, colliding with other electrons and positive ions in the lattice. This constitutes charge in motion, but it doesn’t constitute net movement of charge, because the electrons move randomly. If there’s no net motion of charge, there’s no current. However, if we were to create a potential difference between the ends of the wire, meaning if we set up an electric field, the electrons would experience an electric force, and they would start to drift through the wire. This is current. Although the electric field would travel through the wire at nearly the speed of light, the electrons themselves would still have to make their way through a crowd of atoms and other free electrons, so their drift speed, vd, would be relatively slow, on the order of a millimeter per second.
A Note on Convention for this Chapter
In this chapter, we will use the standard convention of referring to a change in potential as a voltage, V. In the previous chapter, we referred to these changes in potential as δV. Make note of this as you move forward so as to not be confused.
To measure the current, we have to measure how much charge crosses a plane per unit time. If an amount of charge of magnitude ∆Q crosses an imaginary plane in a time interval ∆t, then the average current is
Because current is charge per unit time, it’s expressed in coulombs per second. One coulomb per second is an ampere (abbreviated A), or amp. So 1 C/s = 1 A.
Although the charge carriers that constitute the current within a metal are electrons, the direction of the current is taken to be the direction that positive charge carriers would move. (This is explicitly stated on the AP Physics 2 Exam.) So, if the conduction electrons drift to the right, we’d say the current points toward the left.
RESISTANCE
Let’s say we had a copper wire and a glass fiber that had the same length and cross-sectional area, and that we hooked up the ends of the metal wire to a source of potential difference and measured the resulting current. If we were to do the same thing with the glass fiber, the current would probably be too small to measure, but why? Well, the glass provided more resistance to the flow of charge. If the potential difference is V and the current is I, then the resistance is
R = or V = IR
This is known as Ohm’s Law. Not all devices are ohmic, but many are. Notice for the same voltage if the current is large, the resistance is low, and if the current is small, then resistance is high. Because resistance is voltage divided by current, it is expressed in volts per amp. One volt per amp is one ohm (Ω, omega). So, 1 V/A = 1 Ω.
RESISTIVITY
The resistance of an object depends on two things: the material it’s made of and its shape. For example, think again of the copper wire and glass fiber of the same length and area. They have the same shape, but their resistances are different because they’re made of different materials. Glass has a much greater intrinsic resistance than copper does; it has a greater resistivity. Each material has its own characteristic resistivity, and resistance depends on how the material is shaped. For a wire of length L and cross-sectional area A made of a material with resistivity ρ, resistance is given by:
R =
This equation only applies to shapes with constant cross sectional area, such as a cylinder or a rectangular prism, and will not apply to those with a varying cross sectional area, such as a sphere or a cone.
The resistivity of copper is around 10–8 Ω∙m, while the resistivity of glass is much greater, around 1012 Ω∙m.
Example 1 A wire of radius 1 mm and length 2 m is made of platinum (resistivity = 1 × 10–7 Ω•m). If a voltage of 9 V is applied between the ends of the wire, what will be the resulting current?
Solution. First, the resistance of the wire is given by the equation
Then, from I = V/R, we get
VOLTAGE
Now that we know how to measure current, the next question is, What causes it? Why does an electron drift through the circuit? One answer is to say that there’s an electric field inside the wire, and since negative charges move in the direction opposite to the electric field lines, electrons would drift opposite the electric field.
Another (equivalent) answer to the question is that there’s a difference in electric potential—otherwise known as a potential difference or, most simply, a voltage—between the ends of the wire. Negative charges naturally move from regions of higher potential to lower potential.
Voltage across two points along a conductor creates a current.
It is not uncommon to see the voltage that creates a current referred to as electromotive force (emf) ε, since it is the cause that sets the charges into motion in a preferred direction. The voltage is provided in a circuit by some source, which is typically a battery for direct current circuits.
ELECTRIC CIRCUITS
An electric current is maintained when the terminals of a voltage source (a battery, for example) are connected by a conducting pathway, in what’s called a circuit. If the current always travels in the same direction through the pathway, it’s called a direct current.
To try to imagine what’s happening in a circuit in which a steady-state current is maintained, let’s follow one of the charge carriers drifting through the pathway. (Remember we’re pretending that the charge carriers are positive.) The charge is introduced by the positive terminal of the battery and enters the wire, where it’s pushed by the electric field. It encounters resistance, bumping into the relatively stationary atoms that make up the metal’s lattice and setting them into greater motion. So the electrical potential energy that the charge had when it left the battery is turning into heat. By the time the charge reaches the negative terminal, all of its original electrical potential energy is lost. In order to keep the current going, the voltage source must do positive work on the charge, forcing it to move from the negative terminal toward the positive terminal. The charge is now ready to make another journey around the circuit.
Energy and Power
When a carrier of positive charge q drops by an amount V in potential, it loses potential energy in the amount qV. If this happens in time t, then the rate at which this energy is transformed is equal to (qV)/t = (q/t)V. But q/t is equal to the current, I, so the rate at which electrical energy is transferred is given by the equation
P = IV
This equation works for the power delivered by a battery to the circuit as well as for resistors. The power dissipated in a resistor, as electrical potential energy is turned into heat, is given by P = IV, but, because of the relationship V = IR, we can express this in two other ways:
Resistors become hot when current passes through them.
CIRCUIT ANALYSIS
We will now develop a way of specifying the current, voltage, and power associated with each element in a circuit. Our circuits will contain three basic elements: batteries, resistors, and connecting wires. As we’ve seen, the resistance of an ordinary metal wire is negligible; resistance is provided by devices that control the current: resistors. All the resistance of the system is concentrated in the resistors, which are symbolized in a circuit diagram by this symbol:
Batteries are denoted by the symbol:
where the longer line represents the positive (higher potential) terminal, and the shorter line is the negative (lower potential) terminal. Sometimes a battery is denoted by more than one pair of such lines:
Here’s a simple circuit diagram:
The electric potential (V) of the battery is indicated, as is the resistance (R) of the resistor. As stated before, before the electrons return to the battery, all their energy is lost in the circuit. Since the resistor is the only thing in this circuit, it has to have the same potential difference as the battery, just with the opposite sign. For now, this sign isn’t important, but it will be later on when we cover Kirchhoff’s rules. Determining the current in this case is straightforward. The equation V = IR gives us
I =
COMBINATIONS OF RESISTORS
Two common ways of combining resistors within a circuit is to place them either in series (one after the other),
or in parallel (that is, side-by-side):
In order to simplify the circuit, our goal is to find the equivalent resistance of combinations. Resistors are said to be in series if they all share the same current and if the total voltage drop across them is equal to the sum of the individual voltage drops.
In this case, then, if V denotes the voltage drop across the combination, we have
This idea can be applied to any number of resistors in series (not just two):
Resistors are said to be in parallel if they all share the same voltage drop, and the total current entering the combination is split among the resistors. Imagine that a current I enters the combination. It splits; some of the current, I1, would go through R1, and the remainder, I2, would go through R2 , such that the voltage is the same for each resistor.
So if V is the voltage drop across the combination, we have
This idea can be applied to any number of resistors in parallel (not just two): The reciprocal of the equivalent resistance for resistors in parallel is equal to the sum of the reciprocals of the individual resistances:
Example 2 Calculate the equivalent resistance for the following circuit:
Solution. First find the equivalent resistance of the two parallel resistors:
This resistance is in series with the 4 Ω resistor, so the overall equivalent resistance in the circuit is R = 4 Ω + 2 Ω = 6 Ω.
The reason for us considering these equivalent resistances is because the easiest problem to solve is the simple, one-resistor problem presented originally. You can represent many combinations of resistors in series or parallel as a single equivalent resistor, and then use Ohm’s Law to solve for the current.
Example 3 Determine the current through each resistor, the voltage drop across each resistor, and the power given off (dissipated) as heat in each resistor of this circuit:
Solution. You might want to redraw the circuit each time we replace a combination of resistors by its equivalent resistance. From our work in the preceding example, we have
From diagram 3, which has just one resistor, we can figure out the current:
Now we can work our way back to the original circuit (diagram 1). In going from 3 to 2, we are going back to a series combination, and what do resistors in series share? That’s right, the same current. So, we take the current, I = 2 A, back to diagram 2. The current through each resistor in diagram 2 is 2 A.
Since we know the current through each resistor, we can figure out the voltage drop across each resistor using the equation V = IR. The voltage drop across the 4 Ω resistor is (2 A)(4 Ω) = 8 V, and the voltage drop across the 2 Ω resistor is (2 A)(2 Ω) = 4 V. Notice that the total voltage drop across the two resistors is 8 V + 4 V = 12 V, which matches the emf of the battery.
Now for the last step; going from diagram 2 back to diagram 1. Nothing needs to be done with the 4 Ω resistor; nothing about it changes in going from diagram 2 to 1, but the 2 Ω resistor in diagram 2 goes back to the parallel combination. And what do resistors in parallel share? The same voltage drop. So we take the voltage drop, V = 4 V, back to diagram 1. The voltage drop across each of the two parallel resistors in diagram 1 is 4 V.
Since we know the voltage drop across each resistor, we can figure out the current through each resistor by using the equation I = V/R. The current through the 3 Ω resistor is (4 V)/(3 Ω) = A, and the current through the 6 Ω resistor is (4 V)/ (6 Ω) = A. Note that the current entering the parallel combination (2 A) equals the total current passing through the individual resistors . Again this was expected.
Finally, we will calculate the power dissipated as heat by each resistor. We can use any of the equivalent formulas: P = IV, P = I 2R, or P = V 2/R.
So, the resistors are dissipating a total of
16 W + W + W = 24 W
If the resistors are dissipating a total of 24 J every second, then they must be provided with that much power. This is easy to check: P = IV = (2 A)(12 V) = 24 W.
Example 4 For the following circuit,
(a) In which direction will current flow and why?
(b) What’s the overall emf?
(c) What’s the current in the circuit?
(d) At what rate is energy consumed by, and provided to, this circuit?
Solution.
(a) The battery V1 wants to send current clockwise, while the battery V2 wants to send current counterclockwise. Since V2 > V1, the battery whose emf is V2 is the more powerful battery, so the current will flow counterclockwise.
(b) Charges forced through V1 will lose, rather than gain, 4 V of potential, so the overall emf of this circuit is V2 – V1 = 8 V.
(c) Since the total resistance is 3 Ω + 1 Ω = 4 Ω, the current will be I = (8 V)/(4 Ω) = 2 A. Note that we use the overall emf, and not either of the individual battery emfs.
(d) V2 will provide energy at a rate of P2 = IV2 = (2 A)(12 V) = 24 W, while V1 will absorb at a rate of P1 = IV1 = (2 A)(4 V) = 8 W. Finally, energy will be dissipated in these resistors at a rate of I2R1 + I2R2 = (2 A)2(3 Ω) + (2 A)2(1 Ω) = 16 W. Once again, energy is conserved; the power delivered (24 W) equals the power taken (8 W + 16 W = 24 W).
Example 5 All real batteries contain internal resistance, r. Determine the current in the following circuit when the switch S is closed:
Solution. Before the switch is closed, there is no complete conducting pathway from the positive terminal of the battery to the negative terminal, so no current flows through the resistors. However, once the switch is closed, the resistance of the circuit is 2 Ω + 3 Ω + 5 Ω = 10 Ω, so the current in the circuit is I = (20 V)/(10 Ω) = 2 A. Often the battery and its internal resistance are enclosed in a dashed box:
In this case, a distinction can be made between the emf of the battery and the actual voltage it provides once the current has begun. Since I = 2 A, the voltage drop across the internal resistance is Ir = (2 A)(2 Ω) = 4 V, so the effective voltage provided by the battery to the rest of the circuit—called the terminal voltage—is lower than the ideal voltage. It is V = VB – Ir = 20 V – 4 V = 16 V.
Example 6 A student has three 30 Ω resistors and an ideal 90 V battery. (A battery is ideal if it has a negligible internal resistance.) Compare the current drawn from—and the power supplied by—the battery when the resistors are arranged in parallel versus in series.
Solution. Resistors in series always provide an equivalent resistance greater than any of the individual resistances, and resistors in parallel always provide an equivalent resistance smaller than their individual resistances. So, hooking up the resistors in parallel will create the smallest resistance and draw the greatest total current:
In this case, the equivalent resistance is
and the total current is I = V/RP = (90 V)/(10 Ω) = 9 A. (You could verify that 3 A of current would flow in each of the three branches of the combination.) The power supplied by the battery will be P = IV = (9 A)(90 V) = 810 W.
If the resistors are in series, the equivalent resistance is RS = 30 Ω + 30 Ω + 30 Ω = 90 Ω, and the current drawn is only I = V/RS = (90 V)/(90 Ω) = 1 A. The power supplied by the battery in this case is just P = IV = (1 A)(90 V) = 90 W.
Example 7 A voltmeter is a device used to measure the voltage between two points in a circuit. An ammeter is used to measure current. Determine the readings on the voltmeter (denoted ) and the ammeter (denoted ) in the circuit below.
Solution. We consider the ammeter to be ideal; this means it has negligible resistance, so it doesn’t alter the current that it’s trying to measure. Similarly, we consider the voltmeter to have an extremely high resistance, so it draws negligible current away from the circuit.
Our first goal is to find the equivalent resistance in the circuit. The 600 Ω and 300 Ω resistors are in parallel; they’re equivalent to a single 200 Ω resistor. This is in series with the battery’s internal resistance, r, and R3. The overall equivalent resistance is therefore R = 50 Ω + 200 Ω + 150 Ω = 400 Ω , so the current supplied by the battery is I = V/R (2400 V)/(400 Ω) = 6 A. At the junction marked J, this current splits. Since R1 is twice R2, half as much current will flow through R1 as through R2; the current through R1 is I1 = 2 A, and the current through R2 is I2 = 4 A. The voltage drop across each of these resistors is I1R1 = I2R2 = 1,200 V (matching voltages verify the values of currents I1 and I2). Since the ammeter is in the branch that contains R2, it will read I2 = 4 A.
The voltmeter will read the voltage drop across R3, which is V3 = IR3 = (6 A) (150 Ω) = 900 V. So the potential at point b is 900 V lower than at point a.
Example 8 The diagram below shows a point a at potential V = 20 V connected by a combination of resistors to a point (denoted G) that is grounded. The ground is considered to be at potential zero. If the potential at point a is maintained at 20 V, what is the current through R3?
Solution. R1 and R2 are in parallel; their equivalent resistance is RP, where
RP is in series with R3, so the equivalent resistance is:
R = RP + R3 = (2.4 Ω) + (8 Ω) = 10.4 Ω
and the current that flows through R3 is
Kirchhoff’s Rules
When the resistors in a circuit cannot be classified as either in series or in parallel, we need another method for analyzing the circuit. The rules of Gustav Kirchhoff (pronounced “Keer koff”) can be applied to any circuit:
The Loop Rule. The sum of the potential differences (positive and negative) that traverse any closed loop in a circuit must be zero.
The Junction Rule. The total current that enters a junction must equal the total current that leaves the junction. (This is also known as the Node Rule.)
The Loop Rule just says that, starting at any point, by the time we get back to that same point by following any closed loop, we have to be back to the same potential. Therefore, the total drop in potential must equal the total rise in potential. Put another way, the Loop Rule says that all the decreases in electrical potential energy (for example, caused by resistors in the direction of the current) must be balanced by all the increases in electrical potential energy (for example, caused by a source of emf from the negative to positive terminal). So the Loop Rule is basically a re-statement of the Law of Conservation of Energy.
Similarly, the Junction Rule simply says that the charge (per unit time) that goes into a junction must equal the charge (per unit time) that comes out. This is basically a statement of the Law of Conservation of Charge. In practice, the Junction Rule is straightforward to apply.
The most important things to remember about the Loop Rule can be summarized as follows:
• First, you choose a direction for the loop—either clockwise or counterclockwise, it doesn’t matter.
• When the loop goes across a resistor in the same direction as the current, the potential drops by IR.
• When the loop goes across a resistor in the opposite direction from the current, the potential increases by IR.
• When the loop goes from the negative to the positive terminal of a source of emf, the potential increases by V.
• When the loop goes from the positive to the negative terminal of a source of emf, the potential decreases by V.
Example 9 Use Kirchhoff’s Rules to determine the current through R2 in the following circuit:
Solution. First, let’s label some points in the circuit.
The points c and f are junctions (nodes). We have two nodes and three branches: one branch is fabc, another branch is cdef, and the third branch is cf. Each branch has one current throughout. If we label the current in fabc I1 and the current in branch cdef I2 (with the directions as shown in the diagram below), then the current in branch cf must be I1 – I2, by the Junction Rule: I1 comes into c, and a total of I2 + (I1 – I2) = I1 comes out.
Now pick a loop, say, abcfa, in the counterclockwise direction. Starting at a, we go to b, then across R1 in the direction of the current, so the potential drops by I1R1. Then we move to c, then up through R2 in the direction of the current, so the potential drops by (I1 – I2)R2. Then we reach f, turn left and travel through V1 from the negative to the positive terminal, so the potential increases by V1. We now find ourselves back at a. By the Loop Rule, the total change in potential around this closed loop must be zero, and
–I1R1 – (I1 – I2)R2 + V1 = 0
Since we have two unknowns (I1 and I2), we need two equations, so now pick another loop; let’s choose cdefc, again in the counterclockwise direction. From c to d, we travel across the resistor in the direction of the current, so the potential drops by I2R3. From e to f, we travel through V2 from the positive to the negative terminal, so the potential drops by V2. Heading down from f to c, we travel across R2 but in the direction opposite to the current, so the potential increases by (I1 – I2)R2. At c, our loop is completed, so
–I2R3 – V2 + (I1 – I2)R2 = 0
Substituting in the given numerical values for R1, R2, R3, V1, and V2, and simplifying, these two equations become
3I1 – 2I2 = 2
2I1 – 5I2 = 1
Solving this pair of simultaneous equations, we get
I1 = A = 0.73 A and I2 = A = 0.09 A
So the current through R2 is I1 – I2 = A = 0.64 A.
The choice of directions of the currents at the beginning of the solution was arbitrary. Don’t worry about trying to guess the actual direction of the current in a particular branch. Just pick a direction, stick with it, and obey the Junction Rule. At the end, when you solve for the values of the branch current, a negative value will alert you that the direction of the current is actually opposite to the direction you originally chose for it in your diagram.
COMBINATIONS OF CAPACITORS
Capacitors are often arranged in combination in electric circuits. Here we’ll look at the same two types of arrangements, the parallel combination and the series combination.
A collection of capacitors are said to be in parallel if they all share the same potential difference. The following diagram shows two capacitors wired in parallel:
The top plates are connected by a wire and form a single equipotential; the same is true for the bottom plates. Therefore, the potential difference across one capacitor is the same as the potential difference across the other capacitor.
We want to find the capacitance of a single capacitor that would perform the same function as this combination. If the capacitances are C1 and C2, then the charge on the first capacitor is Q1 = C1V and the charge on the second capacitor is Q2 = C2V. The total charge on the combination is Q1 + Q2, so the equivalent capacitance, CP, must be
So the equivalent capacitance of a collection of capacitors in parallel is found by adding the individual capacitances.
A collection of capacitors are said to be in series if they all share the same charge magnitude. This is similar to resistors in series, except resistors in series would share current, not charge. The following diagram shows two capacitors wired in series:
When a potential difference is applied, as shown, negative charge will be deposited on the bottom plate of the bottom capacitor; this will push an equal amount of negative charge away from the top plate of the bottom capacitor toward the bottom plate of the top capacitor. When the system has reached equilibrium, the charges on all the plates will have the same magnitude:
If the top and bottom capacitors have capacitances of C1 and C2 , respectively, then the potential difference across the top capacitor is V1 = Q/C1, and the potential difference across the bottom capacitor is V2 = Q/C2. The total potential difference across the combination is V1 + V2, which must equal V. Therefore, the equivalent capacitance, CS, must be
We can write this in another form:
In words, the reciprocal of the capacitance of a collection of capacitors in series is found by adding the reciprocals of the individual capacitances.
Example 10 Given that C1 = 2 µF, C2 = 4 µF, and C3 = 6 µF, calculate the equivalent capacitance for the following combination:
Solution. Notice that C2 and C3 are in series, and they are in parallel with C1. That is, the capacitor equivalent to the series combination of C2 and C3 (which we’ll call C2-3) is in parallel with C1. We can represent this as follows:
So, the first step is to find C2-3:
Now this is in parallel with C1, so the overall equivalent capacitance (C1-2-3) is
Substituting in the given numerical values, we get
CAPACITORS AND DIELECTRICS REVISITED
Now let’s examine the case in which a capacitor is first charged up and, while it’s still connected to its voltage source, has a dielectric inserted between its plates. First, since the capacitor is still connected to the battery, the voltage between the plates must match the voltage of the battery. Therefore, V will not change. Next, because the capacitance C increases, the equation Q = CV tells us that the charge Q must increase; in fact, because V doesn’t change and C increases by a factor of κ, we see that Q will increase by a factor of κ. Next, using the equation V = Ed, we see that because V doesn’t change, neither will E. Finally, using the equation U = QV, we conclude that since V doesn’t change and Q increases by a factor of κ, the stored electric potential energy increases by a factor of κ. An important point to notice is that V doesn’t change because the battery will transfer additional charge to the capacitor plates. This increase in Q offsets any momentary decrease in the electric field strength when the dielectric is inserted (because the molecules of the dielectric are polarized, as above) and brings the electric field strength back to its original value. Furthermore, as more charge is transferred to the plates, more electric potential energy is stored.
The following figures summarize the effects on the properties of a capacitor with a dielectric inserted between the plates.
C increases by a factor of κ | C always increases |
Q stays the same | battery disconnected |
V decreases by a factor of κ | Q↑ = C ↓ V |
E decreases by a factor of κ | ↓ V = ↓ Ed |
UPE decreases by a factor of κ | ↓ U = (1/2) ↓ QV |
C increases by a factor of κ | C always increases |
Q stays the same | battery disconnected |
V decreases by a factor of κ | ↑ Q = ↑ CV |
E decreases by a factor of κ | V = Ed |
UPE decreases by a factor of κ | ↑ U = (1/2) ↑ QV |
RESISTANCE–CAPACITANCE (RC) CIRCUITS
Capacitors are typically charged by batteries. Once the switch in the diagram on the left is closed, electrons are attracted to the positive terminal of the battery and leave the top plate of the capacitor. Electrons also accumulate on the bottom plate of the capacitor, and this continues until the voltage across the capacitor plates matches the emf of the battery. When this condition is reached, the current stops and the capacitor is fully charged.
Example 11 Determine the current through and the voltage across each electrical device in the following circuit when
(a) the switch has just been closed
(b) the switch has been thrown for a long period of time
Solution.
(a) When the switch has just been closed, the capacitor is treated like a wire with no resistance. The circuit behaves like a simple parallel circuit. There are 12 V across the 6 Ω resistor so 2 amps of current flow through it. There are 12 V across the 3 Ω resistor so 4 amps flow through it. By Kirchhoff’s Law, the currents add to 6 amps through the battery and the equivalent resistance of the circuit is 2 Ω.
(b) After the switch has been closed for a long period of time, the capacitor behaves like an infinite resistor. There are still 12 V across the 6 Ω resistor so 2 amps of current still flow through it. If the capacitor behaves like an infinite resistor, no current can flow through that branch. There will be no current through the 3 Ω resistor. The capacitor will have 12 V across it and there will be no voltage across the 3 Ω resistor (otherwise current would be flowing through it). The current will be 2 amps through the battery and the equivalent resistance of the circuit is 6 Ω.
Example 12 Determine the current through and the voltage across each electrical device in the following circuit when
(a) the switch has just been closed
(b) the switch has been thrown for a long period of time
Solution.
(a) Immediately after one throws the switch, the capacitor exhibits behavior similar to that of a wire with no resistance through it. The circuit can be redrawn and thought of as
The 6 Ω and 12 Ω resistors are in parallel and can be thought of as 4 Ω (from 1/Req = 1/R1 + 1/R2 ). Similarly, the 4 Ω and 8 Ω resistors are in parallel and can be thought of as 2.67 Ω (again, from 1/Req = 1/R1 + 1/R2). The equivalent resistance of these two branches, which are in series with each other, is 6.67 Ω (from Req = R1 + R2).
The current through the battery is therefore 3 amps (from Ohm’s Law, V = IR).
The current through the 6 Ω and 12 Ω (top two resistors) is 3 amps traveling through an effective resistance of 4 Ω. The voltage drop across the top branch is 12 V (from Ohm’s Law). The currents through the 6 Ω and 12 Ω resistors are 2 amps and 1 amp, respectively (again from Ohm’s Law).
The current through the 4 Ω and 8 Ω (bottom two resistors) is also 3 amps through an effective resistance of 2.67 Ω. The voltage drop across the branch is 8 V (from Ohm’s Law or from the fact that the voltage across each series section must sum to the total voltage across the battery). The currents through the 4 Ω and 8 Ω resistors are 2 amps and 1 amp, respectively (again from Ohm’s Law).
(b) When the switch has been closed for a long period of time, the capacitor behaves just like an infinite resistor. The circuit can be redrawn and thought of as
The 6 Ω and 4 Ω resistors are in series and can be thought of as a 10 Ω resistor (from Req = R1 + R2). Similarly, the 12 Ω and 8 Ω resistors can be thought of as 20 Ω (again, from Req = R1 + R2). Since the branches are in parallel, the voltage is 20 volts across each path and the current can be calculated by Ohm’s Law (V = IR); 2 amps go through the left branch (the 6 Ω and 4 Ω resistor) and 1 amp goes through the right branch (the 12 Ω and 8 Ω resistor).
The current through the battery is therefore 3 amps (Kirchhoff’s Node Rule).
The voltage across each of the 6 Ω, 4 Ω, 12 Ω, and 8 Ω resistors can be obtained from Ohm’s Law. They are 12 V, 8 V, 12 V, and 8 V, respectively.
Chapter 7 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 7 Review Questions.
Section I: Multiple Choice
1. A wire made of brass and a wire made of silver have the same length, but the diameter of the brass wire is 4 times the diameter of the silver wire. The resistivity of brass is 5 times greater than the resistivity of silver. If RB denotes the resistance of the brass wire and RS denotes the resistance of the silver wire, which of the following is true?
(A) RB = RS
(B) RB = RS
(C) RB = RS
(D) RB = RS
2. For an ohmic conductor, doubling the voltage without changing the resistance will cause the current to
(A) decrease by a factor of 4
(B) decrease by a factor of 2
(C) increase by a factor of 2
(D) increase by a factor of 4
3. If a 60-watt light bulb operates at a voltage of 120 V, what is the resistance of the bulb?
(A) 2 Ω
(B) 30 Ω
(C) 240 Ω
(D) 720 Ω
4. A battery whose emf is 40 V has an internal resistance of 5 Ω. If this battery is connected to a 15 Ω resistor R, what will the voltage drop across R be?
(A) 10 V
(B) 30 V
(C) 40 V
(D) 50 V
5.
Determine the equivalent resistance between points a and b.
(A) 0.25 Ω
(B) 0.333 Ω
(C) 1.5 Ω
(D) 2 Ω
6.
Three identical light bulbs are connected to a source of emf, as shown in the diagram above. What will happen if the middle bulb burns out?
(A) The light intensity of the other two bulbs will decrease (but they won’t go out).
(B) The light intensity of the other two bulbs will increase.
(C) The light intensity of the other two bulbs will remain the same.
(D) More current will be drawn from the source of emf.
7.
What is the voltage drop across the 12 Ω resistor in the portion of the circuit shown above?
(A) 24 V
(B) 36 V
(C) 48 V
(D) 72 V
8.
What is the current through the 8 Ω resistor in the circuit shown above?
(A) 0.5 A
(B) 1.0 A
(C) 1.5 A
(D) 3.0 A
9. How much energy is dissipated as heat in 20 s by a 100 Ω resistor that carries a current of 0.5 A?
(A) 50 J
(B) 100 J
(C) 250 J
(D) 500 J
10.
If each of the capacitors in the array shown above is C, what is the capacitance of the entire combination?
(A) C/2
(B) 2C/3
(C) 5C/6
(D) 2C
Section II: Free Response
1. Consider the following circuit:
(a) At what rate does the battery deliver energy to the circuit?
(b) Find the current through the 40 Ω resistor.
(c) (i) Determine the potential difference between points a and b.
(ii) At which of these two points is the potential higher?
(d) Find the energy dissipated by the 100 Ω resistor in 10 s.
(e) Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is 0.45 Ω∙m, determine its radius.
2. Consider the following circuit:
(a) What is the current through each resistor?
(b) What is the potential difference across each resistor?
(c) What is the equivalent resistance of the circuit?
Summary
The resistance of an object can be determined by R = where ρ is the resistivity (a property of the material), ℓ is the length, and A is the cross-sectional area.
The current is the rate at which charge is transferred and given by IAVG = .
Many objects obey Ohm’s Law, which is given by V = IR.
The electrical power in a circuit is given by P = IV or P = I2 R or P = .
This is the same power we’ve encountered in our discussion of energy P = .
Kirchhoff’s Loop Rule tells us that the sum of the potential differences in any closed loop in a circuit must be zero.
Kirchhoff’s Junction Rule (Node Rule) tells us the total current that enters a junction must equal the total current that leaves the junction.
To find the combined capacitance of capacitors in parallel (side by side), simply add their capacitances.
To find the capacitance of capacitors in series (one after another), add their inverses:
Chapter 8
Magnetic Forces and Fields
INTRODUCTION
In Chapter 5, we learned that electric charges are the sources of electric fields and that other charges experience an electric force in those fields. The charges generating the field were assumed to be at rest, because if they weren’t, then another force field would have been generated in addition to the electric field. Electric charges that move are the sources of magnetic fields, and other charges that move can experience a magnetic force in these fields.
MAGNETIC FIELDS
Similar to our discussion about electric fields, the space surrounding a magnet is permeated by a magnetic field. The direction of the magnetic field is defined as pointing out of the north end of a magnet and into the south end of a magnet as illustrated below.
When two magnets get near each other the magnetic fields interfere with each other and can be drawn as follows. Note that, for simplicity’s sake, only the field lines closest to the poles are shown.
Notice there is a curve to the above fields. We call the field uniform if the field lines are parallel and of equal strength. It is easy to recognize a uniform magnetic field to the right, left, top of the page, or bottom of the page. But you will also see a field going into or out of the page. A field into the page looks as if there were a north pole of a magnet above the page pointing down at the south pole of a magnet that is below the page. It is represented by an area with X’s going into the page. A field coming out of the page looks as if there were a north pole of a magnet below the page pointing up at the south pole of a magnet that is above the page. It is represented by an area with dots (•) coming out of the page.
BAR MAGNETS
A permanent bar magnet creates a magnetic field that closely resembles the magnetic field produced by a circular loop of current-carrying wire:
By convention, the magnetic field lines emanate from the end of the magnet designated the north pole (N) and then curl around and re-enter the magnet at the end designated the south pole (S). The magnetic field decreases in strength as distance from the magnet increases. The magnetic field created by a permanent bar magnet is due to the electrons, which have an intrinsic spin as they orbit the nuclei. They are literally charges in motion, which is the ultimate source of any and all magnetic fields. If a piece of iron is placed in an external magnetic field (for example, one created by a current-carrying solenoid), the individual magnetic dipole moments of the electrons will be forced to more or less line up. Because iron is ferromagnetic, these now-aligned magnetic dipole moments tend to retain this configuration, thus permanently magnetizing the bar and causing it to produce its own magnetic field.
As with electric charges, like magnetic poles repel each other, while opposite magnetic poles attract each other.
However, while you can have a positive electric charge all by itself, you can’t have a single magnetic pole all by itself: the existence of a lone magnetic pole has never been confirmed. That is, there are no magnetic monopoles; magnetic poles always exist in pairs. If you break a bar magnet into two pieces, it does not produce one piece with just an N and another with just an S; it produces two separate, complete magnets, each with an N-S pair.
THE MAGNETIC FORCE ON A MOVING CHARGE
If a particle with charge q moves with velocity v through a magnetic field B, it will experience a magnetic force, FB:
FB = qv × B
with magnitude:
FB = |q|vB sin θ
where θ is the angle between v and B. From this equation, we can see that if the charge is at rest, then v = 0 immediately gives us FB = 0. This tells us that magnetic forces only act on moving charges. Also, if v is parallel (or antiparallel) to B, then FB = 0 since, in either of these cases, sin θ = 0. So, only charges that cut across the magnetic field lines will experience a magnetic force. Furthermore, the magnetic force is maximized when v is perpendicular to B, since if θ = 90°, then sin θ is equal to 1, its maximum value.
The direction of FB is always perpendicular to both v and B and depends on the sign of the charge q and the direction of v × B (which can be found by using the right-hand rule).
If q is positive, use your right hand and the right-hand rule.
If q is negative, you need your thumb to point in the direction of the product of q and v, not just the direction of the velocity.
Whenever you use the right-hand rule, follow these steps:
1. Orient your hand so that your thumb points in the direction of the velocity v. If the charge is negative, turn your thumb by 180 degrees.
2. Point your fingers in the direction of B.
3. The direction of FB will then be perpendicular to your palm.
Think of your palm pushing with the force FB; the direction it pushes is the direction of FB.
Note that there are fundamental differences between the electric force and magnetic force on a charge. First, a magnetic force acts on a charge only if the charge is moving; the electric force acts on a charge whether it moves or not. Second, the direction of the magnetic force is always perpendicular to the magnetic field, while the electric force is always parallel (or antiparallel) to the electric field.
Example 1 For each of the following charged particles moving through a magnetic field, determine the direction of the force acting on the charge.
Solution.
a) If you point your fingers to the top of the page and thumb to the right of the page, your palm should point out of the page. The force is out of the page.
b) If you point your fingers into the page and thumb to the top of the page your palm should point to the left of the page. The force is to the left of the page.
c) If you point your fingers into the page and thumb to the right of the page (remember, when the charge is negative, your thumb points in the opposite direction from the velocity), your palm should point to the bottom of the page. The force of a positive charge would be toward the bottom of the page, but because this is a negative charge, the force points up.
The SI unit for the magnetic field is the tesla (abbreviated T) which is one newton per ampere-meter.
Example 2 A charge +q = +6 × 10–6 C moves with speed v = 4 × 105 m/s through a magnetic field of strength B = 0.4 T, as shown in the figure below. What is the magnetic force experienced by q?
Solution. The magnitude of FB is
FB = qvB sin θ = (6 × 10–6 C)(4 × 105 m/s)(0.4 T) sin 30° = 0.48 N
By the right-hand rule, the direction is into the plane of the page, which is symbolized by ×.
Example 3 A particle of mass m and charge +q is projected with velocity v (in the plane of the page) into a uniform magnetic field B that points into the page. How will the particle move?
Solution. Since v is perpendicular to B, the particle will feel a magnetic force of strength qvB, which will be directed perpendicular to v (and to B) as shown:
Since FB is always perpendicular to v, the particle will undergo uniform circular motion; FB will provide the centripetal force. Notice that, because FB is always perpendicular to v, the magnitude of v will not change, just its direction. Magnetic forces alone cannot change the speed of a charged particle, they can only change its direction of motion. The radius of the particle’s circular path is found from the equation FB = FC:
Example 4 A particle of charge –q is shot into a region that contains an electric field, E, crossed with a perpendicular magnetic field, B. If E = 2 × 104 N/C and B = 0.5 T, what must be the speed of the particle if it is to cross this region without being deflected?
Solution. If the particle is to pass through undeflected, the electric force it feels has to be canceled by the magnetic force. In the diagram above, the electric force on the particle is directed upward (since the charge is negative and E is downward), and the magnetic force is directed downward by the right-hand rule. So FE and FB point in opposite directions, and in order for their magnitudes to balance, qE must equal qvB, so v must equal E/B, which in this case gives
Example 5 A particle with charge +q, traveling with velocity v, enters a uniform magnetic field B, as shown below. Describe the particle’s subsequent motion.
Solution. If the particle’s velocity were parallel to B, then it would be unaffected by B. If v were perpendicular to B, then it would undergo uniform circular motion (as we saw in Example 2). In this case, v is neither purely parallel nor perpendicular to B. It has a component (v1) that’s parallel to B and a component (v2) that’s perpendicular to B.
Component v1 will not be changed by B, so the particle will continue upward in the direction of B. However, the presence of v2 will create circular motion. The superposition of these two types of motion will cause the particle’s trajectory to be a helix; it will spin in circular motion while traveling upward with the speed v1 = v sin φ:
THE MAGNETIC FORCE ON A CURRENT-CARRYING WIRE
Since magnetic fields affect moving charges, they should also affect current- carrying wires. After all, a wire that contains a current contains charges that move. Remember, a current is the flow of positive charges.
Let a wire of length ℓ be immersed in magnetic field B. If the wire carries a current I, then the force on the wire is
FB = Iℓ × B
with magnitude
FB = BIℓ sin θ
where θ is the angle between ℓ and B. Here, the direction of ℓ is the direction of the current, I. The direction of FB can be found using the right-hand rule and by letting your thumb point in the direction in which the current flows. Remember, a current is the flow of positive charges.
Example 6 A U-shaped wire of mass m is lowered into a magnetic field B that points out of the plane of the page. How much current I must pass through the wire in order to cause the net force on the wire to be zero?
Solution. The total magnetic force on the wire is equal to the sum of the magnetic forces on each of the three sections of wire. The force on the first section (the right, vertical one), FB1, is directed to the left (applying the right-hand rule, and the force on the third piece (the left, vertical one), FB3, is directed to the right. Since these pieces are the same length, these two oppositely directed forces have the same magnitude, Iℓ1B = Iℓ3B, and they cancel. So the net magnetic force on the wire is the magnetic force on the middle piece. Since I points to the left and B is out of the page, the right-hand rule tells us the force is upward.
Since the magnetic force on the wire is Iℓ2B, directed upward, the amount of current must create an upward magnetic force that exactly balances the downward gravitational force on the wire. Because the total mass of the wire is m, the resultant force (magnetic + gravitational) will be zero if
Iℓ2B = mg ⇒
Example 7 A rectangular loop of wire that carries a current I is placed in a uniform magnetic field, B, as shown in the diagram below and is free to rotate. What torque does it experience?
Solution. Ignoring the tiny gap in the vertical left-hand wire, we have two wires of length ℓ1 and two of length ℓ2. There is no magnetic force on either of the sides of the loop of length ℓ2, because the current in the top side is parallel to B and the current in the bottom side is antiparallel to B. The magnetic force on the right-hand side points out of the plane of the page, while the magnetic force on the left-hand side points into the plane of the page.
Each of these two forces exerts a torque that tends to turn the loop in such a way that the right-hand side rises out of the plane of the page and the left-hand side rotates into the page. Relative to the axis shown above (which cuts the loop in half), the torque of FB1 is
τ1 = rFB1 sin θ = ( ℓ2) (I ℓ1B) sin 90° = Iℓ1ℓ2B
and the torque of FB2 is
τ2 = rFB2 sin θ = ( ℓ2) (Iℓ1B) sin 90° = Iℓ1ℓ2B
Since both these torques rotate the loop in the same direction, the net torque on the loop is
τ1 + τ2 = Iℓ1ℓ2B
MAGNETIC FIELDS CREATED BY CURRENT-CARRYING WIRES
In the previous section, we examined the force that a current-carrying wire experiences when it is subjected to an external magnetic field. The source of the magnetic field in the previous section was unspecified. As we said at the beginning of this chapter, the sources of magnetic fields are electric charges that move; they may spin, circulate, move through space, or flow down a wire. For example, consider a long, straight wire that carries a current I. The current generates a magnetic field in the surrounding space, of magnitude
where r is the distance from the wire. The symbol µ0 denotes a fundamental constant called the permeability of free space. Its value is:
μ0 = 4 π × 10−7 N/A2 = 4π × 10−7 T · m/A
The magnetic field lines are actually circles whose centers are on the wire. The direction of these circles is determined by a variation of the right-hand rule. Imagine grabbing the wire in your right hand with your thumb pointing in the direction of the current. Then the direction in which your fingers curl around the wire gives the direction of the magnetic field line.
Right-Hand Rule for the Magnetic Field Created by a Current-Carrying Wire:
1. Put your thumb in direction of current or in direction of a positive traveling charge.
2. Grab the wire/path.
3. As the fingers curl around your thumb, it represents the magnetic field going around the wire/path.
Example 8 The diagram below shows a proton moving with a speed of 2 × 105 m/s, initially parallel to, and 4 cm from, a long, straight wire. If the current in the wire is 20 A, what’s the magnetic force on the proton?
Solution. Above the wire (where the proton is), the magnetic field lines generated by the current-carrying wire point out of the plane of the page, so v0 × B points downward. Since the proton’s charge is positive, the magnetic force FB = q(v0 × B) is also directed down, toward the wire.
The strength of the magnetic force on the proton is
Example 9 The diagram below shows a pair of long, straight, parallel wires, separated by a small distance, r. If currents I1 and I2 are established in the wires, what is the magnetic force per unit length they exert on each other?
Solution. To find the force on Wire 2, consider the current in Wire 1 as the source of the magnetic field. Below Wire 1, the magnetic field lines generated by Wire 1 point into the plane of the page. Therefore, the force on Wire 2, as given by the equation FB2 = I2(ℓ2 × B1), points upward.
The magnitude of the magnetic force per unit length felt by Wire 2, due to the magnetic field generated by Wire 1, is found this way:
By Newton’s Third Law, this is the same force that Wire 1 feels due to the magnetic field generated by Wire 2. The force is attractive because the currents point in the same direction; if one of the currents were reversed, then the force between the wires would be repulsive.
Chapter 8 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 8 Review Questions.
Section I: Multiple Choice
1. Which of the following is/are true concerning magnetic forces and fields? Select two answers.
(A) The magnetic field lines due to a current-carrying wire radiate away from the wire.
(B) The kinetic energy of a charged particle can be increased by a magnetic force.
(C) A charged particle can move through a magnetic field without feeling a magnetic force.
(D) A moving charged particle generates a magnetic field.
2. The velocity of a particle of charge +4.0 × 10–9 C and mass 2 × 10–4 kg is perpendicular to a 0.1-tesla magnetic field. If the particle’s speed is 3 × 104 m/s, what is the acceleration of this particle due to the magnetic force?
(A) 0.0006 m/s2
(B) 0.006 m/s2
(C) 0.06 m/s2
(D) 0.6 m/s2
3. In the figure below, what is the direction of the magnetic force FB?
(A) Downward, in the plane of the page
(B) Upward, in the plane of the page
(C) Out of the plane of the page
(D) Into the plane of the page
4. In the figure below, what must be the direction of the particle’s velocity, v ?
(A) Downward, in the plane of the page
(B) Upward, in the plane of the page
(C) Out of the plane of the page
(D) Into the plane of the page
5. Due to the magnetic force, a positively charged particle executes uniform circular motion within a uniform magnetic field, B. If the charge is q and the radius of its path is r, which of the following expressions gives the magnitude of the particle’s linear momentum?
(A) qBr
(B) qB/r
(C) q/(Br)
(D) B/(qr)
6. A straight wire of length 2 m carries a 10-amp current. How strong is the magnetic field at a distance of 2 cm from the wire?
(A) 1 × 10–5 T
(B) 2 × 10–5 T
(C) 1 × 10–4 T
(D) 2 × 10–4 T
7. Two long, straight wires are hanging parallel to each other and are 1 cm apart. The current in Wire 1 is 5 A, and the current in Wire 2 is 10 A, in the same direction. Which of the following best describes the magnetic force per unit length felt by the wires?
(A) The force per unit length on Wire 1 is twice the force per unit length on Wire 2.
(B) The force per unit length on Wire 2 is twice the force per unit length on Wire 1.
(C) The force per unit length on Wire 1 is 0.0003 N/m, away from Wire 2.
(D) The force per unit length on Wire 1 is 0.001 N/m, toward Wire 2.
8. In the figure below, what is the magnetic field at the Point P, which is midway between the two wires?
(A) 2µ0I/(πd), into the plane of the page
(B) µ0I/(2πd), out of the plane of the page
(C) µ0I/(2πd), into the plane of the page
(D) Zero
9. Here is a section of a wire with a current moving to the right. Where is the magnetic field strongest and pointing INTO the page?
(A) A
(B) B
(C) C
(D) D
10. What is the direction of force acting on the current-carrying wire as shown below?
(A) To the bottom of the page
(B) Into the page
(C) Out of the page
(D) To the right of the page
Section II: Free Response
1. The diagram below shows a simple mass spectrograph. It consists of a source of ions (charged atoms) that are accelerated (essentially from rest) by the voltage V and enter a region containing a uniform magnetic field, B. The polarity of V may be reversed so that both positively charged ions (cations) and negatively charged ions (anions) can be accelerated. Once the ions enter the magnetic field, they follow a semicircular path and strike the front wall of the spectrograph, on which photographic plates are constructed to record the impact. Assume that the ions have mass m.
(a) What is the acceleration of an ion of charge q just before it enters the magnetic field?
(b) Find the speed with which an ion of charge q enters the magnetic field.
(c) (i) Which semicircular path, 1 or 2, would a cation follow?
(ii) Which semicircular path, 1 or 2, would an anion follow?
(d) Determine the mass of a cation entering the apparatus in terms of y, q, B, and V.
(e) Once a cation of charge q enters the magnetic field, how long does it take to strike the photographic plate?
(f) What is the work done by the magnetic force in the spectrograph on a cation of charge q?
2. A particle accelerator has a collision that results in a photon, an anti-bottom quark, and a charm quark. The magnetic field is 6.00 × 10–8 T and can be described as into the page. A photon has no charge and has an upper theoretical mass of 3.6 × 10–52 kg. The charm quark has a mass of 2.23 × 10–27 kg, a charge of 1.07 × 10–19 C, and a velocity of 40.1 m/s. The anti-bottom quark has a mass of 7.49 × 10–27 kg and orbits with a radius of 92.7 m at a velocity of 41.5 m/s in a clockwise manner.
(a) What is the orbital radius of the photon?
(b) What is the orbital radius of the charm quark?
(c) What is the charge of the anti-bottom quark?
Summary
Charges moving though a magnetic field experience a force whose magnitude is given by FB = qvB sin θ and whose direction is given by the right-hand rule.
Because the force is always perpendicular to the direction of velocity, the charge may experience uniform circular motion. It would then follow all the appropriate circular motion relationships and orbit in a radius given by r = .
Because wires have charges moving though them, a wire will experience a force if placed in a magnetic field. This is expressed by FB = BIℓ sin θ. Also, a current-carrying wire will produce a magnetic field whose strength is given by B = , where µ0 = 4π × 10–7 T⋅m/A, I is the current through the wire, and r is the radial distance from the wire.
Chapter 9
Electromagnetic Induction
INTRODUCTION
In Chapter 8, we learned that electric currents generate magnetic fields, and we will now see how magnetism can generate electric currents.
MOTIONAL EMF
The figure below shows a conducting wire of length ℓ, moving with constant velocity v in the plane of the page through a uniform magnetic field B that’s perpendicular to the page. The magnetic field exerts a force on the moving conduction electrons in the wire. With B pointing into the page, the direction of v is to the right, so the magnetic force, FB , for positive charges, would be upward by the right-hand rule, so FB on these electrons (which are negatively charged) is downward.
As a result, electrons will be pushed to the lower end of the wire, which will leave an excess of positive charge at its upper end. This separation of charge creates a uniform electric field, E, within the wire (pointing downward).
A charge q in the wire feels two forces: an electric force, FE = qE, and a magnetic force, FB = qvBsin θ = qvB, because θ = 90°.
If q is negative, FE is upward and FB is downward; if q is positive, FE is downward and FB is upward. So, in both cases, the forces act in opposite directions. Once the magnitude of FE equals the magnitude of FB, the charges in the wire are in electromagnetic equilibrium. This occurs when qE = qvB; that is, when E = vB.
The presence of the electric field creates a potential difference between the ends of the rod. Since negative charge accumulates at the lower end (which we’ll call point a) and positive charge accumulates at the upper end (point b), point b is at a higher electric potential. The potential difference Vba is equal to Eℓ and, since E = vB, the potential difference can be written as vBℓ.
Now, imagine that the rod is sliding along a pair of conducting rails connected at the left by a stationary bar. The sliding rod now completes a rectangular circuit, and the potential difference Vba causes current to flow.
Induced Current
An induced current can be created three different ways:
1. Changing the area of the loop of wire in a stationary magnetic field
2. Changing the magnetic field strength through a stationary circuit
3. Changing the angle between the magnetic field and the wire loop
Each of these three changes causes a change in the flux, as we will see in the next section. The motion of the sliding rod through the magnetic field creates an electromotive force, called motional emf:
ε = Bℓv
The existence of a current in the sliding rod causes the magnetic field to exert a force on it. Using the formula FB = BIℓ, the fact that ℓ points upward (in the direction of the current) and B is into the page, tells us that the direction of FB on the rod is to the left. An external agent must provide this same amount of force to the right to maintain the rod’s constant velocity and keep the current flowing. The power that the external agent must supply is P = Fv = IℓBv, and the electrical power delivered to the circuit is P = IVba = Iε = IvBℓ. Notice that these two expressions are identical. The energy provided by the external agent is transformed first into electrical energy and then thermal energy as the conductors making up the circuit dissipate heat.
This relationship between current in a coil of wire and magnetic fields sets the basis for Faraday’s Law.
FARADAY’S LAW OF ELECTROMAGNETIC INDUCTION
Electromotive force can be created by the motion of a conducting wire through a magnetic field, but there is another way to generate an emf from a magnetic field.
Faraday’s discovery found that a current is induced when the magnetic flux passing through the coil or loop of wire changes. Magnetic flux helps us define what that amount of field passing into the loop means.
Imagine holding a loop of wire in front of a fan as shown:
The amount of air that flows through the loop depends on the area of the loop as well as its orientation (tilt angle θ).
The most effective airflow is when the loop is completely perpendicular, as in the situation to the left. The least effective is when the airflow and loop are in the situation to the right.
We can apply this idea to a magnetic field passing through a loop. The magnetic flux, ΦB, through an area A is equal to the product of A and the magnetic field parallel to the area vector. The area vector points normal to the surface.
ΦB = B • A = BA cos θ
Magnetic flux measures the density of magnetic field lines that cross through an area. (Note that the direction of A is taken to be perpendicular to the plane of the loop.)
Example 1 The figure below shows two views of a circular loop of radius 3 cm placed within a uniform magnetic field, B (magnitude 0.2 T).
(a) What’s the magnetic flux through the loop?
(b) What would be the magnetic flux through the loop if the loop were rotated 45°?
(c) What would be the magnetic flux through the loop if the loop were rotated 90°?
Solution.
(a) Since B is parallel to A, the magnetic flux is equal to BA:
ΦB = BA = B • πr2 = (0.2 T) • π(0.03 m)2 = 5.7 × 10–4 T•m2
The SI unit for magnetic flux, the tesla•meter2, is called a weber (abbreviated Wb). So ΦB = 5.7 × 10–4 Wb.
(b) Since the angle between B and A is 45°, the magnetic flux through the loop is
ΦB = BA cos 45° = B • πr2 cos 45°= (0.2 T) • π(0.03 m)2 cos 45° = 4.0 × 10–4 Wb
(c) If the angle between B and A is 90°, the magnetic flux through the loop is zero, since cos 90° = 0.
The concept of magnetic flux is crucial, because changes in magnetic flux induce emf. According to Faraday’s Law of Electromagnetic Induction, the magnitude of the emf induced in a circuit is equal to the rate of change of the magnetic flux through the circuit. This can be written mathematically in the form
This induced emf can produce a current, which will then create its own magnetic field. The direction of the induced current is determined by the polarity of the induced emf and is given by Lenz’s Law: The induced current will always flow in the direction that opposes the change in magnetic flux that produced it. If this were not so, then the magnetic flux created by the induced current would magnify the change that produced it, and energy would not be conserved. Lenz’s Law can be included mathematically with Faraday’s Law by the introduction of a minus sign; this leads to a single equation that expresses both results:
Example 2 The circular loop of Example 1 rotates at a constant angular speed through 45° in 0.5 s.
(a) What’s the induced emf in the loop?
(b) In which direction will current be induced to flow?
Solution.
(a) As we found in Example 1, the magnetic flux through the loop changes when the loop rotates. Using the values we determined earlier, Faraday’s Law gives
(b) The original magnetic flux was 5.7 × 10–4 Wb upward, and was decreased to 4.0 × 10–4 Wb. So the change in magnetic flux is –1.7 × 10–4 Wb upward, or, equivalently, ∆ΦB = 1.7 × 10–4 Wb, downward. To oppose this change we would need to create some magnetic flux upward. The current would be induced in the counterclockwise direction (looking down on the loop), because the right-hand rule tells us that then the current would produce a magnetic field that would point up.
The current will flow only while the loop rotates, because emf is induced only when magnetic flux is changing. If the loop rotates 45° and then stops, the current will disappear.
Example 3 Again consider the conducting rod that’s moving with constant velocity v along a pair of parallel conducting rails (separated by a distance ℓ), within a uniform magnetic field, B:
Find the induced emf and the direction of the induced current in the rectangular circuit.
Solution. The area of the rectangular loop is ℓx, where x is the distance from the left-hand bar to the moving rod:
Because the area is changing, the magnetic flux through the loop is changing, which means that an emf will be induced in the loop. To calculate the induced emf, we first write ΦB = BA = Bℓx, then since ∆x/∆t = v, we get
We can figure out the direction of the induced current from Lenz’s Law. As the rod slides to the right, the magnetic flux into the page increases. How do we oppose an increasing into-the-page flux? By producing out-of-the-page flux. In order for the induced current to generate a magnetic field that points out of the plane of the page, the current must be directed counterclockwise (according to the right-hand rule).
Note that the magnitude of the induced emf and the direction of the current agree with the results we derived earlier, in the section on motional emf.
This example also shows how a violation of Lenz’s Law would lead directly to a violation of the Law of Conservation of Energy. The current in the sliding rod is directed upward, as given by Lenz’s Law, so the conduction electrons are drifting downward. The force on these drifting electrons—and thus, the rod itself—is directed to the left, opposing the force that’s pulling the rod to the right. If the current were directed downward, in violation of Lenz’s Law, then the magnetic force on the rod would be to the right, causing the rod to accelerate to the right with ever-increasing speed and kinetic energy, without the input of an equal amount of energy from an external agent.
Example 4 A permanent magnet creates a magnetic field in the surrounding space. The end of the magnet at which the field lines emerge is designated the north pole (N), and the other end is the south pole (S):
(a) The figure below shows a bar magnet moving down, through a circular loop of wire. What will be the direction of the induced current in the wire?
(b) What will be the direction of the induced current in the wire if the magnet is moved as shown in the following diagram?
Solution.
(a) The magnetic flux down, through the loop, increases as the magnet is moved. By Lenz’s Law, the induced emf will generate a current that opposes this change. How do we oppose a change of more flux downward? By creating flux upward. So, according to the right-hand rule, the induced current must flow counterclockwise (because this current will generate an upward-pointing magnetic field):
(b) In this case, the magnetic flux through the loop is upward and, as the south pole moves closer to the loop, the magnetic field strength increases so the magnetic flux through the loop increases upward. How do we oppose a change of more flux upward? By creating flux downward. Therefore, in accordance with the right-hand rule, the induced current will flow clockwise (because this current will generate a downward-pointing magnetic field):
Example 5 A square loop of wire 2 cm on each side contains 5 tight turns and has a total resistance of 0.0002 Ω. It is placed 20 cm from a long, straight, current-carrying wire. If the current in the straight wire is increased at a steady rate from 20 A to 50 A in 2 s, determine the magnitude and direction of the current induced in the square loop. (Because the square loop is at such a great distance from the straight wire, assume that the magnetic field through the loop is uniform and equal to the magnetic field at its center.)
Solution. At the position of the square loop, the magnetic field due to the straight wire is directed out of the plane of the page and its strength is given by the equation B = (µ0/2π)(I/r). As the current in the straight wire increases, the magnetic flux through the turns of the square loop changes, inducing an emf and current. There are N = 5 turns; each loop contributes the same flux, so the total flux becomes the number of loops N times the flux of each individual loop, ΦB. Faraday’s Law becomes εavg = –N(∆ΦB/∆t), and
Substituting the given numerical values, we get
The magnetic flux through the loop is out of the page and increases as the current in the straight wire increases. To oppose an increasing out-of-the-page flux, the direction of the induced current should be clockwise, thereby generating an into-the-page magnetic field (and flux).
The value of the current in the loop will be
Chapter 9 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 9 Review Questions.
Section I: Multiple Choice
1. A metal rod of length L is pulled upward with constant velocity v through a uniform magnetic field B that points out of the plane of the page.
What is the potential difference between points a and b ?
(A) 0
(B) v BL, with point b at the higher potential
(C) vBL, with point a at the higher potential
(D) vBL, with point b at the higher potential
2. A conducting rod of length 0.2 m and resistance 10 ohms between its endpoints slides without friction along a U-shaped conductor in a uniform magnetic field B of magnitude 0.5 T perpendicular to the plane of the conductor, as shown in the diagram below.
If the rod is moving with velocity v = 3 m/s to the left, what is the magnitude and direction of the current induced in the rod?
Current | Direction | |
(A) | 0.03 A | down |
(B) | 0.03 A | up |
(C) | 0.3 A | down |
(D) | 0.3 A | up |
3. In the figure below, a small, circular loop of wire (radius r) is placed on an insulating stand inside a hollow solenoid of radius R. The solenoid has n turns per unit length and carries a counterclockwise current I. If the current in the solenoid is decreased at a steady rate of a amps/s, determine the induced emf, ε, and the direction of the induced current in the loop. Note that B = μ0 n I for a solenoid.
(A) ε = μ0πr2a; induced current is clockwise
(B) ε = μ0πr2a; induced current is counterclockwise
(C) ε = μ0πR2a; induced current is clockwise
(D) ε = μ0πR2a; induced current is counterclockwise
4. In the figures below, a permanent bar magnet is below a loop of wire. It is pulled upward with a constant velocity through the loop of wire as shown in Figure b.
Which of the following best describes the direction(s) of the current induced in the loop (looking down on the loop from above)?
(A) Always clockwise
(B) Always counterclockwise
(C) First clockwise, then counterclockwise
(D) First counterclockwise, then clockwise
5. A square loop of wire (side length = s) surrounds a long, straight wire such that the wire passes through the center of the square.
If the current in the wire is I, determine the current induced in the square loop.
(A)
(B)
(C)
(D) 0
Section II: Free Response
1. A rectangular wire is pulled through a uniform magnetic field of 2 T going into the page as shown. The resistor has a resistance of 20 Ω.
(a) What is the voltage across the resistor as the wire is pulled horizontally at a velocity of 1 m/s and it just enters the field?
(b) What is the current through the circuit in the above case and in what direction does it flow?
(c) What would the voltage be if the wire were rotated 90 degrees and pulled horizontally at the same velocity?
(d) What would the velocity have to be in order to maintain the same voltage as in part (a) but with the orientation of part (c)?
Summary
An electromotive force is produced as a conducting wire’s position changes or as the magnetic field changes. This idea is summarized by ε = Bℓv.
The flux (φ) tells the amount of something (for this chapter the magnetic field) that goes through a surface. The magnetic flux depends on the strength of the magnetic field, the surface area through which the field passes, and the angle between the two. This idea is summed up by φB = BA cos θ.
Faraday’s Law of Induction says that if the wire is formed in a loop, an electromagnetic force is produced if the magnetic flux changes with time. This idea can be summarized by εavg = −.
Chapter 10
Optics
INTRODUCTION
Light (or visible light) makes up only a small part of the entire spectrum of electromagnetic waves, which ranges from radio waves to gamma rays. Most waves require a material medium for their transmission, but electromagnetic waves can propagate through empty space. Electromagnetic waves consist of time-varying electric and magnetic fields that oscillate perpendicular to each other and to the direction of propagation of the wave. Through a vacuum, all electromagnetic waves travel at a fixed speed:
c = 3.00 × 108 m/s
regardless of their frequency. Like all the waves, we can say v = fλ.
ELECTROMAGNETIC WAVES
Back in AP Physics 1, we saw that if we oscillate one end of a long rope, we generate a wave that travels down the rope and has the same frequency as that of the oscillation.
You can think of an electromagnetic wave in a similar way: An oscillating electric charge generates an electromagnetic (EM) wave, which is composed of oscillating electric and magnetic fields. These fields oscillate with the same frequency at which the electric charge that created the wave oscillated. The fields oscillate in phase with each other, perpendicular to each other and the direction of propagation. For this reason, electromagnetic waves are transverse waves. The direction in which the wave’s electric field oscillates is called the direction of polarization of the wave.
Unlike waves on a rope or sound waves, electromagnetic waves do not require a material medium to propagate; they can travel through empty space (vacuum). When an EM wave travels through a vacuum, its speed is constant, c = 3 × 108 m/s.
EM waves obey the classic wave equation:
v = λf
THE ELECTROMAGNETIC SPECTRUM
Electromagnetic waves can be categorized by their frequency (or wavelength); the full range of waves is called the electromagnetic (or EM) spectrum. Types of waves include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and -rays (gamma rays) and, although they’ve been delineated in the spectrum below, there’s no universal agreement on all the boundaries, so many of these bands overlap. You should be familiar with the names of the major categories, and, in particular, memorize the order of the colors within the visible spectrum (which, as you can see, accounts for only a tiny sliver of the full EM spectrum). In order of increasing wave frequency, the colors are red, orange, yellow, green, blue, and violet, which is commonly remembered as ROYGBV (“roy-gee-biv”). The wavelengths of the colors in the visible spectrum are usually expressed in nanometers. For example, electromagnetic waves whose wavelengths are between 577 nm and 597 nm are seen as yellow light.
Example 1 What’s the frequency range for green light?
Solution. According to the spectrum, light is green if its wavelength is between 4.92 × 10–7 m and 5.77 × 10–7 m. Using the equation v = fλ, we find that the upper end of this wavelength range corresponds to a frequency of
while the lower end corresponds to
So the frequency range for green light is
5.20 × 1014 Hz ≤ fgreen ≤ 6.10 × 1014 Hz
Example 2 How would you classify electromagnetic radiation that has a wavelength of 1 cm?
Solution. According to the electromagnetic spectrum presented above, electromagnetic waves with λ = 10–2 m are microwaves.
INTERFERENCE AND DIFFRACTION
As we learned in AP Physics 1, waves experience interference when they meet, and whether they interfere constructively or destructively depends on their relative phase. If they meet in phase (crest meets crest), they combine constructively, but if they meet out of phase (crest meets trough), they combine destructively. The key to the interference patterns we’ll study in the next section rests on this observation. In particular, if waves that have the same wavelength meet, then the difference in the distances they’ve traveled determine whether the waves are in phase. Assuming that the waves are coherent (which means that their phase difference remains constant over time and does not vary), if the difference in their path lengths, ∆ℓ, is a whole number of wavelengths—0, ±λ, ±2λ, etc.—they’ll arrive in phase at the meeting point. On the other hand, if this difference is a whole number plus one-half a wavelength—± λ, ±(1 + )λ, ±(2 + )λ, etc.—then they’ll arrive exactly out of phase. That is,
Young’s Double-Slit Interference Experiment
The following figure shows light incident on a barrier that contains two narrow slits (perpendicular to the plane of the page), separated by a distance d. On the right is a screen whose distance from the barrier, L, is much greater than d. The question is, what will we see on the screen? You might expect that we’ll just see two bright narrow strips of light, directly opposite the slits in the barrier. As reasonable as this may sound, it doesn’t take into account the wave nature of light.
When a wave encounters an aperture whose width is comparable to its wavelength, the wave will fan out after it passes through. The alteration in the straight-line propagation of a wave when it encounters a barrier is called diffraction. In the set-up above, the waves will diffract through the slits, and spread out and interfere as they travel toward the screen.
You can clearly see points of interference in the figure above. Look for solid lines intersecting other solid lines—these are points of constructive interference. Look for solid lines intersecting dashed lines—these are points of destructive interference.
The screen will show the results of this interference: There will be bright bands (bright fringes) centered at those points at which the waves interfere constructively, alternating with dark fringes, where the waves interfere destructively. Let’s determine the locations of these fringes.
In the figure below, we’ve labeled the slits S1 and S2. A point P on the screen is selected, the path lengths from S1 and S2 to P are called ℓ1 and ℓ2, respectively, and the angle that the line from the midpoint of the slits to P makes with the horizontal is θ. Segment S1Q is perpendicular to line S2P. Because L is so much larger than d, the angle that line S2P makes with the horizontal is also approximately θ, which tells us that ∠S2S1Q is approximately θ and that the path difference, ∆ℓ = ℓ2 – ℓ1, is nearly equal to S2Q.
Because S2Q = d sin θ, we get ∆ℓ = d sin θ. Now, using what we learned earlier about how constructive or destructive interference depends on ∆ℓ, we can write:
To locate the positions of, say, the bright fringes on the screen, we use the fact that x = L tan θ. If θ is small, then tan θ ≈ sin θ, so we can write x = L sin θ (we can tell this from the figure). Since sin θ = mλ/d for bright fringes, we get
Also, the intensity of the bright fringes decreases as m increases in magnitude. The bright fringe directly opposite the midpoint of the slits—the central maximum—will have the greatest intensity when m = 0. The bright fringes with m = 1 will have a lower intensity, those with m = 2 will be fainter still, and so on. If more than two slits are cut in the barrier, the interference pattern becomes sharper, and the distinction between dark and bright fringes becomes more pronounced. Barriers containing thousands of tiny slits per centimeter—called diffraction gratings—are used precisely for this purpose.
Example 3 For the experimental set-up we’ve been studying, assume that d = 1.5 mm, L = 6.0 m, and that the light used has a wavelength of 589 nm.
(a) How far above the center of the screen will the second brightest maximum appear?
(b) How far below the center of the screen is the third dark fringe?
(c) What would happen to the interference pattern if the slits were moved closer together?
Solution.
(a) The central maximum corresponds to m = 0 (x0 = 0). The first maximum above the central one is labeled x1 (since m = 1). The other bright fringes on the screen are labeled accordingly.
The value of x1 is
(b) The first dark fringe occurs when the path difference is 0.5λ, the second dark fringe occurs when the path difference is 1.5λ, and the third dark fringe occurs when the path difference is 2.5λ, so
(c) Since xm = mλL/d, a decrease in d would cause an increase in xm. That is, the fringes would become larger; the interference pattern would be more spread out.
Single-Slit Diffraction
A diffraction pattern will also form on the screen if the barrier contains only one slit. The central maximum will be very pronounced, but lower-intensity maxima will also be seen because of interference from waves arriving from different locations within the slit itself. The width of the central maximum will become wider as the width of the slit is decreased.
For a circular pinhole aperture, the diffraction pattern will consist of a central, bright circular disk surrounded by rings of decreasing intensity.
REFLECTION AND REFRACTION
Imagine a beam of light directed toward a smooth transparent surface. When it hits this surface, some of its energy will be reflected off the surface and some will be transmitted into the new medium. We can figure out the directions of the reflected and transmitted beams by calculating the angles that the beams make with the normal to the interface. In the following figure, an incident beam strikes the boundary of another medium; it could be a beam of light in air striking a piece of glass. Notice all angles are measured from the normal.
The angle that the incident beam makes with the normal is called the angle of incidence, or θ1. The angle that the reflected beam makes with the normal is called the angle of reflection, θ′1, and the angle that the transmitted beam makes with the normal is called the angle of refraction, θ2. The incident, reflected, and transmitted beams of light all lie in the same plane.
The relationship between θ1 and θ′1 is pretty easy; it is called the Law of Reflection:
θ1 = θ′1
The Law of Reflection basically states the Angle of Reflection is equal to the Angle of Incidence. In order to describe how θ1 and θ2 are related, we first need to talk about a medium’s index of refraction.
When light travels through empty space (vacuum), its speed is c = 3.00 × 108 m/s; this is one of the fundamental constants of nature. But when light travels through a material medium (such as water or glass), it’s constantly being absorbed and re-emitted by the atoms that compose the material and, as a result, its apparent speed, v, is some fraction of c. The reciprocal of this fraction,
is called the medium’s index of refraction. For example, since the speed of light in water is v = 2.25 × 108 m/s, the index of refraction of water is
Note that n has no units; it’s also never less than 1, because light always travels slower in a medium than in a vacuum.
The equation that relates θ1 and θ2 involves the index of refraction of the incident medium (n1) and the index of refraction of the refracting medium (n2); it’s called Snell’s Law:
n1 sin θ1 = n2 sin θ2
If n2 > n1, then Snell’s Law tells us that θ2 < θ1; that is, the beam will bend (refract) toward the normal as it enters the medium. On the other hand, if n2 < n1, then θ2 > θ1, and the beam will bend away from the normal.
Example 4 A beam of light in air is incident upon a piece of glass, striking the surface at an angle of 30°. If the index of refraction of the glass is 1.5, what are the angles of reflection and refraction?
Solution. If the light beam makes an angle of 30° with the surface, then it makes an angle of 60° with the normal; this is the angle of incidence. By the Law of Reflection, then, the angle of reflection is also 60°. We use Snell’s Law to find the angle of refraction. The index of refraction of air is close to 1, so we can say that n = 1 for air.
n1 sin θ1 = n2 sin θ2
(1) sin 60° = 1.5 sin θ2
sin θ2 = 0.5774
θ2 = 35°
Note that θ2 < θ1, as we would expect since the refracting medium (glass) has a greater index than does the incident medium (air).
Example 5 A fisherman drops a flashlight into a lake that’s 10 m deep. The flashlight sinks to the bottom where its emerging light beam is directed almost vertically upward toward the surface of the lake, making a small angle (θ1) with the normal. How deep will the flashlight appear to be to the fisherman? (Use the fact that tan θ is almost equal to sin θ if θ is small.)
Solution. Take a look at the figure below.
Since the refracting medium (air) has a lower index than the incident medium (water), the beam of light will bend away from the normal as it emerges from the water. As a result, the fisherman will think that the flashlight is at a depth of only y1, rather than its actual depth of y2 = 10 m. By simple trigonometry, we know that
So,
Snell’s Law tells us that n1 sin θ1 = sin θ2 (since n2 = nair = 1), so
Using the approximations sin θ1 ≈ tan θ1 and sin θ2 ≈ tan θ2, we can write
which means
Because y2 = 10 m and n1 = nwater = 1.33,
DISPERSION OF LIGHT
One thing we learned when we studied waves is that wave speed is independent of frequency. For a given medium, different frequencies give rise to different wavelengths, because the equation v = f λ must always be satisfied and v doesn’t vary. But when light travels through a material medium, it displays dispersion, which is a variation in wave speed with frequency (or wavelength). So, the definition of the index of refraction, n = c/v, should be accompanied by a statement of the frequency of the light used to measure v, since different frequencies have different speeds and different indices. A piece of glass may have the following indices for visible light:
Note that as the wavelength decreases, refractive index increases. In general, higher frequency waves have higher indices of refraction. Most lists of refractive index values are tabulated using yellow light of wavelength 589 nm (frequency 5.1 × 1014 Hz).
Although the variation in the values of the refractive index across the visible spectrum is pretty small, when white light (which is a combination of all the colors of the visible spectrum) hits a glass prism, the beam is split into its component colors.
Why? Because each color has its own index, Snell’s Law tells us that each color will have its own angle of refraction. Therefore, each color emerges from the prism at a slightly different angle, so the light disperses into its component colors.
Example 6 In the figure above, assume that the incident beam strikes the prism with an angle of incidence of 60.00°. If the top angle of the prism has a measure of 60.00°, what are the angles of refraction of red light and violet light when the beam emerges from the other side? (Use n = 1.502 for red light and 1.530 for violet light.)
Solution. The beam refracts twice, once when it enters the prism and again when it exits.
Snell’s Law tells us that
sin 60.00° = n sin θ2 and n sin θ3 = sin θ4
where n is the index of the glass. We’ll ignore the minuscule variation in n for light in air, and just use nair = 1 for all colors. The relationship between θ2 and θ3 follows from the fact that the sum of the measures of the angles in any triangle is 180°, so
(90° – θ2) + 60° + (90° – θ3) = 180°
which implies that θ3 = 60° – θ2. We can now make the following table of values:
In general, the differences in n are so small that, unless you are specifically addressing dispersion, n will be given as a single number, regardless of color.
TOTAL INTERNAL REFLECTION
When a beam of light strikes the boundary to a medium that has a lower index of refraction, the beam bends away from the normal. As the angle of incidence increases, the angle of refraction becomes larger. At some point, when the angle of incidence reaches a critical angle, θc, the angle of refraction becomes 90°, which means the refracted beam is directed along the surface.
For angles of incidence that are greater than θc, there is no angle of refraction; the entire beam is reflected back into the original medium. This phenomenon is called total internal reflection (sometimes abbreviated TIR).
Total internal reflection occurs when:
1) n1 > n2
and
2) θ1 > θc, where θc = sin–1 (n2/n1)
Notice that total internal reflection cannot occur if n1 < n2. This is because the largest output of sin θ is 1, so the largest input of sin–1 (x) is 1. If n1 > n2, then total internal reflection is a possibility; it will occur if the angle of incidence is large enough, that is, if it’s greater than the critical angle, θc.
Example 7 What is the critical angle for total internal reflection between air and water? In which of these media must light be incident for total internal reflection to occur?
Solution. First, total internal reflection can only occur when the light is incident in the medium that has the greater refractive index and strikes the boundary to a medium that has a lower index. So, in this case, total internal reflection can occur only when the light source is in the water and the light is incident upon the water/air surface. The critical angle is found as follows:
Total internal reflection will occur if the light from the water strikes the water/air boundary at an angle of incidence greater than 49°.
Example 8 How close must the fisherman be to the flashlight in Example 5 in order to see the light it emits?
Solution. In order for the fisherman to see the light, the light must be transmitted into the air from the water; that is, it cannot undergo total internal reflection. The figure below shows that, within a circle of radius x, light from the flashlight will emerge from the water. Outside this circle, the angle of incidence is greater than the critical angle, and the light would be reflected back into the water, rendering it undetectable by the fisherman above.
Because the critical angle for total internal reflection at a water/air interface is 49° (as we found in the preceding example), we can solve for x:
tan 49° = ⇒ x = ytan 49° = (10 m)tan 49° = 11.5 m
Example 9 The refractive index for the glass prism shown below is 1.55. In order for a beam of light to experience total internal reflection at the right-hand face of the prism, the angle θ1 must be smaller than what value?
Solution. Total internal reflection will occur at the glass/air boundary if θ3 is greater than the critical angle, θc, which we can calculate this way:
Because θ3 = 60° – θ2, total internal reflection will take place if θ2 is smaller than 20°. Now, by Snell’s Law, θ2 = 20° if
nair sin θ1 = nglass sin θ2
sin θ1 = 1.55 sin 20°
θ1 = 32°
Therefore, total internal reflection will occur at the right-hand face of the prism if θ1 is smaller than 32°.
MIRRORS
A mirror is an optical device that forms an image by reflecting light. We’ve all looked into a mirror and seen images of nearby objects, and the purpose of this section is to analyze these images mathematically. We begin with a plane mirror, which is flat, and is the simplest type of mirror. Then we’ll examine curved mirrors; we’ll have to use geometrical methods or algebraic equations to analyze the patterns of reflection from these.
Plane Mirrors
The figure below shows an object (denoted by a vertical, bold arrow) in front of a flat mirror. Light that’s reflected off of the object strikes the mirror and is reflected back to our eyes. The directions of the rays reflected off the mirror determine where we perceive the image to be.
There are four questions we’ll answer about the image formed by a mirror:
(1) Where is the image?
(2) Is the image real or is it virtual?
(3) Is the image upright or is it inverted?
(4) What is the height of the image (compared with that of the object)?
When we look at ourselves in a mirror, it seems like our image is behind the mirror, and, if we take a step back, our image also takes a step back. The Law of Reflection can be used to show that the image seems as far behind the mirror as the object is in front of the mirror. This answers question (1).
An image is said to be real if light rays actually focus at the image. A real image can be projected onto a screen. For a flat mirror, light rays bounce off the front of the mirror; so, of course, no light focuses behind it. Therefore, the images produced by a flat mirror are not real; they are virtual. This answers question (2).
When we look into a flat mirror, our image isn’t upside down; flat mirrors produce upright images, and question (3) is answered.
Finally, the image formed by a flat mirror is neither magnified nor diminished (minified) relative to the size of the object. This answers question (4).
Spherical Mirrors
A spherical mirror is a mirror that’s curved in such a way that its surface forms part of a sphere.
The center of this imaginary sphere is the mirror’s center of curvature, and the radius of the sphere is called the mirror’s radius of curvature, R. Halfway between the mirror and the center of curvature, C, is the focus (or focal point), F. The intersection of the mirror’s optic axis (its axis of symmetry) with the mirror itself is called the vertex, V, and the distance from V to F is called the focal length, f, equal to one-half of the radius of curvature:
If the mirror had a parabolic cross-section, then any ray parallel to the axis would be reflected by the mirror through the focal point. Spherical mirrors do this for incident light rays near the axis (paraxial rays) because, in the region of the mirror that’s close to the axis, the shapes of a parabolic mirror and a spherical mirror are nearly identical.
The previous two figures illustrate a concave mirror, a mirror whose reflective side is caved in toward the center of curvature. The following figure illustrates the convex mirror, which has a reflective side curving away from the center of curvature. We will call F the virtual focus in this case.
RAY TRACING FOR MIRRORS
One method of answering the four questions listed above concerning the image formed by a mirror involves a geometric approach called ray tracing. Representative rays of light are sketched in a diagram that depicts the object and the mirror; the point at which the reflected rays intersect (or appear to intersect) is the location of the image. Some rules governing rays are:
Concave mirrors:
• An incident ray parallel to the axis is reflected through the focus.
• An incident ray that passes through the focus is reflected parallel to the axis.
• An incident ray that strikes the vertex is reflected at an equal angle to the axis.
Convex mirrors:
• An incident ray parallel to the axis is reflected away from the virtual focus.
• An incident ray directed toward the virtual focus is reflected parallel to the axis.
• An incident ray that strikes the vertex is reflected at an equal angle to the axis.
Example 10 The figure below shows a concave mirror and an object (the bold arrow). Use a ray diagram to locate the image of the object.
Solution. It only takes two distinct rays to locate the image:
Example 11 The figure below shows a convex mirror and an object (the arrow). Use a ray diagram to locate the image of the object.
Solution.
The ray diagrams of the preceding examples can be used to determine the location, orientation, and size of the image. The nature of the image—that is, whether it’s real or virtual—can be determined by seeing on which side of the mirror the image is formed. If the image is formed on the same side of the mirror as the object, then the image is real, but if the image is formed on the opposite side of the mirror, it’s virtual. Another way of looking at this is: if you had to trace lines back to form an image, that image is virtual. Therefore, the image in Example 10 is real, and the image in Example 11 is virtual.
Using Equations to Answer Questions About the Image
While ray diagrams can answer our questions about images completely, the fastest and easiest way to get information about an image is to use two equations and some simple conventions. The first equation, called the mirror equation, is
where so is the object’s distance from the mirror, si is the image’s distance from the mirror, and f is the focal length of the mirror. The value of so is always positive for a real object, but si can be positive or negative. The sign of si tells us whether the image is real or virtual: If si is positive, the image is real; and if si is negative, the image is virtual.
The second equation is called the magnification equation:
This gives the magnification; the height of the image, hi, is|M| times the height of the object, ho. If M is positive, then the image is upright relative to the object; if M is negative, it’s inverted relative to the object. Because so is always positive, we can come to two important conclusions. If si is positive, then M is negative, so real images are always inverted, and, if si is negative, then M is positive, so virtual images are always upright.
Finally, to distinguish mathematically between concave and convex mirrors, we always write the focal length f as a positive value for concave mirrors and a negative value for convex mirrors. With these two equations and their accompanying conventions, all four questions about an image can be answered.
Example 12 An object of height 4 cm is placed 30 cm in front of a concave mirror whose focal length is 10 cm.
(a) Where’s the image?
(b) Is it real or virtual?
(c) Is it upright or inverted?
(d) What’s the height of the image?
Solution.
(a) With so = 30 cm and f = 10 cm, the mirror equation gives
The image is located 15 cm in front of the mirror.
(b) Because si is positive, the image is real.
(c) Real images are inverted.
(d)
The –2 cm also confirms the image is inverted.
Example 13 An object of height 4 cm is placed 20 cm in front of a convex mirror whose focal length is –30 cm.
(a) Where’s the image?
(b) Is it real or virtual?
(c) Is it upright or inverted?
(d) What’s the height of the image?
Solution.
(a) With so = 20 cm and f = –30 cm, the mirror equation gives us:
So, the image is located 12 cm behind the mirror.
(b) Because si is negative, the image is virtual.
(c) Virtual images are upright.
(d)
[The fact that the magnification is positive tells us that the image is upright, as we said in part (c).]
Example 14 Show how the statements made earlier about plane mirrors can be derived from the mirror and magnification equations.
Solution. A plane mirror can be considered a spherical mirror with an infinite radius of curvature (and an infinite focal length). If f = ∞, then 1/f = 0, so the mirror equation becomes
So, the image is as far behind the mirror as the object is in front. Also, since so is always positive, si is negative, so the image is virtual. The magnification is
and the image is upright and has the same height as the object. The mirror and magnification equations confirm our description of images formed by plane mirrors.
Example 15 Show why convex mirrors can only form virtual images.
Solution. Because f is negative and so is positive, the mirror equation
immediately tells us that si cannot be positive (if it were, the left-hand side would be the sum of two positive numbers, while the right-hand side would be negative). Since si must be negative, the image must be virtual.
Example 16 An object placed 60 cm in front of a spherical mirror forms a real image at a distance of 30 cm from the mirror.
(a) Is the mirror concave or convex?
(b) What’s the mirror’s focal length?
(c) Is the image taller or shorter than the object?
Solution.
(a) The fact that the image is real tells us that the mirror cannot be convex, since convex mirrors form only virtual images. The mirror is concave.
(b) With so = 60 cm and si = 30 cm (si is positive since the image is real), the mirror equation tells us that
Note that f is positive, which confirms that the mirror is concave.
(c) The magnification is
Since the absolute value of M is less than 1, the mirror makes the object look smaller (minifies the height of the object). The image is only half as tall as the object (and is upside down, because M is negative).
Example 17 A concave mirror with a focal length of 25 cm is used to create a real image that has twice the height of the object. How far is the image from the mirror?
Solution. Since hi (the height of the image) is twice ho (the height of the object) the value of the magnification is either +2 or –2. To figure out which, we just notice that the image is real; real images are inverted, so the magnification, M, must be negative. Therefore, M = –2, so
Substituting this into the mirror equation gives us
THIN LENSES
A lens is an optical device that forms an image by refracting light. We’ll now talk about the equations and conventions that are used to analyze images formed by the two major categories of lenses: converging and diverging.
A converging lens—like the bi-convex one shown below—converges parallel paraxial rays of light to a focal point on the far side. (This lens is bi-convex; both of its faces are convex. Converging lenses all have at least one convex face.) Because parallel light rays actually focus at F, this point is called a real focus. Its distance from the lens is the focal length, f.
A diverging lens—like the bi-concave one shown below—causes parallel paraxial rays of light to diverge away from a virtual focus, F, on the same side as the incident rays. (Diverging lenses all have at least one concave face.)
RAY TRACING FOR LENSES
Just as is the case with mirrors, representative rays of light can be sketched in a diagram along with the object and the lens; the point at which the reflected rays intersect (or appear to intersect) is the location of the image. The rules that govern these rays are as follows:
Converging lenses:
• An incident ray parallel to the axis is refracted through the real focus.
• Incident rays pass undeflected through the optical center, O (the central point within the lens where the axis intersects the lens).
Diverging lenses:
• An incident ray parallel to the axis is refracted away from the virtual focus.
• Incident rays pass undeflected through the optical center, O.
Example 18 The figure below shows a converging lens and an object (denoted by the bold arrow). Use a ray diagram to locate the image of the object.
Solution.
Example 19 The figure below shows a diverging lens and an object. Use a ray diagram to locate the image of the object.
Solution.
The nature of the image—that is, whether it’s real or virtual—is determined by the side of the lens upon which the image is formed. If the image is formed on the side of the lens that’s opposite the object, then the image is real, and if the image is formed on the same side of the lens as the object, then it’s virtual. Another way of looking at this is: If you had to trace lines back to form an image, that image is virtual. Therefore, the image in Example 18 is real, while the image in Example 19 is virtual.
USING EQUATIONS TO ANSWER QUESTIONS ABOUT THE IMAGE
Lenses and mirrors use the same equations, notation, and sign conventions, with the following note. Converging optical devices (+ f) are concave mirrors and convex lenses. Diverging optical devices (– f) are convex mirrors and concave lenses.
Example 20 An object of height 11 cm is placed 44 cm in front of a converging lens with a focal length of 24 cm.
(a) Where’s the image?
(b) Is it real or virtual?
(c) Is it upright or inverted?
(d) What’s the height of the image?
Solution.
(a) With so = 44 cm and f = 24 cm, the lens equation gives us:
So, the image is located 53 cm from the lens, on the opposite side from the object.
(b) Because si is positive, the image is real.
(c) Real images are inverted.
(d)
The negative hi reaffirms that we have an inverted image.
Example 21 An object of height 11 cm is placed 48 cm in front of a diverging lens with a focal length of ‒24.5 cm.
(a) Where’s the image?
(b) Is it real or virtual?
(c) Is it upright or inverted?
(d) What’s the height of the image?
Solution.
(a) With so = 48 cm and f = –24.5 cm, the lens equation gives us:
The image is 16 cm from the lens, on the same side as the object.
(b) Because si is negative, the image is virtual.
(c) Virtual images are upright.
(d)
These results are illustrated in Example 19.
Example 22 The power of a lens, P, is defined as 1/f, where f is the lens’s focal length (in meters). Lens power is expressed in diopters (abbreviated D), where 1 D = 1 m–1. When an object is placed 10 cm from a converging lens, the real image formed is twice the height of the object. What’s the power of this lens?
Solution. Since hi, the height of the image, is twice ho, the height of the object, the value of the magnification is either +2 or –2. To decide, we simply notice that the image is real; real images are inverted, so the magnification, m, must be negative. Since m = –2, we find that
Because so = 10 cm, it must be true that si = 20 cm, so the lens equation gives us:
Expressing the focal length in meters and then taking the reciprocal gives the lens power in diopters:
Chapter 10 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 10 Review Questions.
Section I: Multiple Choice
1. What is the wavelength of an X-ray whose frequency is 1.0 × 1018 Hz ?
(A) 3.3 × 10–11 m
(B) 3.0 × 10–10 m
(C) 3.3 × 10–9 m
(D) 3.0 × 10–8 m
2. In Young’s double-slit interference experiment, what is the difference in path length of the light waves from the two slits at the center of the first bright fringe above the central maximum?
(A) λ
(B) λ
(C) λ
(D) λ
3. A beam of light in air is incident upon the smooth surface of a piece of flint glass, as shown:
If the reflected beam and refracted beam are perpendicular to each other, what is the index of refraction of the glass?
(A)
(B)
(C)
(D) 2
4. When green light (wavelength = 500 nm in air) travels through diamond (refractive index = 2.5), what is its wavelength?
(A) 200 nm
(B) 300 nm
(C) 500 nm
(D) 1250 nm
5. A beam of light traveling in Medium 1 strikes the interface to another transparent medium, Medium 2. If the speed of light is less in Medium 2 than in Medium 1, the beam will
(A) refract toward the normal
(B) refract away from the normal
(C) undergo total internal reflection
(D) have an angle of reflection smaller than the angle of incidence
6. If a clear liquid has a refractive index of 1.45 and a transparent solid has an index of 2.90 then, for total internal reflection to occur at the interface between these two media, which of the following must be true?
incident beam originates in | at an angle of incidence greater than | |
(A) | The solid | 30° |
(B) | The liquid | 30° |
(C) | The liquid | 60° |
(D) Total internal reflection cannot occur. |
7. An object is placed 60 cm in front of a concave spherical mirror whose focal length is 40 cm. Which of the following best describes the image?
Nature of image | Distance from mirror | |
(A) | Virtual | 24 cm |
(B) | Real | 24 cm |
(C) | Virtual | 120 cm |
(D) | Real | 120 cm |
8. An object is placed 60 cm from a spherical convex mirror. If the mirror forms a virtual image 20 cm from the mirror, what’s the magnitude of the mirror’s radius of curvature?
(A) 7.5 cm
(B) 15 cm
(C) 30 cm
(D) 60 cm
9. The image created by a converging lens is projected onto a screen that’s 60 cm from the lens. If the height of the image is 1/4 the height of the object, what’s the focal length of the lens?
(A) 36 cm
(B) 45 cm
(C) 48 cm
(D) 72 cm
10. Which of the following is true concerning a bi-concave lens? (A bi-concave lens has both surfaces concave.)
(A) Its focal length is positive.
(B) It cannot form real images.
(C) It cannot form virtual images.
(D) It can magnify objects.
Section II: Free Response
1. Two trials of a double-slit interference experiment are set up as follows. The slit separation is d = 0.50 mm, and the distance to the screen, L, is 4.0 m.
(a) What is the purpose of the first (single-slit) barrier? Why not use two light sources, one at each slit at the second barrier? Explain briefly.
In the first trial, white light is used.
(b) What is the vertical separation on the screen (in mm) between the first-order maxima for red light (λ = 750 nm) and violet light (λ = 400 nm)?
(c) Locate the nearest point to the central maximum where an intensity maximum for violet light (λ = 400 nm) coincides with an intensity maximum for orange-yellow light (λ = 600 nm).
In the second trial, the entire region between the double-slit barrier and the screen is filled with a large slab of glass of refractive index n = 1.5, and monochromatic green light (λ = 500 nm in air) is used.
(d) What is the separation between adjacent bright fringes on the screen?
2. In order to determine the criteria for constructive and destructive interference, the following rules are used:
i) When light strikes the boundary to a medium with a higher refractive index than the incident medium, it undergoes a 180° phase change upon reflection (this is equivalent to a shift by one-half wavelength).
ii) When light strikes the boundary to a medium with a lower refractive index than the incident medium, it undergoes no phase change upon reflection.
These rules can be applied to the two situations described below.
A thin soap film of thickness T, consisting of a mixture of water and soap (refractive index = 1.38), has air on both sides. Incident sunlight is reflected off the front face and the back face, causing interference.
(a) Which beam, 1 or 2, suffers a 180° phase change upon reflection?
(b) Since the beams are out of phase, destructive interference will occur if the difference in their path lengths, ∆ℓ ≈ 2T for near-normal incidence, is equal to a whole number of wavelengths (wavelength as measured in the soap film). What is the criterion for constructive interference? Write your answer as an algebraic equation.
3. An object of height 5 cm is placed 40 cm in front of a spherical concave mirror. An image is formed 72 cm behind the mirror.
(a) Is the image real or virtual?
(b) Is the image upright or inverted?
(c) What’s the height of the image?
(d) What is the mirror’s radius of curvature?
(e) In the figure below, sketch the mirror, labeling its vertex and focal point, and then construct a ray diagram (with a minimum of two rays) showing the formation of the image.
Summary
All electromagnetic waves travel at the speed of light (c = 3 × 108 m/s). They obey the wave equation. When monochromatic coherent light goes through double slits (or a diffraction grating) there will be constructive interference when the path difference is given by PD = mλ or d sin θ = mλ where m = 0, 1, 2….There will be destructive interference for PD = (m + )λ or d sin θ = (m + )λ. For small angles we can say xm = , where xm is the distance the mth maximum is located from the central maximum and shows where constructive interference occurs.
Geometric optics always measures angles from the normal line. Topics include the fact that the angle of incidence is equal to the angle of reflection and that these angles lie in the same plane.
Snell’s Law states when light enters a new medium (material) it changes speeds and may change directions. This is stated in the formula
n1 sin θ1 = n2 sin θ2
where n is the index of refraction, which is a ratio of the speed of light in a vacuum to the speed of light in the substance (n = ). The index of refraction is always greater than 1 and has no units. When going from a high index of refraction to a lower index of refraction, light may have total internal reflection if the angle is larger than the critical angle (sin θc, = ).
For both thin lenses and curved mirrors you can use the thin lens equation:
Where f is positive for a convex lens or a concave mirror and f is negative for a concave lens or a convex mirror s0 is always positive.
For a lens, if si is negative the lens image is virtual and located on the same side of the lens as the object. If si is positive, the lens image is real and located on the opposite side of the lens from the object.
For a mirror, if si is negative, the image is virtual and located on the opposite side of the mirror from the object. If si is positive, the image is real and located on the same side of the mirror as the object.
The magnification of a lens is given by .
A diverging optical device could be either a convex mirror or a concave lens. The thing these two have in common is that no matter where the object is located they form images that are virtual and upright. The sign of hi will always be positive and the sign of si will always be negative. The sign of the magnification will be positive and its value will be less than one. The image will always be located between the focal length and the optical device.
A converging optical device could be either a concave mirror or a convex lens. The thing these two have in common is that when so is outside the focal length they form images that are real and inverted. The sign of hi will be negative and the sign of si will be positive. The sign of the magnification will be negative and its absolute value can be any number greater than zero. The other thing they have in common is that when so is inside the focal length they form images that are virtual and upright. The sign of hi will be positive and the sign of si will be negative. The sign of the magnification will be positive and its value will be greater than one.
Chapter 11
Modern Physics
INTRODUCTION
The subject matter of the previous chapters was developed in the seventeenth, eighteenth, and nineteenth centuries, but as we delve into the physics of the very small, we enter the twentieth century.
PHOTONS AND THE PHOTOELECTRIC EFFECT
Max Planck first proposed the idea of light being emitted as individual packets of constant energy, called quanta. This is where the name quantum mechanics comes from. The particle nature of light was pioneered by Einstein in 1905, when he showed light transferred energy like a particle, and Arthur Compton in 1923, when he showed light has momentum and can undergo elastic collisions. A quantum of electromagnetic energy is known as a photon. Light behaves like a stream of photons, and this is illustrated by the photoelectric effect.
When a piece of metal is illuminated by electromagnetic radiation (specifically visible light, ultraviolet light, or X-rays), the energy absorbed by electrons near the surface of the metal can liberate them from their bound state, and these electrons can fly off. The released electrons are known as photoelectrons. In this case, the classical, wave-only theory of light would predict three results:
(1) There would be a significant time delay between the moment of illumination and the ejection of photoelectrons. This is because the metal would have to heat up to the point that the thermal energy was enough to overcome the binding energy.
(2) Increasing the intensity of the light would cause the electrons to leave the metal surface with greater kinetic energy. This is because the energy carried by a wave is related to its intensity.
(3) Photoelectrons would be emitted regardless of the frequency of the incident energy, as long as the intensity was high enough.
Surprisingly, none of these predictions was observed. Photoelectrons were ejected within just a few billionths of a second after illumination, disproving prediction (1). Secondly, increasing the intensity of the light did not cause photoelectrons to leave the metal surface with greater kinetic energy. Although more electrons were ejected as the intensity was increased, there was a maximum photoelectron kinetic energy; prediction (2) was false. And, for each metal, there was a certain threshold frequency, f0: If light of frequency lower than f0 were used to illuminate the metal surface, no photoelectrons were ejected, regardless of how intense the incident radiation was; prediction (3) was also false. Clearly, something was wrong with the wave-only theory of light.
Einstein explained these observations by borrowing from Planck’s idea that light came in individual quanta of energy. He called the quanta photons. The energy of a photon is proportional to the frequency of the wave,
E = hf
where h is Planck’s constant (about 6.63 × 10−34 J∙s). A certain amount of energy had to be imparted to an electron on the metal surface in order to liberate it; this was known as the metal’s work function, or φ. If an electron absorbed a photon whose energy E was greater than φ, it would leave the metal with a maximum kinetic energy equal to E – φ. This process could occur very quickly, which accounts for the rapidity with which photoelectrons are produced after illumination. This explains why prediction (1) was incorrect.
In this view, increasing the intensity (and therefore the energy) just means increasing the number of photons and results in the ejection of more photoelectrons—but since the energy of each incident photon is fixed by the equation E = hf, the value of Kmax will still be E – φ. This can be expressed as
Kmax = hf – φ
This accounts for the observation that disproved prediction (2).
Finally, if the incoming photon’s energy was less than phi (or, E = hf < φ), the photon energy would not be enough to liberate electrons. Blasting the metal surface with more photons (that is, increasing the intensity of the incident beam) would also do nothing; none of the photons would have enough energy to eject electrons. This accounts for the observation of a threshold frequency, which we now know is φ/h. This can be expressed as
fo = φ/h
and is why prediction (3) was incorrect. Before we get to some examples, it’s worthwhile to introduce a new unit for energy. The SI unit for energy is the joule, but it’s too large to be convenient in the domains we’re studying now. We’ll use a much smaller unit, the electronvolt (abbreviated eV). The eV is equal to the energy gained (or lost) by an electron accelerated through a potential difference of one volt. Using the equation ∆UE = qV, we find that
1 eV = (1 e)(1 V) = (1.6 × 10–19 C)(1 V) = 1.6 × 10−19 J
In terms of electronvolts, the value of Planck’s constant is 4.14 × 10–15 eV∙s.
Example 1 The work function, φ, for aluminum is 4.08 eV.
(a) What is the threshold frequency required to produce photoelectrons from aluminum?
(b) Classify the electromagnetic radiation that can produce photoelectrons.
(c) If light of frequency f = 4.00 × 1015 Hz is used to illuminate a piece of aluminum,
(i) what is Kmax, the maximum kinetic energy of ejected photoelectrons?
(ii) what’s the maximum speed of the photoelectrons? (Electron mass = 9.11 × 10–31 kg.)
(d) If the light described in part (b) were increased by a factor of 2 in intensity, what would happen to the value of Kmax?
Solution.
(a) We know from the statement of the question that, in order for a photon to be successful in liberating an electron from the surface of the aluminum, its energy cannot be less than 4.08 eV. Therefore, the minimum frequency of the incident light—the threshold frequency—must be
(b) Based on the electromagnetic spectrum given in the previous chapter, the electromagnetic radiation used to produce photoelectrons from aluminum must be at least in the ultraviolet region of the EM spectrum.
(c) (i) The maximum kinetic energy of photoelectrons is found from the equation
(ii) Using the above result and K = mv2, we can find vmax:
(d) Nothing will happen. Increasing the intensity of the illuminating radiation will cause more photons to impinge on the metal surface, thereby ejecting more photoelectrons, but their maximum kinetic energy will remain the same. The only way to increase Kmax would be to increase the frequency of the incident energy.
THE BOHR MODEL OF THE ATOM
In the years immediately following Rutherford’s announcement of his nuclear model of the atom, a young physicist, Niels Bohr, would add an important piece to the atomic puzzle. Rutherford told us where the positive charge of the atom was located; Bohr would tell us about the electrons.
For fifty years it had been known that atoms in a gas discharge tube emit and absorb light only at specific wavelengths. The light from a glowing gas, passed through a prism to disperse the beam into its component wavelengths, produces patterns of sharp lines called atomic spectra. The visible wavelengths that appear in the emission spectrum of hydrogen had been summarized by the Balmer formula
where R is the Rydberg constant (about 1.1 × 107 m–1). The formula worked—that is, it fit the observational data—but it had no theoretical basis. So, the question was, why do atoms emit (or absorb) radiation only at certain discrete wavelengths?
Bohr’s model of the atom explains the spectroscopists’ observations. Using the simplest atom, hydrogen (with only one electron), Bohr postulated that the electron would orbit the nucleus only at certain discrete radii. When the electron is in one of these special orbits, it does not radiate away energy (as the classical theory would predict). However, if the electron absorbs a certain amount of energy, it is excited to a higher orbit, one with a greater radius. After a short time in this excited state, it returns to a lower orbit, emitting a photon in the process. Since each allowed orbit—or energy level—has a specific radius (and corresponding energy), the photons emitted in each jump have only specific wavelengths.
When an excited electron drops from energy level n = j to a lower one, n = i, the transition causes a photon of energy to be emitted, and the energy of the photon is the difference between the two energy levels:
Eemitted photon = |ΔE| = Ej − Ei
The wavelength of this photon is
Example 2
The first five energy levels of an atom are shown in the diagram below:
(a) If the atom begins in the n = 3 level, what photon energies could be emitted as it returns to its ground state?
(b) What could happen if this atom, while in an undetermined energy state, were bombarded with a photon of energy 10 eV?
Solution.
(a) If the atom is in the n = 3 state, it could return to ground state by a transition from 3 → 1, or from 3 → 2 and then 2 → 1. The energy emitted in each of these transitions is simply the difference between the energies of the corresponding levels:
(b) Since no two energy levels in this atom are separated by 10 eV, the atom could not absorb a 10 eV photon, and as a result, nothing would happen. This atom would be transparent to light of energy 10 eV.
Example 3
(a) Using the energy level diagram for hydrogen below, how much energy must a ground-state electron in a hydrogen atom absorb to be excited to the n = 4 energy level?
(b) With the electron in the n = 4 level, what wavelengths are possible for the photon emitted when the electron drops to a lower energy level? In what regions of the EM spectrum do these photons lie?
Solution.
(a) The ground-state energy level (n = 1) for hydrogen is –13.6 eV, and E4 = −0.85 eV.
Therefore, in order for an electron to make the transition from E1 to E4, it must absorb energy in the amount E4 – E1 = (–0.85 eV) – (–13.6 eV) = 12.8 eV.
(b) An electron in the n = 4 energy level can make several different transitions: It can drop to n = 3, n = 2, or all the way down to the ground state, n = 1. The energies of the n = 2 and n = 3 levels are
The following diagram shows the electron dropping from n = 4 to n = 3:
There are three possible values for the energy of the emitted photon, E4→3, E4→ 2, or E4→1:
From the equation E = hf = hc/λ, we get λ = hc/E, so
Note that λ4→2 is in the visible spectrum; this wavelength corresponds to the color blue-green; λ4→3 is an ultraviolet wavelength, and λ4→3 is infrared.
WAVE-PARTICLE DUALITY
Light and other electromagnetic waves exhibit wave-like characteristics through interference and diffraction. However, as we saw in the photoelectric effect, light also behaves as if its energy were granular, composed of particles. This is wave-particle duality: Electromagnetic radiation propagates like a wave but exchanges energy like a particle.
Since an electromagnetic wave can behave like a particle, can a particle of matter behave like a wave? In 1924, the French physicist Louis de Broglie proposed that the answer is “yes.” His conjecture, which has since been supported by experiment, is that a particle of mass m and speed v—and thus, linear momentum p—has an associated wavelength, which is called its de Broglie wavelength:
λ =
Particles in motion can display wave characteristics and behave as if they had a wavelength.
Since the value of h is so small, ordinary macroscopic objects do not display wave-like behavior. For example, a baseball (mass = 0.15 kg) thrown at a speed of 40 m/s has a de Broglie wavelength of
This is much too small to measure. However, with subatomic particles, the wave nature is clearly evident. The 1937 Nobel prize was awarded for experiments that revealed that a stream of electrons exhibited diffraction patterns when scattered by crystals—a behavior that’s characteristic of waves.
There’s an interesting connection between the de Broglie wavelength for electrons and Bohr’s model of quantized orbital radii. Bohr postulated that the electron’s orbital angular momentum, mevnrn, had to be an integral multiple of h = . The equation
mevnrn = n
can be rewritten as follows:
In this last form, the equation says that the circumference of the electron’s orbit must be equal to a whole number of wavelengths in order for it to be stable, a restriction that should remind you of sustained standing waves.
Example 4 Electrons in a diffraction experiment are accelerated through a potential difference of 200 V. What is the de Broglie wavelength of these electrons?
Solution. By definition, the kinetic energy of these electrons is 200 eV. Since the relationship between linear momentum and kinetic energy is p = ,
This wavelength is characteristic of X-rays.
RELATIVITY
Another important idea in modern physics is the theory of relativity. Relativity has only two postulates:
1. The results of physical experiments will be the same in any non-accelerating reference frames.
2. The speed of light is constant.
For the AP Exam, you are not required to know any of the mathematical framework surrounding relativity, but only have a qualitative understanding of the consequences of the these postulates.
The major consequence of relativity is that time is not absolute, but that space and time are linked together. To see why this must be the case, imagine a car driving down the highway at 65 mph. In non-relativistic physics, the light from the headlight would be moving at the speed of light plus 65 mph. However, the second postulate of relativity states that this cannot be true, and the light from a headlight of a moving car must also be exactly the speed of light.
As the name indicates, relativity describes the consequences of performing experiments in different reference frames and measuring the same results.
When a clock placed on a fast-moving airplane is compared to a clock at rest on the ground, the clock in the airplane shows that less time has passed than the time recorded by the clock on the ground. This phenomenon is known as time dilation, and it has been demonstrated experimentally using synchronized atomic clocks.
The speed of light is constant according to the theory of relativity. However, time dilation shows there are disagreements about the amount of time that passes in different reference frames. To be consistent with time dilation, there must also be disagreement about distances. This is known as length contraction. A scientist moving along with the object he measures will observe the object to have a different length than a scientist in a different reference frame.
NUCLEAR PHYSICS
The nucleus of the atom is composed of particles called protons and neutrons, which are collectively called nucleons. The number of protons in a given nucleus is called the atom’s atomic number, and is denoted Z, and the number of neutrons (the neutron number) is denoted N. The total number of nucleons, Z + N, is called the mass number (or nucleon number), and is denoted A. The number of protons in the nucleus of an atom defines the element. For example, the element chlorine (abbreviated Cl) is characterized by the fact that the nucleus of every chlorine atom contains 17 protons, so the atomic number of chlorine is 17; but, different chlorine atoms may contain different numbers of neutrons. In fact, about three-fourths of all naturally occurring chlorine atoms have 18 neutrons in their nuclei (mass number = 35), and most of the remaining one-fourth contain 20 neutrons (mass number = 37).
Nuclei that contain the same numbers of protons but different numbers of neutrons are called isotopes. The notation for a nuclide—the term for a nucleus with specific numbers of protons and neutrons—is to write Z and A before the chemical symbol of the element:
The isotopes of chlorine mentioned above would be written as follows:
Cl and Cl
Example 5 How many protons and neutrons are contained in the nuclide Cu?
Solution. The subscript (the atomic number, Z) gives the number of protons, which is 29. The superscript (the mass number, A) gives the total number of nucleons. Since A = 63 = Z + N , we find that N = 63 – 29 = 34.
Example 6 The element neon (abbreviated Ne, atomic number 10) has several isotopes. The most abundant isotope contains 10 neutrons, and two others contain 11 and 12. Write symbols for these three nuclides.
Solution. The mass numbers of these isotopes are 10 + 10 = 20, 10 + 11 = 21, and 10 + 12 = 22. So, we’d write them as follows:
Another common notation—which we also use—is to write the mass number after the name of the element. These three isotopes of neon would be written as neon-20, neon-21, and neon-22.
The Nuclear Force
Why wouldn’t any nucleus that has more than one proton be unstable? After all, protons are positively charged and would therefore experience a repulsive Coulomb force from each other. Why don’t these nuclei explode? And what holds neutrons—which have no electric charge—in the nucleus? These issues are resolved by the presence of another fundamental force, the strong nuclear force, which binds neutrons and protons together to form nuclei. Although the strength of the Coulomb force can be expressed by a simple mathematical formula (it’s inversely proportional to the square of their separation), the nuclear force is much more complicated; no simple formula can be written for the strength of the nuclear force.
Binding Energy
The masses of the proton and neutron are listed below.
proton: mp = 1.6726 × 10–27 kg
neutron: mn = 1.6749 × 10–27 kg
Because these masses are so tiny, a much smaller mass unit is used. With the most abundant isotope of carbon (carbon-12) as a reference, the atomic mass unit (abbreviated amu or simply u) is defined as 1/12 the mass of a 12C atom. The conversion between kg and u is 1 u = 1.6605 × 10–27 kg. In terms of atomic mass units,
mp = 1.00728 u
mn = 1.00867 u
Now consider the deuteron, the nucleus of deuterium, an isotope of hydrogen that contains 1 proton and 1 neutron. The mass of a deuteron is 2.01356 u, which is a little less than the sum of the individual masses of the proton and neutron. The difference between the mass of any bound nucleus and the sum of the masses of its constituent nucleons is called the mass defect, ∆m. In the case of the deuteron (symbolized d), the mass defect is
What happened to this missing mass? It was converted to energy when the deuteron was formed. In 1905, Einstein gave us the famous equation:
E = mc2
which tells us how much energy mass contains. Because of this, the mass-difference gives us the amount of energy needed to break the deuteron into a separate proton and neutron. Since this tells us how strongly the nucleus is bound, it is called the binding energy of the nucleus.
EB = (∆m)c2
Using E = mc2, the energy equivalent of 1 atomic mass unit is
In terms of electronvolts, then, the binding energy of the deuteron is
EB (deuteron) = 0.00239 u × = 2.23 MeV B
Since the deuteron contains 2 nucleons, the binding-energy-per-nucleon is
= 1.12 MeV/nucleon
This is the lowest value of all nuclides. The highest, 8.8 MeV/nucleon, is for an isotope of nickel, 62Ni. Typically, when nuclei smaller than nickel are fused to form a single nucleus, the binding energy per nucleon increases, which tells us that energy is released in the process. On the other hand, when nuclei larger than nickel are split, binding energy per nucleon again increases, releasing energy. This is the basis of nuclear fission.
Example 7 What is the maximum wavelength of EM radiation that could be used to photodisintegrate a deuteron?
Solution. The binding energy of the deuteron is 2.23 MeV, so a photon would need to have at least this much energy to break the deuteron into a proton and neutron. Since E = hf and f = c/λ,
Example 8 The atomic mass of Al is 26.9815 u. What is its nuclear binding energy per nucleon? (Mass of electron = 0.0005486 u.)
Solution. The nuclear mass of Al is equal to its atomic mass minus the mass of its electrons. Since an aluminum atom has 13 protons, it must also have 13 electrons. So,
Now, the nucleus contains 13 protons and 27 – 13 = 14 neutrons, so the total mass of the individual nucleons is
and, the mass defect of the aluminum nucleus is
∆m = M – m = 27.2160 u – 26.9744 u = 0.2416 u
Converting this mass to energy, we can see that
EB = 0.2416 u × = 225 MeV
so the binding energy per nucleon is
= 8.3 MeV/nucleon
NUCLEAR REACTIONS
Natural radioactive decay provides one example of a nuclear reaction. Other examples of nuclear reactions include the bombardment of target nuclei with subatomic particles to artificially induce radioactivity, such as the emission of a particle or the splitting of the nucleus (this is nuclear fission), and the nuclear fusion of small nuclei at extremely high temperatures. In all cases of nuclear reactions that we’ll study, nucleon number and charge must be conserved. In order to balance nuclear reactions, we write p or H for a proton and n for a neutron. Gamma-ray photons can also be produced in nuclear reactions; they have no charge or nucleon number and are represented as γ.
Alpha Decay
When a nucleus undergoes alpha decay, it emits an alpha particle, which consists of two protons and two neutrons and is the same as the nucleus of a helium-4 atom. An alpha particle can be represented as
Very large nuclei can shed nucleons quickly by emitting one or more alpha particles, for example, radon-222 (Rn) is radioactive and undergoes alpha decay.
This reaction illustrates two important features of a nuclear reaction.
(1) Mass number is conserved.
(2) Charge is conserved.
The decaying nuclide is known as the parent, and the resulting nuclide is known as the daughter. (Here, radon-222 is the parent nuclide and polonium-218 is the daughter.) Alpha decay decreases the mass number by 4 and the atomic number by 2. Therefore, alpha decay looks like the following:
Beta Decay
There are three subcategories of beta (β) decay, called β–, β+, and electron capture (EC).
β−Decay
When the neutron-to-proton ratio is too large, the nucleus undergoes β– decay, which is the most common form of beta decay. β– decay occurs when a neutron transforms into a proton and releases an electron. The expelled electron is called a beta particle. The transformation of a neutron into a proton and an electron (and another particle, the electron-antineutrino, ) is caused by the action of the weak nuclear force, another of nature’s fundamental forces. A common example of a nuclide that undergoes β– decay is carbon-14, which is used to date archaeological artifacts.
The reaction is balanced, since 14 = 14 + 0 and 6 = 7 + (–1).
β+Decay
When the neutron-to-proton ratio is too small, the nucleus will undergo β+ decay. In this form of beta decay, a proton is transformed into a neutron and a positron, e (the electron’s antiparticle), plus another particle, the electron-neutrino, νe, which are then both ejected from the nucleus. An example of a positron emitter is fluorine-17.
Electron Capture Another way in which a nucleus can increase its neutron-to-proton ratio is to capture an orbiting electron and then cause the transformation of a proton into a neutron. Beryllium-7 undergoes this process.
Gamma Decay
In each of the decay processes defined above, the daughter was a different element than the parent. Radon becomes polonium as a result of α decay, carbon becomes nitrogen as a result of β– decay, fluorine becomes oxygen from β+ decay, and beryllium becomes lithium from electron capture. By contrast, gamma decay does not alter the identity of the nucleus; it just allows the nucleus to relax and shed energy. Imagine that potassium-42 undergoes β– decay to form calcium-42.
The asterisk indicates that the daughter calcium nucleus is left in a high-energy, excited state. For this excited nucleus to drop to its ground state, it must emit a photon of energy, a gamma ray, symbolized by γ.
Let’s sum up the three types of radiation:
Type of Radiation | What Happens? | Charge |
Alpha | Particle—a helium nucleus, containing 2 protons and 2 neutrons, is released from the nucleus | Positive 2+ |
Beta | Particle—a highly energetic electron is released from the nucleus. It has negligible mass. When a beta particle is produced, a neutron divides into a proton and an electron. The electron is ejected from the nucleus. That ejected electron is the beta particle. | Negative 1− |
Gamma | Wave—a photon of energy is emitted | No Charge |
Example 9 A mercury-198 nucleus is bombarded by a neutron, which causes a nuclear reaction:
What’s the unknown product particle, X?
Solution. In order to balance the superscripts, we must have 1 + 198 = 197 + A, so A = 2, and the subscripts are balanced if 0 + 80 = 79 + Z, so Z = 1:
Therefore, X must be a deuteron, H (or just d).
DISINTEGRATION ENERGY
Nuclear reactions not only produce new nuclei and other subatomic product particles, they also involve the absorption or emission of energy. Nuclear reactions must conserve total energy, so changes in mass are accompanied by changes in energy according to Einstein’s equation
∆E = (∆m)c2
A general nuclear reaction is written
A + B → C + D + Q
where Q denotes the disintegration energy. If Q is positive, the reaction is exothermic (or exoergic) and the reaction can occur spontaneously; if Q is negative, the reaction is endothermic (or endoergic) and the reaction cannot occur spontaneously. The energy Q is calculated as follows:
Q = [(mA + mB) – (mC + mD)]c2 = (∆m)c2
For spontaneous reactions—ones that liberate energy—most of the energy is revealed as kinetic energy of the least massive product nuclei.
Example 10 The process that powers the Sun—and upon which all life on Earth is dependent—is the fusion reaction:
(a) Show that this reaction releases energy.
(b) How much energy is released per proton?
Use the fact that mα = 4.0015 u and ignore the mass of the electron-neutrino, νe.
Solution.
(a) We need to find the mass difference between the reactants and products:
Since ∆m is positive, the reaction is exothermic: Energy is released.
(b) Converting the mass difference to energy gives
Since four protons went into the reaction, the energy liberated per proton is
Example 11 Can the following nuclear reaction occur spontaneously?
(The mass of the nitrogen nucleus is 13.9992 u, and the mass of the oxygen nucleus is 16.9947 u.)
Solution. We first figure out the mass equivalent of the disintegration energy:
Since ∆m is negative, this reaction is nonspontaneous; energy must be supplied in order for this reaction to proceed. But how much?
Chapter 11 Review Questions
Solutions can be found in Chapter 12.
Click here to download a PDF of Chapter 11 Review Questions.
Section I: Multiple Choice
1. What’s the energy of a photon whose wavelength is 2.07 nm ?
(A) 60 eV
(B) 600 eV
(C) 960 eV
(D) 6000 eV
2. A metal whose work function is 6.0 eV is struck with light of frequency 7.2 × 1015 Hz. What is the maximum kinetic energy of photoelectrons ejected from the metal’s surface?
(A) 7 eV
(B) 13 eV
(C) 19 eV
(D) 24 eV
3. An atom with one electron has an ionization energy of 25 eV. How much energy will be released when the electron makes the transition from an excited energy level, where E = –16 eV, to the ground state?
(A) 9 eV
(B) 11 eV
(C) 16 eV
(D) 25 eV
4. The single electron in an atom has an energy of –40 eV when it’s in the ground state, and the first excited state for the electron is at –10 eV. What will happen to this electron if the atom is struck by a stream of photons, each of energy 15 eV ?
(A) The electron will absorb the energy of one photon and become excited halfway to the first excited state, then quickly return to the ground state, emitting a 15 eV photon in the process.
(B) The electron will absorb the energy of one photon and become excited halfway to the first excited state, then quickly absorb the energy of another photon to reach the first excited state.
(C) The electron will absorb two photons and be excited to the first excited state.
(D) Nothing will happen.
5. What is the de Broglie wavelength of a proton whose linear momentum has a magnitude of 3.3 × 10–23 kg∙m/s ?
(A) 0.0002 nm
(B) 0.002 nm
(C) 0.02 nm
(D) 0.2 nm
6. A partial energy-level diagram for an atom is shown below. What photon energies could this atom emit if it begins in the n = 3 state?
(A) 5 eV only
(B) 3 eV or 7 eV only
(C) 2 eV, 3 eV, or 7 eV
(D) 3 eV, 4 eV, or 7 eV
7. The ground-state energy level for He+ is –54.4 eV. How much energy must the electron in the ground state of He+ absorb in order to be excited to the next higher energy level?
(A) 13.6 eV
(B) 27.2 eV
(C) 40.8 eV
(D) 68.0 eV
8. What would happen to the energy of a photon if its wavelength were reduced by a factor of 2 ?
(A) It would decrease by a factor of 4.
(B) It would decrease by a factor of 2.
(C) It would increase by a factor of 2.
(D) It would increase by a factor of 4.
9. In an exothermic nuclear reaction, the difference in mass between the reactants and the products is m, and the energy released is Q. In a separate exothermic nuclear reaction in which the mass difference between reactants and products is m/4, how much energy will be released?
(A) Q/4
(B) Q/2
(C) (Q/4)c2
(D) (Q/2)c2
10. What’s the missing particle in the following nuclear reaction?
(A) Proton
(B) Neutron
(C) Electron
(D) Positron
11. What’s the missing particle in the following nuclear reaction?
(A) Proton
(B) Electron
(C) Positron
(D) Gamma
Section II: Free Response
1. The Bohr model of electron energy levels can be applied to any one-electron atom, such as doubly ionized lithium (Li2+). The energy levels for the electron are given by the equation
where Z is the atomic number. The emission spectrum for Li2+ contains four spectral lines corresponding to the following wavelengths:
11.4 nm, 13.5 nm, 54.0 nm, 72.9 nm
(a) What’s the value of Z for Li2+?
(b) Identify which energy-level transitions give rise to the four wavelengths cited.
(c) Can the emission spectrum for Li2+ contain a line corresponding to a wavelength between 54.0 nm and 72.9 nm? If so, calculate its wavelength. If not, briefly explain.
(d) What is the next shortest wavelength in the emission spectrum (closest to, but shorter than, 11.4 nm)?
Summary
The energy available to liberate electrons near the surface of a metal (the photoelectric effect) is proportional to the frequency of the incident photon. This idea is expressed in E = hf where h is Planck’s constant. (h = 6.63 × 10–34 J⋅s).
The work function (φ) indicates the amount of energy needed to liberate the electron. There is a minimum frequency needed to liberate the electrons: f0 = . The kinetic energy of the emitted electron is Kmax = hf − φ.
Particles in motion have wavelike properties: λ = where p is the momentum of the particle. Combining this with E = hf and c = fλ yields E = pc.
Standard notation for an element is X, where A is the mass number, Z is the number of protons in the nucleus, and A = Z+N, where N is the number of neutrons in the nucleus.
A nuclear reaction produces new nuclei, other subatomic particles, and the absorption or emission of energy. The change in mass between the reactants and the products tells how much energy is released (exothermic or +Q) or how much energy is needed to produce the reaction (endothermic or –Q) in the general equation A + B → C + D + Q, where A and B are reactants and C and D are products. This energy released or absorbed is given by ΔE = (Δm)c2.
Chapter 12
Solutions to Chapter Review Questions
CHAPTER 3 REVIEW QUESTIONS
Section I: Multiple Choice
1. D Since Point X is 5 m below the surface of the water, the pressure due to the water at X, PX, is ρ ghx = ρ g(5 m), where ρ is the density of water. Because Point Y is 4 m below the surface of the water, the pressure due to the water at Y, pY, is ρ g(4 m). Therefore,
⇒ 4PX = 5PY
2. C Because the top of the box is at a depth of D – z below the surface of the liquid, the pressure on the top of the box is
P = Patm + ρ gh = Patm + ρ g(D – z)
The area of the top of the box is A = xy, so the force on the top of the box is
F = PA = [Patm + ρ g(D ‒ z)]xy
3. C Let P0 be the pressure of the gas at the surface of the liquid. Then the pressures at points Y and Z are, respectively,
PY = P0 + ρ g(1 m)
PZ = P0 + ρ g(3 m)
Subtracting the first equation from the second, we get PZ – PY = ρ g(2 m). Because we’re given the values of PY and PZ, we know that PZ – PY = 16,000 Pa, so we can write
16,000 Pa = ρ g(2 m) ⇒ ρ g = 8,000
Substituting this into the equation for PY, we find that
PV + P0 + ρ g(1 m) ⇒ P0 = PY ‒ ρ g(1 m) = (13,000 Pa) ‒ (8,000) (1 m) = 5,000 Pa
4. A The density of the plastic cube is
This is 4/5 the density of the water, so 4/5 of the cube’s volume is submerged; this means that 1/5 of the cube’s volume is above the surface of the water.
5. B The buoyant force on the Styrofoam block is Fbouy = ρLVg, and the weight of the block is Fg = mSg = ρSVg. Because ρL > ρS, the net force on the block is upward and has magnitude Fnet = Fbuoy ‒ Fg = (ρL ‒ ρS)Vg
Therefore, by Newton’s Second Law, we have
6. B The buoyant force acting on the ball is
Fbuoy = ρwater Vsub g = ρwater Vg = (1000)(5 × 10‒3)m3 (10)= 50 N
The weight of the ball is
FE = ρball Vg = (400)(5 × 10‒3 m3 (10)= 20 N
Because the upward force on the ball, Fbuoy, balances the total downward force, Fg + FT, the tension in the string is
FT = Fbuoy – Fg = 50 N – 20 N = 30 N
7. A If the object weighs 100 N less when completely submerged in water, the buoyant force must be 100 N; therefore
Fbuoy = ρwater Vsub g = ρwater Vg = 100 N ⇒
Now that we know the volume of the object, we can figure out its weight:
Fg = mg = ρobject Vg = (2000)(10‒2 m3) (10)= 200 N
8. A The cross-sectional diameter at Y is 3 times the cross-sectional diameter at X, so the cross- sectional area at Y is 32 = 9 times that at X. The Continuity Equation tells us that the flow speed, v, is inversely proportional to the cross-sectional area, A. So, if A is 9 times greater at Point Y than it is at X, then the flow speed at Y is 1/9 the flow speed at X; that is, vY = (1/9)vX = (1/9) (6 m/s) = 2/3 m/s.
9. D Each side of the rectangle at the bottom of the conduit is 1/4 the length of the corresponding side at the top. Therefore, the cross-sectional area at the bottom is (1/4)2 = 1/16 the cross-sectional area at the top. The Continuity Equation tells us that the flow speed, v, is inversely proportional to the cross-sectional area, A. So, if A at the bottom is 1/16 the value of A at the top, then the flow speed at the bottom is 16 times the flow speed at the top.
10. D We’ll apply Bernoulli’s Equation to a point at the pump (Point 1) and at the nozzle (the exit point, Point 2). We’ll choose the level of Point 1 as the horizontal reference level; this makes y1 = 0 and y2 = 1 m. Now, because the cross-sectional diameter decreases by a factor of 10 between Points 1 and 2, the cross-sectional area decreases by a factor of 102 = 100, so flow speed must increase by a factor of 100; that is, v2 = 100v1 = 100(0.4 m/s) = 40 m/s. Because Point 2 is exposed to the air, the pressure there is Patm. Bernoulli’s Equation becomes
P1 + ρv = Patm + ρ gy2 + ρv
Therefore,
Section II: Free Response
1. (a) The pressure at the top surface of the block is Ptop = Patm + ρ1 gh. Since the area of the top of the block is A = xy, the force on the top of the block has magnitude
Ftop = Pbottom A = (Patm + ρL gh)gy
The pressure at the bottom of the block is Pbottom = Patm + ρL g(h + z). Since the area of the bottom face of the block is also A = xy, the force on the bottom surface of the block has magnitude
Fbottom = Pbottom A = [Patm + ρL g(h + z)]xy
These forces are sketched below:
(b) Each of the other four faces of the block (left and right, front and back) is at an average depth of h + z, so the average pressure on each of these four sides is
Psides = Patm + ρLg(h + z)
The left and right faces each have area A = xz, so the magnitude of the average force on this pair of faces is
Fleft and right = Psides A = [Patm + ρLg(h + z)]xz
The front and back faces each have area A = yz, so the magnitude of the average force on this pair of faces is
Ffront and back = Psides A = [Patm + ρLg(h + z)]yz
These forces are sketched below:
(c) The four forces sketched in part (b) add up to zero, so the total force on the block due to the pressure is the sum of Fon top and Fon bottom ; because Fon bottom > Fon top , this total force points upward and its magnitude is
Fon bottom − Fon top = [Patm + ρLg(h + z)]xy − (Patm + ρLgh)xy = ρLgxyz
(d) By Archimedes’ Principle, the buoyant force on the block is upward and has magnitude
Fbuoy = ρLVsubg = ρLVg = ρLxyzg
This is the same as the result we found in part (c).
(e) The weight of the block is
Fg = mg = ρBVg = ρBxyzg
If FT is the tension in the string, then the total upward force on the block, FT + Fbuoy, must balance the downward force, Fg; that is, FT + Fbuoy = Fg, so
FT = FE − Fbuoy = ρBxyzg − ρLxyzg = xyzg(ρB − ρL)
2. (a) See the section on Torricelli’s Theorem for the derivation of the efflux speed from the hole; applying Bernoulli’s Equation to a point on the surface of the water in the tank (Point 1) and a point at the hole (Point 2), the assumption that v1 ≈ 0 leads to the result
(b) The initial velocity of the water, as it emerges from the hole, is horizontal. Since there’s no initial vertical velocity, the time t required to drop the distance y = D – h to the ground is found as follows:
Therefore, the horizontal distance the water travels is
(c) The second hole would be at a depth of h/2 below the surface of the water, so the horizontal distance it travels—from the edge of the tank to the point where it hits the ground—is given by the same formula we found in part (b) except it will have h/2 in place of h; that is,
If both streams land at the same point, then the value of x from part (b) is the same as x2:
(d) Once again, we apply Bernoulli’s Equation to a point on the surface of the water in the tank (Point 1) and to a point at the hole (Point 2). We’ll choose the ground level as our horizontal reference level; then y1 = D and y2 = D – h. If v1 is the flow speed of Point 1—that is, the speed with which the water level in the tank drops—and v2 is the efflux speed from the hole, then, by the Continuity Equation, A1v1 = A2v2 , where A1 and A2 are the cross-sectional areas at Points 1 and 2, respectively. Therefore, v1 = (A2/A1)v2. Bernoulli’s Equation then becomes
P1 + ρgD + ρ = P2 + ρg(D − h) + ρ
Since P1 = P2 = Patm , these terms cancel out; and substituting v1 = (A2/A1)v2, we have
Now, since A1 = πR2 and A2 = πr2, this final equation can be written as
[Note that if r ≪ R, then (r/R)4 ≈ 0, and the equation above reduces to v2 = , as in part (a).]
3. (a) Point X is at a depth of h1 below Point 1, where the pressure is P1. Therefore, the hydrostatic pressure at X is PX = P1 + ρFgh1.
(b) Point Y is at a depth of h2 + d below Point 2, where the pressure is P2. The column of static fluid above Point Y contains fluid of density of depth h2 and fluid of density ρF of depth d. Therefore, the hydrostatic pressure at Y is PY = P2 + ρFgh2 + ρVgd.
(c) First, notice that Points 1 and 2 are at the same horizontal level; therefore, the heights y1 and y2 are the same, and the terms ρFgy1 and ρFgy2 will cancel out of the equation. Bernoulli’s Equation then becomes
P1 + ρF = P2 + ρF
By the Continuity Equation, we have A1v1 = A2v2, so v1 = (A2/A1)v2. Therefore,
(d) In parts (a) and (b) above, we found that PX = P1 + ρF gh1 and PY = P2 + ρF gh2+ρvgd. Since PX = PY, we have
P1 + ρFgh1 = P2 + ρFgh2 + ρVgd
so
P1 − P2 = ρFg(h2 − h1) + ρVgd
= ρFg(−d) + ρVgd
= (ρV − ρF)gd
(e) In parts (c) and (d), we found two expressions for P1 – P2. Setting them equal to each other gives
The flow rate in the pipe is
Since
we see that f is proportional to , as desired.
CHAPTER 4 REVIEW QUESTIONS
Section I: Multiple Choice
1. B Because the average kinetic energy of a molecule of gas is directly proportional to the temperature of the sample, the fact that the gases are at the same temperature—since they’re in the same container at thermal equilibrium—tells us that the molecules have the same average kinetic energy. The ratio of their kinetic energies is therefore equal to 1.
2. D From the Ideal Gas Law, we know that P = nRT/V. If both T and V are doubled, the ratio T/V remains unchanged, so P remains unchanged.
3. A The work done on the gas during a thermodynamic process is equal to the area of the region in the P–V diagram above the V-axis and below the path the system takes from its initial state to its final state. Since the area below path 1 is the greatest, the work done on the gas during the transformation along path 1 is the greatest.
4. C During an isothermal change, ∆U is always zero.
5. B Because the gas is confined, n remains constant, and because we’re told the volume is fixed, V remains constant as well. Since R is a universal constant, the Ideal Gas Law, PV = nRT, tells us that P and T are proportional. Therefore, if T increases by a factor of 2, then so does P.
6. D Neither (A) nor (B) can be correct. Using PV = nRT, both containers have the same V, n is the same, P is the same, and R is a universal constant. Therefore, T must be the same for both samples. The kinetic theory of gases predicts that the rms speed of the gas molecules in a sample of molar mass M and temperature T is
Hydrogen has a smaller molar mass than does helium, so vrms for hydrogen must be greater than vrms for helium (because both samples are at the same T).
7. A By convention, work done on the gas sample is designated as positive, so in the First Law of Thermodynamics, ∆U = Q + W, we must write W = +320 J. Therefore, Q = ∆U – W = 560 J – 320 J = +240 J. Positive Q denotes heat in.
8. C No work is done during the step from state a to state b because the volume doesn’t change. Therefore, the work done from a to c is equal to the work done from b to c. Since the pressure remains constant (this step is isobaric), we find that
W = –P∆V = –(3.0 × 105 Pa)[(10 – 25) × 10–3 m3] = 4,500 J
9. C Choice (A) is wrong because “no heat exchanged between the gas and its surroundings” is the definition of adiabatic, not isothermal. Choice (B) cannot be correct since the step described in the question is isothermal; by definition, the temperature does not change. This also eliminates (D) and supports (C). If the sample could be brought back to its initial state and have a 100% conversion of heat to work, that would violate the Second Law. The Second Law states that heat cannot be completely converted to work with no other change taking place. In this case, there are changes taking place: The pressure decreases and the volume increases.
10. C The Second Law of Thermodynamics indicates that energy will flow from a hot object to a cool object, making the final temperature of the hot object cooler and the cold object warmer.
Section II: Free Response
1. (a) First, let’s calculate ∆Uacb. Using path acb, the question tells us that Q = +70 J and W = –30 J (W is negative here because it is the system that does the work). The First Law, ∆U = Q + W, tells us that ∆Uacb = +40 J. Because ∆Ua→b doenot depend on the path taken from a to b, we must have ∆Uab = +40 J, and ∆Uba = –∆Uab = –40 J. Thus, –40 J = Qba + Wba, where Qba and Wba are the values along the curved path from b to a. Since Qba = –60 J, it follows that Wba = +20 J. Therefore, the surroundings do 20 J of work on the system.
(b) Again, using the fact that ∆Ua→b does not depend on the path taken from a to b, we know that ∆Uadb = +40 J, as computed above. Writing ∆Uadb = QWadb + Wadb, if Wadb= –10 J, it follows that Qadb = +50 J. That is, the system absorbs 50 J of heat.
(c) For the process db, there is no change in volume, so Wdb = 0. Therefore, ∆Udb= Qdb + Wdb = Qdb. Now, since Uab = +40 J, the fact that Ua = 0 J implies that Ub = 40 J, so ∆Udb = Ub – Ud = 40 J – 30 J = 10 J. Thus, Qdb = 10 J. Now let’s consider the process ad. Since Wadb = Wad + Wdb = Wad + 0 = Wad, the fact that Wadb = –10 J [computed in part (c)] tells us that Wad = –10 J. Because ∆Uad = Ud – Ua = 30 J, it follows from ∆Uad = Qad + Wad that Qad = ∆Uad – Wad = 30 J – (–10 J) = 40 J.
(d) The process adbca is cyclic, so ∆U is zero. Because this cyclic process is traversed counterclockwise in the P–V diagram, we know that W is positive. Then, since ∆U = Q + W, it follows that Q must be negative.
2. (a) (i) Use the Ideal Gas Law:
(ii) Since state b is on the isotherm with state a, the temperature of state b must also be 870 K.
(iii) Use the Ideal Gas Law:
(b) (i) Since step ab takes place along an isotherm, the temperature of the gas does not change, so neither does the internal energy; ∆Uab = 0.
(ii) By the First Law of Thermodynamics, ∆Ubc = Qbc + Wbc. Since step bc is isobaric (constant pressure), we have
Qbc nCP∆Tbc = (0.4 mol)(29.1 J/mol · K)(220 K − 870 K) = −7,600 J
Next,
Wbc = −P∆Vbc = −(0.6 × 105 Pa)[(12 − 48) × 10−3 m3] = 2,200 J
Therefore,
∆Ubc = Qbc + Wbc = (–7,600 J) + 2,200 J = –5,400 J
(iii) By the First Law of Thermodynamics, ∆Uca = Qca + Wca. Since step ca is isochoric (constant volume), we have
Qca = nCV∆Tca = (0.4 mol)(20.8 J/mol · K)(870 K − 220 K) = 5,400 J
and Wca = 0. Therefore, ∆Uca = Qca + Wca = (5,400 J) + 0 = 5,400 J.
Note that part (iii) could have been answered as follows:
∆Uaa = ∆Uab + ∆Ubc + ∆Uca
0 = 0 + ∆Ubc + ∆Uca
∆Uca = −∆Ubc
= −(−5,400 J)
= 5,400 J
(c) Using the equation given, we find that
(d) The total work done over the cycle is equal to the sum of the values of the work done over each step:
Wcycle = Wab + Wbc + Wca
= Wab + Wbc
= (−4,000 J) + (2,200 J)
= −1,800 J
(e) (i) and (ii) By the First Law of Thermodynamics,
∆Uab = Qab + Wab
0 = Qab + Wab
Qab = −Wab
= 4,000 J [from part (c)]
Since Q is positive, this represents heat absorbed by the gas.
(f) The maximum possible efficiency is the efficiency of a Carnot engine:
CHAPTER 5 REVIEW QUESTIONS
Section I: Multiple Choice
1. D Electrostatic force obeys an inverse-square law: FE ∝ 1/r2. Therefore, if r increases by a factor of 3, then FE decreases by a factor of 32 = 9.
2. C The strength of the electric force is given by kq2/r2, and the strength of the gravitational force is Gm2/r2. Since both of these quantities have r2 in the denominator, we simply need to compare the numerical values of kq2 and Gm2. There’s no contest: Since
kq2 = (9 × 109 N·m2/C2)(1 C)2 = 9 × 109 N·m2
and
Gm2 = (6.7 × 10–11 N·m2/kg2)(1 kg)2 = 6.7 × 10–11 N·m2
we see that kq2 ≫ Gm2, so FE is much stronger than FG.
3. C If the net electric force on the center charge is zero, the electrical repulsion by the +2q charge must balance the electrical repulsion by the +3q charge:
4. D Since P is equidistant from the two charges, and the magnitudes of the charges are identical, the strength of the electric field at P due to +Q is the same as the strength of the electric field at P due to –Q. The electric field vector at P due to +Q points away from +Q, and the electric field vector at P due to –Q points toward –Q. Since these vectors point in the same direction, the net electric field at P is (E to the right) + (E to the right) = (2E to the right).
5. C The acceleration of the small sphere is
As r increases (that is, as the small sphere is pushed away), a decreases. However, since a is always positive, the small sphere’s speed, v, is always increasing.
6. B Since FE (on q) = qE, it must be true that FE (on −2q) = –2qE = –2FE.
7. C All excess electric charge on a conductor resides on the outer surface.
Section II: Free Response
1. (a) From the figure below, we have F1−2 = F1/cos 45°.
Since the net force on +Q is zero, we want F1-2 = F3. If s is the length of each side of the square, then:
Given that is the square root of 2 and cancelling like terms yields: q =
(b) No. If q = Q/2, as found in part (a), then the net force on –q is not zero.
This is because F1−2 is greater than F4, resulting in a net force toward the center of the square, as the following calculations show:
but
(c) By symmetry, E1 = E2 and E3 = E4, so the net electric field at the center of the square is zero:
2. (a) The magnitude of the electric force on Charge 1 is
The direction of F1 is directly away from Charge 2; that is, in the +y direction, so
(b) The electric field vectors at the origin due to Charge 1 and due to Charge 2 are
Therefore, the net electric field at the origin is
(c) No. The only point on the x-axis where the individual electric field vectors due to each of the two charges point in exactly opposite directions is the origin (0, 0). But at that point, the two vectors are not equal and thus do not cancel.
Therefore, at no point on the x-axis could the total electric field be zero.
(d) Yes. There will be a Point P on the y-axis between the two charges,
where the electric fields due to the individual charges will cancel each other out.
Disregarding the value y = (−4 −3)a (because it would place the point P below Charge 2 on the y-axis, where the electric field vectors do not point in opposite directions), we have that E = 0 at the point P = (0, –y) = (0, (4 − 3)a).
(e) Use the result of part (b) with Newton’s Second Law:
CHAPTER 6 REVIEW QUESTIONS
Section I: Multiple Choice
1. C By definition, ∆UE = –WE, so if WE is negative, then ∆UE is positive. This implies that the potential energy, UE, increases.
2. B The work required to assemble the configuration is equal to the potential energy of the system:
3. B Use the definition ∆V = –WE/q. If an electric field accelerates a negative charge doing positive work on it, then WE > 0. If q < 0, then –WE/q is positive. Therefore, ∆V is positive, which implies that V increases.
4. D By definition,
VA→B = ∆UE/q, so VB – VA = ∆UE/q
5. B Because E is uniform, the potential varies linearly with distance from either plate (∆V = Ed). Since Points 2 and 4 are at the same distance from the plates, they lie on the same equipotential. (The equipotentials in this case are planes parallel to the capacitor plates.)
6. B By definition, WE = –q∆V, which gives
WE = −q(VB m VA) = −(−0.05 C)(100 V − 200 V) = −5 J
Note that neither the length of the segment AB nor that of the curved path from A to B is relevant.
Section II: Free Response
1. (a) Labeling the four charges as given in the diagram, we get
(b) Let Ei denote the electric field at the center of the square due to Charge i. Then by symmetry, E1 = E3, E2 = E4, and E1 = E2 = E3 = E4. The horizontal components of the four individual field vectors cancel, leaving only a downward-pointing electric field of magnitude Etotal = 4E1 cos 45°:
(c) The potential at the center of the square is zero:
(d) At every point on the center horizontal line shown, r1 = r4 and r2 = r3, so V will equal zero (just as it does at the center of the square):
(e) The work done by the electric force as q is displaced from Point A to Point B is given by the equation WE = –q∆VA→B = –q(VB – VA) = –qVB (since VA = 0).
2. (a) The capacitance is C = ε0A/d. Since the plates are rectangular, the area A is equal to Lw, so C = ε0Lw/d.
(b) and (c) Since the electron is attracted upward, the top plate must be the positive plate:
(d) The acceleration of the electron is a = FE/m = qE/m = eE/m, vertically upward. Therefore, applying Big Five #3 for vertical motion, ∆y = v0yt + ayt2, we get
∆y = (1)
To find t, notice that v0x = v0 remains constant (because there is no horizontal acceleration). Therefore, the time necessary for the electron to travel the horizontal distance L is t = L/v0. In this time, ∆y is d/2, so Equation (1) becomes
(e) Substituting the result of part (d) into the equation ∆V = Ed gives
Since Q = C∆V (by definition), the result of part (a) now gives
(f) Applying the equation UE = C(∆V)2, we get
3. (a) Outside the sphere, the sphere behaves as if all the charge were concentrated at the center. Inside the sphere, the electrostatic field is zero:
(b) On the surface and outside the sphere, the electric potential is (πε0)(Q/r). Within the sphere, V is constant (because E = 0) and equal to the value on the surface. Therefore,
(c) See diagrams:
CHAPTER 7 REVIEW QUESTIONS
Section I: Multiple Choice
1. A Let ρS denote the resistivity of silver and let AS denote the cross-sectional area of the silver wire. Then
2. C The equation I = V/R implies that increasing V by a factor of 2 will cause I to increase by a factor of 2.
3. C Use the equation P = V2/R:
4. B The current through the circuit is
Therefore, the voltage drop across R is V = IR = (2 A)(15 Ω) = 30 V.
5. D The 12 Ω and 4 Ω resistors are in parallel and are equivalent to a single 3 Ω resistor, because . This 3 Ω resistor is in series with the top 3 Ω resistor, giving an equivalent resistance in the top branch of 3 + 3 = 6 Ω. Finally, this 6 Ω resistor is in parallel with the bottom 3 Ω resistor, giving an overall equivalent resistance of 2 Ω, because .
6. C If each of the identical bulbs has resistance R, then the current through each bulb is ε/R. This is unchanged if the middle branch is taken out of the parallel circuit. (What will change is the total amount of current provided by the battery.)
7. B The three parallel resistors are equivalent to a single 2 Ω resistor, because . This 2 Ω resistance is in series with the given 2 Ω resistor, so their equivalent resistance is 2 Ω + 2 Ω = 4 Ω. Therefore, three times as much current will flow through this equivalent 4 Ω resistance in the top branch as through the parallel 12 Ω resistor in the bottom branch, which implies that the current through the bottom branch is 3 A, and the current through the top branch is 9 A. The voltage drop across the 12 Ω resistor is therefore V = IR = (3 A)(12 Ω) = 36 V.
8. D Since points a and b are grounded, they’re at the same potential (call it zero).
Traveling from b to a across the battery, the potential increases by 24 V, so it must decrease by 24 V across the 8 Ω resistor as we reach point a. Thus, I = V/R = (24 V)/(8 Ω) = 3 A.
9. D The equation P = I2R gives
P = (0.5 A)2(100 Ω) = 25 W = 25 J/s
Therefore, in 20 s, the energy dissipated as heat is
E = Pt = (25 J/s)(20 s) = 500 J
10. D
Capacitors 1 and 2 are in series, so their equivalent capacitance is C1-2 = C/2. (This is obtained from the equation 1/C1-2 = 1/C1 + 1/C2 = 1/C + 1/C = 2/C.) Capacitors 4 and 5 are also in series, so their equivalent capacitance is C4-5 = C/2. The capacitances C1-2, C3, and C4-5 are in parallel, so the overall equivalent capacitance is (C/2) + C + (C/2) = 2C.
Section II: Free Response
1. (a) The two parallel branches, the one containing the 40 Ω resistor and the other a total of 120 Ω, is equivalent to a single 30 Ω resistance. This 30 Ω resistance is in series with the three 10 Ω resistors, giving an overall equivalent circuit resistance of 10 Ω + 10 Ω + 30 Ω + 10 Ω = 60 Ω. Therefore, the current supplied by the battery is I = V/R = (120V)/(60) = 2 A, so it must supply energy at a rate of P = IV = (2 A)(120 V) = 240 W.
(b) Since three times as much current will flow through the 40 Ω resistor as through the branch containing 120 Ω of resistance, the current through the 40 Ω resistor must be 1.5 A.
(c) (i) Va – Vb = IR20 + IR100 = (0.5 A)(20 Ω) + (0.5 A)(100 Ω) = 60 V.
(ii) Point a is at the higher potential (current flows from high to low potential).
(d) Because energy is equal to power multiplied by time, we get
E = Pt = I2Rt = (0.5 A)2(100 Ω)(10 s) = 250 J
(e) Using the equation R = ρL/A, with A = πr2, we find
2. (a) There are many ways to solve this problem. If you notice that each of the three branches of the parallel section is 60 Ω, then they must all have the same current flowing through them. The currents through the 20 Ω, 40 Ω, and other 60 Ω resistors are all 0.5 A.
If you had not noticed this, you would have used Ohm’s Law to determine the voltage across the resistors and proceeded from there (see part b below). Because 0.5 A goes through each of the three pathways, Kirchhoff’s Junction Rule tells us that the current that must have come though the 10 Ω resistor is 1.5 A.
(b) The voltage across the 60 Ω resistor is given by Ohm’s Law. Because V = IR, V = (0.5 A)(60 Ω) = 30 V. All three parallel branches must have the same voltage across them, so the other 60 Ω resistor also has 30 V across it and the combination of the 20 Ω and 60 Ω resistor must also be 30 V. To determine the voltage across the 20 Ω and 40 Ω resistor you can rely on the previously solved currents of 0.5 A and Ohm’s Law to yield V = IR = (0.5 A)(20 Ω) = 10 V and V = (0.5 A)(40 Ω) = 20 V.
You also could have used the ratio of the resistors. That is, we know the two voltages must sum to 30 V and the voltage drop across the 40 Ω must be twice the amount across the 20 Ω. The voltage across the 10 Ω resistor can be found using Ohm’s Law: V = (1.5 A)(10 Ω) = 15 V.
(c) The equivalent resistance of the circuit can be solved either by adding the resistances or by using Ohm’s Law.
If you want to add resistances, start by summing the 20 Ω and 40 Ω resistors to get 60 Ω. Then add the three parallel branches using or , which becomes 20 Ω. Then adding this section in series to the 10 Ω resistor to get Rp = R1+R2 = 10 Ω + 20 Ω = 30 Ω.
You could have also realized that the total voltage drop across the battery is 45 V (15 V across the 10 Ω resistor and 30 V across the parallel branch). Using Ohm’s Law again Req = or Req = = 30 Ω.
CHAPTER 8 REVIEW QUESTIONS
Section I: Multiple Choice
1. C, D Choice (A) is false. The magnetic field lines due to a current-carrying wire encircle the wire in closed loops. Choice (B) is false, since the magnetic force is always perpendicular to the charged particle’s velocity vector, it cannot do work on the charged particle; therefore, it cannot change the particle’s kinetic energy. Choice (C) is true. If the charged particle’s velocity is parallel (or anti-parallel) to the magnetic field lines, then the particle will feel no magnetic force. Choice (D) is also true. Magnetic fields are created by moving charges, F = qvB sin θ.
2. C The magnitude of the magnetic force is FB = qvB, so the acceleration of the particle has magnitude
3. B By the right-hand rule, the direction of v × B is into the plane of the page. Since the particle carries a negative charge, the magnetic force it feels will be out of the page.
4. C Since FB is always perpendicular to v, v cannot be upward or downward in the plane of the page; this eliminates (A) and (B). Because the charge is positive, the direction of FB will be the same as the direction of v × B. In order for v × B to be downward in the plane of the page, the right-hand rule implies that v must be out of the plane of the page.
5. A The magnetic force provides the centripetal force on the charged particle. Therefore,
6. C The strength of the magnetic field at a distance r from a long, straight wire carrying a current I is given by the equation B = (µ0/2π)(I/r). Therefore,
7. D By Newton’s Third Law, neither (A) nor (B) can be correct. Also, as we learned in Chapter 8, Example 9, if two parallel wires carry current in the same direction, the magnetic force between them is attractive; this eliminates (C). Therefore, the answer must be (D). The strength of the magnetic field at a distance r from a long, straight wire carrying a current I1 is given by the equation B1 = (µ0/2π)(I1/r). The magnetic force on a wire of length ℓ carrying a current I through a magnetic field B is I(ℓ × B), so the force on Wire #2 (FB2) due to the magnetic field of Wire #1 (B1) is FB2 = I2ℓB1 = I2ℓ which implies
8. D The strength of the magnetic field at a distance r from a long, straight wire carrying a current I is given by the equation B = (µ0/2π)(I/r). Therefore, the strength of the magnetic field at Point P due to either wire is B = (µ0/2π)(I/d). By the right-hand rule, the direction of the magnetic field at P due to the top wire is into the plane of the page and the direction of the magnetic field at P due to the bottom wire is out of the plane of the page. Since the two magnetic field vectors at P have the same magnitude and opposite directions, the net magnetic field at Point P is zero.
9. C Use the right-hand rule for wires. If you point your thumb to the right and wrap your fingers along the wire, you will note that the magnetic field goes into the page when you are below the wire and comes out of the page above the wire. This allows us to eliminate (A) and (B). Because B = , the closer we are to the wire the stronger the magnetic field. Choice (C) is closer, so it is the correct answer.
10. C Magnetic fields point from north to south. Therefore, the magnetic field between the two magnets is toward the right of the page. Use the right-hand rule. Because the B field is to the right and the charges through the wire flow to the bottom of the page, the force must be out of the page.
Section II: Free Response
1. (a) The acceleration of an ion of charge q is equal to FE/m. The electric force is equal to qE, where E = V/d. Therefore, a = qV/(dm).
(b) Using a = qV/(dm) and the equation v2 = v02 + 2ad = 2ad, we get
As an alternate solution, notice that the change in the electrical potential energy of the ion from the source S to the entrance to the magnetic-field region is equal to qV; this is equal to the gain in the particle’s kinetic energy.
Therefore,
(c) (i) and (ii) Use the right-hand rule. Since v points to the right and B is into the plane of the page, the direction of v × B is upward. Therefore, the magnetic force on a positively charged particle (cation) will be upward, and the magnetic force on a negatively charged particle (anion) will be downward. The magnetic force provides the centripetal force that causes the ion to travel in a circular path. Therefore, a cation would follow Path 1 and an anion would follow Path 2.
(d) Since the magnetic force on the ion provides the centripetal force,
Now, by the result of part (b),
(e) Since the magnetic force cannot change the speed of a charged particle, the time required for the ion to hit the photographic plate is equal to the distance traveled (the length of the semicircle) divided by the speed computed in part (b):
(f) Since the magnetic force FB is always perpendicular to a charged particle’s velocity vector v, it can do no work on the particle. Thus, the answer is zero.
2. (a) Because a photon has no charge, it will experience no force and therefore travel in a straight line.
(b) From , we get
(c) From , we get . So . To determine the sign of the charge, use the right-hand rule. Because the magnetic field is into the page and the charge spins in a clockwise manner, it would not obey the right-hand rule that works for all positive charges. Therefore, the charge must be negative.
CHAPTER 9 REVIEW QUESTIONS
Section I: Multiple Choice
1. D Since v is upward and B is out of the page, the direction of v × B is to the right. Therefore, free electrons in the wire will be pushed to the left, leaving an excess of positive charge at the right. Therefore, the potential at Point b will be higher than at Point a, by ε = vBL (motional emf).
2. A As we discussed in this chapter, the magnitude of the emf induced between the ends of the rod is ε = BLv = (0.5 T)(0.2 m)(3 m/s) = 0.3 V. Since the resistance is 10 Ω, the current induced will be I = V/R = (0.3 V)/(10 Ω) = 0.03 A. To determine the direction of the current, we can note that since positive charges in the rod are moving to the left and the magnetic field points into the plane of the page, the right-hand rule tells us that the magnetic force, qv × B, points downward. Since the resulting force on the positive charges in the rod is downward, so is the direction of the induced current.
3. A The magnetic field through the loop is B = µ0nI. Since its area is A = πr2, the magnetic flux through the loop is ΦB = BA = (µ0nI)(πr2). If the current changes (with ∆I/∆t = –a), then the magnetic flux through the loop changes, which, by Faraday’s Law, implies that an emf (and a current) will be induced. We get
Since the magnetic flux into the page is decreasing, the direction of the induced current will be clockwise (opposing a decreasing into-the-page flux means that the induced current will create more into-the-page flux).
4. C By definition, magnetic field lines emerge from the north pole and enter at the south pole. Therefore, as the north pole is moved upward through the loop, the upward magnetic flux increases. To oppose an increasing upward flux, the direction of the induced current will be clockwise (as seen from above) to generate some downward magnetic flux. Now, as the south pole moves away from the center of the loop, there is a decreasing upward magnetic flux, so the direction of the induced current will be counterclockwise.
5. D Since the current in the straight wire is steady, there is no change in the magnetic field, no change in magnetic flux, and, therefore, no induced emf or current.
Section II: Free Response
1. (a) ε = Bℓv ⇒ (2 T)(0.4 m)(1 m/s) or 0.8 V.
(b) V = IR becomes I = or I = = 0.04 A. The direction is given by Lenz’s Law. Current flows in order to oppose the change in the magnetic flux. Because there is suddenly a new flux “in,” current flows to produce an outward flux. This would be in the counterclockwise direction.
(c) ε = Bℓv ⇒ (2 T)(0.2 m)(1 m/s) or 0.4 V
(d) In order to get the 0.8 V you would need or v = 2.0 m/s. You also could have noticed that if you cut the length in half, you have to compensate by doubling the speed.
CHAPTER 10 REVIEW QUESTIONS
Section I: Multiple Choice
1. B From the equation λf = c, we find that
2. C Since the fringe is bright, the waves must interfere constructively. This implies that the difference in path lengths must be a whole number times the wavelength, eliminating (A) and (B). The central maximum is equidistant from the two slits, so ∆ℓ = 0 there. At the first bright fringe above the central maximum, we have ∆ℓ = λ.
3. C First, eliminate (A) and (B): The index of refraction is never smaller than 1. Refer to the following diagram:
Since the reflected and refracted beams are perpendicular to each other, we have θ2 = 30°. Snell’s Law then becomes
n1 sin θ1 = n2 sin θ2
1 · sin 60° = n2 sin 30°
= n2 ·
= n2
4. A The frequency is unchanged, but because the speed of light in diamond is less than in air, the wavelength of the light in diamond is shorter than its wavelength in air:
5. A If the speed of light is less in Medium 2 than in Medium 1, then Medium 2 must have the higher index of refraction; that is, n2 > n1. Snell’s Law then implies that θ2 < θ1: The beam will refract toward the normal upon transmission into Medium 2.
6. A The critical angle for total internal reflection is computed as follows:
n1 sin θc = n2 sin 90°
sin θc = ⇒ θc = 30°
Total internal reflection can happen only if the incident beam originates in the medium with the higher index of refraction, and therefore the beam must originate in the solid. The refracted ray must strikes the interface of the other medium at an angle of incidence greater than the critical angle.
7. D If so = 60 cm and f = 40 cm, the mirror equation tells us that
And, since si is positive, the image is real.
8. D Because the image is virtual, we must write the image distance, si, as a negative quantity: si = –20 cm. Now, with so = 60 cm, the mirror equation gives
The focal length is half the radius of curvature, so
f = ⇒ R = 2f = 2(−30 cm) = −60 cm ⇒ |R| = 60 cm
9. C Since the image is projected onto a screen, it must be real, and therefore inverted. The magnification must be negative, so
Because si = 60 cm, the object distance, so, must be 240 cm. Therefore,
10. B A bi-concave lens is a diverging lens. Diverging lenses (like convex mirrors) have negative focal lengths and therefore cannot form real images. (Note that (D) is false; diverging lenses and convex mirrors always form diminished, virtual images, as you can verify using the mirror and magnification equations.)
Section II: Free Response
1. (a) Interference effects can be observed only if the light is coherent. Using two independent light sources at the slits in Barrier 2 would generate incoherent light waves. The set-up shown guarantees that the light reaching the two slits will be coherent.
(b) Maxima are located at positions given by the equation xm = mλL/d, where m is an integer (y0 = 0 is the central maximum). The first-order maximum for red light occurs at
and the first-order maximum for violet light occurs at
Therefore, the vertical separation of these maxima on the screen is ∆x = 6.0 mm – 3.2 mm = 2.8 mm.
(c) Maxima are located at positions given by the equation xm = mλL/d, where m is an integer. We therefore want to solve the equation
The smallest integers that satisfy this equation are m = 3 and n = 2. That is, the third-order maximum for violet light coincides with the second-order maximum for orange-yellow light. The position on the screen (relative to the central maximum at x = 0) of these maxima is
(d) Within the glass, the wavelength is reduced by a factor of n = 1.5 from the wavelength in air. Therefore, the difference in path lengths, d sin θ, must be equal to m(λ/n) in order for constructive interference to occur. This implies that the maxima are located at positions given by xm = mλL/(nd). The distance between adjacent bright fringes is therefore
2. (a) Beam 1 undergoes a 180° phase change when it reflects off the soap film. The refracted portion of the incident beam does not undergo a phase change upon reflection at the film/air boundary, because it strikes the boundary to a medium whose index is lower; therefore, Beam 2 does not suffer a 180° phase change.
(b) We’re told that the criterion for destructive interference in this case is ∆ℓ = 2T = m(λ/n), where n is the refractive index of the soap film and m is a whole number. Therefore, the criterion for constructive interference must be
Δℓ = 2T = (m + )
Alternatively, Δℓ = 2T = , where k is an odd whole number.
3. (a) Since the image is formed behind the mirror, it is virtual.
(b) Virtual images are upright.
(c) Writing si = –72 cm (negative because the image is virtual), the magnification is
so the height of the image is hi = |m| · ho = 1.8(5 cm) = 9 cm.
(d) The mirror equation can be used to solve for the mirror’s focal length:
Therefore, the mirror’s radius of curvature is R = 2f = 2(90 cm) = 180 cm.
(e)
CHAPTER 11 REVIEW QUESTIONS
Section I: Multiple Choice
1. B Combining the equation E = hf with f = c/λ gives us
2. D The energy of the incident photons is
E = hf = (4.14 × 10–15 eV ∙ s)(7.2 × 1015 Hz) = 30 eV
Since E > φ, photoelectrons will be produced, with maximum kinetic energy
Kmax = E – φ = 30 eV – 6 eV = 24 eV
3. A If the atom’s ionization energy is 25 eV, then the electron’s ground-state energy must be –25 eV. Making a transition from the –16 eV energy level to the ground state will cause the emission of a photon of energy
∆E = (–16 eV) – (–25eV) = 9 eV
4. D The gap between the ground-state and the first excited state is
–10 eV – (–40 eV) = 30 eV
Therefore, the electron must absorb the energy of a 30 eV photon (at least) in order to move even to the first excited state. Since the incident photons have only 15 eV of energy, the electron will be unaffected.
5. C The de Broglie wavelength of a particle whose momentum is p is λ = h/p. For this proton, we find that
6. D If the atom begins in the n = 3 state, it could lose energy by making any of the following transitions: 3 → 2, 3 → 1, or 3 → 2 → 1. The 3 → 2 transition would result in the emission of –5 eV – (–8 eV) = 3 eV; the 3 → 1 transition would emit a –5 eV – (–12 eV) = 7 eV photon; and the 2 → 1 transition would result in the emission of –8 eV – (–12 eV) = 4 eV. Therefore, if the atom is initially in the n = 3 state, it could emit photons of energy 3 eV, 4 eV, or 7 eV.
7. C The energy of the nth level is given by the equation En = E1/n2, where E1 is the ground-state energy. Therefore, the energy of level n = 2 (the next level above the ground state) is E2 = (–54.4 eV)/4 = –13.6 eV. Therefore, the difference in energy between the n = 1 and n = 2 levels is ∆E = –13.6 eV – (–54.4 eV) = 40.8 eV.
8. C The energy of a photon is given by the equation E = hf, or equivalently by E = hc/λ. Therefore, E is inversely proportional to λ. If λ decreases by a factor of 2, then E will increase by a factor of 2.
9. A The equation that relates the mass difference m and the disintegration energy Q is Einstein’s mass–energy equivalence formula, Q = mc2. Because c2 is a constant, we see that Q is proportional to m. Therefore, if m decreases by a factor of 4, then so will Q.
10. B In order to balance the mass number (the superscripts), we must have 2 + 63 = 64 + A, so A = 1. In order to balance the charge (the subscripts), we need 1 + 29 = 30 + Z, so Z = 0. A particle with a mass number of 1 and no charge is a neutron, n.
11. D In order to balance the mass number (the superscripts), we must have 196 + 1 = 197 + A, so A = 0. In order to balance the charge (the subscripts), we need 78 + 0 = 78 + Z, so Z = 0. The only particle listed that has zero mass number and zero charge is a gamma-ray photon, γ.
Section II: Free Response
1. (a) If ionizing lithium twice leaves the atom with just one electron, then it must have originally had three electrons. Since neutral atoms have the same number of protons as electrons, Z must be 3 for lithium.
(b) First, use the formula given for En to determine the values of the first few electron energy levels:
E1 = (−13.6 eV) = −122.4 eV
E2 = (−13.6 eV) = −30.6 eV
E3 = (−13.6 eV) = −13.6 eV
E4 = (−13.6 eV) = −7.65 eV
Differences between electron energy levels equal the energies of the emitted photons. Since E = hf = hc/λ, we have λ = hc/E. Since the E’s are in electronvolts and the wavelengths in nanometers, it is particularly helpful to note that
hc = (4.14 × 10−15 eV · s)(3.00 × 108 × 109 nm/s) = 1,240 eV · nm
The goal here is to match the energy-level differences with the given wavelengths, using the equation λ = (1,240 eV∙nm)/E.
Using this diagram, the four wavelengths given correspond to the following photon energies and energy-level transitions:
(c) No. In order for a spectral line to have a wavelength between 54.0 nm and 72.9 nm, there must be an energy-level transition with an energy between 17 eV and 23 eV. As the diagram above shows, no such intermediate transition is possible.
(d) The 11.4 nm line corresponds to a photon energy of 109 eV. Therefore, the next shortest wavelength must correspond to the next highest possible photon energy. According to the diagram, the next highest available transition energy after 109 eV is 115 eV. This wavelength of the photon emitted in this (4 → 1) transition is
Part VI
Practice Test 2
• Practice Test 2
• Practice Test 2: Answers and Explanations
Practice Test 2
Click here to download a PDF of Practice Test 2.
The Exam
AP® Physics 2 Exam
SECTION I: Multiple-Choice Questions
DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.
At a Glance
Total Time
90 minutes
Number of Questions
50
Percent of Total Grade
50%
Writing Instrument
Pen required
Instructions
Section I of this examination contains 50 multiple-choice questions. Fill in only the ovals for numbers 1 through 50 on your answer sheet.
CALCULATORS MAY BE USED IN BOTH SECTIONS OF THE EXAMINATION.
Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. After you have decided which of the suggested answers is best, completely fill in the corresponding oval on the answer sheet. Give only one answer to each question. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer.
Sample Question
Chicago is a
(A) state
(B) city
(C) country
(D) continent
Sample Answer
Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all the multiple-choice questions.
About Guessing
Many candidates wonder whether or not to guess the answers to questions about which they are not certain. Multiple choice scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers, and no points are awarded for unanswered questions. Because points are not deducted for incorrect answers, you are encouraged to answer all multiple-choice questions. On any questions you do not know the answer to, you should eliminate as many choices as you can, and then select the best answer among the remaining choices.
AP PHYSICS 2
SECTION I
Time—90 minutes
50 Questions
Note: To simplify calculations, you may use g = 10 m/s2 in all problems.
Directions: Each of the questions or incomplete statements below is followed by four suggested answers or completions. Select the one that is best in each case and mark it on your sheet.
1. A positively charged particle enters a region between the plates of a parallel plate capacitor. The particle is moving initially parallel to the plates. What is the correct description of the trajectory of the particle in the region between the plates?
(A) The particle will continue straight.
(B) The particle will move towards the negative plate along a straight-line path.
(C) The particle will move towards the negative plate along a curved path.
(D) The particle will move towards the positive plate along a curved path.
2. Which of the following arguments does not support the assertion that a parallel plate capacitor held with one plate parallel to the floor is an electric analog of the gravitational field near the surface of the Earth?
(A) The voltage is constant along horizontal lines.
(B) The field changes linearly with distance from the lower plate.
(C) Both the gravitational field lines and electric field lines are vertical.
(D) Both the electrical and gravitational force are governed by inverse square laws.
3. An electromagnetic wave moves along the positive z-axis. The wave is polarized so that the magnetic field is confined in the y-axis. Where is there a non-zero electric field from this wave?
(A) The electric field exists everywhere.
(B) The electric field is only along the x-axis.
(C) The electric field is only along the y-axis.
(D) The electric field is only along the z-axis.
4. Three neutral conducting spheres sit on insulating bases. The spheres are separated by a very large distance. The sphere in the center is given a positive charge. The spheres are brought close together, as shown above, but not allowed to come into contact with one another. Which is the correct description of the net charge on each sphere?
(A) All three spheres are positively charged.
(B) The center sphere is positively charged and the two outer spheres are negatively charged.
(C) The center sphere is positively charged and the other two are neutral.
(D) All three spheres are neutral.
Questions 5–7 refer to the following diagram.
5. The circuit above contains a battery, a switch, three identical light bulbs, and a capacitor. The capacitor is initially uncharged. The instant the switch is closed, which of the bulbs are lit?
(A) All three bulbs are unlit.
(B) Only bulb 1 is lit.
(C) Only bulb 2 is lit.
(D) Bulb 1 and bulb 3 are lit.
6. After the capacitor is fully charged, which bulbs are lit?
(A) All three bulbs are unlit.
(B) Bulb 1 and bulb 3 are lit.
(C) Only bulb 2 is lit.
(D) All three bulbs are lit.
7. When the capacitor is fully charged, which components have equal voltage drops across them?
(A) The capacitor and the battery
(B) The capacitor and bulb 2
(C) Bulb 1 and bulb 3
(D) All three of the above selections have equal voltage drops across them.
8. A ray of light enters the pair of tanks as shown above. The light rays in the water and glass tank are shown at the left, with the light exiting the pair of tanks at a point Y. The light ray in the air tank is identical to the ray in the water tank, entering at the same height and same angle. Both sets of tanks use the same glass. How will the ray exiting the second glass tank compare to height Y?
(A) The ray will not exit the second tank due to total internal reflection.
(B) The exiting ray will be closer to the top of the tank than Y.
(C) The exiting ray will be at the same height as Y.
(D) The exiting ray will be farther from the top of the tank than Y.
9. A mirror may be used to create a real image under which of the following conditions?
(A) Use a concave mirror and place the object between the mirror and the focal point.
(B) Use a concave mirror and place the object beyond focal point.
(C) Use a convex mirror and place the object between the mirror and the focal point.
(D) Use a convex mirror and place the object beyond the focal point.
10. The pressure at location A in a pipe is known. The speed of the fluid in the pipe and the pipe diameter are also known at position A. Position B is another location in the pipe. Which pieces of data are required to calculate the speed of the fluid at position B?
(A) The diameter at B only
(B) The diameter at B and the height of B relative to A
(C) The diameter at B, the height of B relative to A and the pressure at B
(D) The diameter at B, the pressure at B, the height of B relative to A and the fluid density
11. The tube shown above carries water. At some point, the tube splits, as shown above. Three points within the tube are labeled A, B, and C. The greatest diameter is at A and the smallest diameter is at B. How do the pressures in the pipes compare at the three points?
(A) PA > PC > PB
(B) PA = PB = PC
(C) PB = PC > PA
(D) PB > PC > PA
12. A paper-filled three ring binder sits in a room for a long time. A student touches the metal rings of the binder and the paper. She expects the two objects to feel the same temperatures, but she observes that the paper feels warmer than the metal. Which of the following correctly explains her observation?
(A) The objects were in thermal equilibrium, but she must have felt the paper first, as when you feel multiple objects at the same temperature, the ones felt first will feel warmest.
(B) The objects were in thermal equilibrium, but the energy flow occurs more slowly between her fingers and the paper than her fingers and the metal, resulting in the paper feeling warmer.
(C) The objects were in thermal equilibrium and the paper warms up quickly when touched. Her perception is not instantaneous, but is the average temperature over a period of time and higher for the paper.
(D) The paper was warmer than the metal because temperature is related to kinetic energy of the molecules, and the more massive metal molecules will have less kinetic energy than the paper molecules.
13. A hot cup of water is placed within an insulating container. Later, the water has cooled. Which of the following explains the phenomenon?
(A) The molecules in the water are fast-moving initially and transfer energy in the air in the container, resulting in the air heating up as the water cools.
(B) The molecules in the air are fast-moving initially and transfer energy in the water in the container, resulting in the water cooling down as the air heats up.
(C) As time goes on, the entropy of the system increases, which results in a decrease in energy and the loss of temperature in the cup.
(D) The total heat in the container must remain constant unless there is leakage with the outside surroundings, so the water can only cool down if the surroundings heat up.
14. A very large sheet of metal has a net negative charge. An electron is placed above the center of the sheet. Which of the following correctly describes the force on the sheet of charge from the electron?
(A) The force pulls the sheet upwards.
(B) The force pushes the sheet downwards.
(C) The electron causes a torque on the sheet, but no net upward or downward force exists.
(D) There is no force on the sheet because the electron is a test charge.
15. An experiment is conducted on a circuit consisting of a battery and several wires. The wires are all made of the same metal. It is found that as the cross-sectional area of the wire increases, the current measured coming out of the battery increases linearly. What can be concluded about the wire from this data?
(A) The battery has an internal resistance, which can be found from the slope of the area vs. current graph.
(B) The battery has an internal resistance, which can be found from the y-intercept of the area vs. current graph.
(C) The wires have the same length, which can be found from the slope of the area vs. current graph.
(D) The wires have the same length, which can be found from the y-intercept of the area vs. current graph.
16. An insulating rod separates two conducting sphere as shown above. Point A is midway between the spheres. Point B lies on the axis of the rod. Which of the following arrangements of charges would result in a net torque on the rod?
(A) Charge both spheres negatively and place a positive charge at point A.
(B) Charge one sphere negatively and the other positively and place a positive charge at point A.
(C) Charge both spheres negatively and place a positive charge at point B.
(D) Charge one sphere negatively and the other positively and place a positive charge at point B.
17. The process of charging by induction requires which physical property of a system?
(A) Charge polarization
(B) Uneven charge distribution
(C) Zero net charge
(D) Contact between a conductor and insulator
18. To determine the direction of the force on a charge that is moving near a current carrying wire, which data do you need?
(A) Only the direction the charge is moving.
(B) The sign of the charge and its direction of motion.
(C) The sign of the charge, the direction of motion and the direction of the current.
(D) The sign of the charge, the direction of motion, the direction of the current and the position of the charge relative to the wire.
19. A positive charge of 2Q and a negative charge of 4Q are arranged at positions as shown above. What is the correct ranking of the magnitude of the electric potential at points along the x-axis?
(A) V(x = 4 m) = V(x = 2 m) > V(x = 1 m)
(B) V(x = 4 m) > V(x = 1 m) > V(x = 2 m)
(C) V(x = 1 m) > V(x = 2 m) > V(x = 4 m)
(D) V(x = 4 m) > V(x = 2 m) > V(x = 1 m)
20. A large piece of wood has the same weight as a small rock. Both the wood and rock are placed into a pool of liquid. The rock sinks while the wood floats. How do the buoyant forces on the wood and rock compare?
(A) The buoyant forces are the same on each.
(B) The buoyant force on the wood is greater than on the rock.
(C) The buoyant force on the wood is less than on the rock.
(D) The buoyant forces cannot be compared without the density of the liquid.
21. An experiment is conducted to determine the power output from a circuit as various voltages are supplied to the circuit. The circuit is set up so that it draws a constant current. What is the value of the current in the circuit that produced the graph above?
(A) 0.00 A
(B) 0.25 A
(C) 1.0 A
(D) 6.0 A
22. The products of several radioactive decays are being studied. Each particle starts with the same speed and enters into a region with a uniform magnetic field directed perpendicular to the initial velocity of the particles. Which observation could be made?
(A) A positron and an electron both turn in the same direction, but the electron turns with a larger radius.
(B) An alpha particle and an electron both turn in the same direction, but the alpha particle turns with a larger radius.
(C) An alpha particle and an electron both turn in opposite directions, but the alpha particle turns with a larger radius.
(D) A positron and an electron both turn in the opposite directions, but the electron turns with a larger radius.
23. A hypothetical atom contains an undetermined number of energy levels above the ground state. A gas of this atom is entirely in the ground state. Light with a broad spectrum is shined upon the gas and the spectrum of the light is recorded after shining through the gas. Three wavelength of light are observed to have diminished intensity. What can be concluded from this?
(A) There must be exactly two energy levels above the ground state.
(B) There must be exactly three energy levels above the ground state.
(C) There must be either exactly two or exactly three energy levels above the ground state.
(D) There are an undetermined number of energy levels above the ground state, but there are at least three levels.
24. An unknown nucleus goes through a decay process. Three He alpha particles are emitted, along with two β− particles. The resulting daughter nucleus is Pb. What was the atomic number of the unknown nucleus that started the process?
(A) 84
(B) 86
(C) 88
(D) 90
25. The wave function of a quantum object vs. position is graphed above. Which of the following correctly ranks the probabilities of observing the particle at the listed positions?
(A) P(x = 0.5 m) > P(x = 2.25 m) > P(x = 1.75 m)
(B) P(x = 0.5 m) > P(x = 1.75 m) > P(x = 1.25 m)
(C) P(x = 3.0 m) > P(x = 1.75 m) > P(x = 2.0 m)
(D) P(x = 3.5 m) > P(x = 3.0 m) > P(x = 1.75 m)
26. Radiothorium-228 decays into Radon-200 through two alpha decays, as shown below.
Which equation correctly describes the energy released during this process?
(A) (mTh − mRn)c2
(B) (mTh + 2mHe − mRn)c2
(C) (mTh – 2mHe − mRn)c2
(D) (mRn + 2mHe − mTh)c2
Questions 27 and 28 refer to the following graph.
A thin lens is used to make an image for several different object distances. The image and object distances are used to make the above graph.
27. What is the approximate focal length of the lens?
(A) 1 cm
(B) 10 cm
(C) 25 cm
(D) 100 cm
28. In order to achieve a magnification of 1, how far from the lens should an object be placed?
(A) 2 cm
(B) 20 cm
(C) 50 cm
(D) 200 cm
29. A container has a round lid, with mass M, and makes an airtight seal with the body of the container. Some of the air in the container is pumped out. The container is turned over, but the lid stays shut. Which of the following explanations for this phenomenon is correct?
(A) Pumping the air out of the container decreases the gravitational force on the lid.
(B) Pumping the air out creates an upward force that will always balance with the gravitational force.
(C) The force on the lid downward from the molecules inside is less than the force on the lid from the particles in the air surrounding the container.
(D) The force on the lid downward from the molecules inside is less than the difference between the force on the lid from the particles in the air surrounding the container and the gravitational force on the lid.
30. A mercury thermometer is placed in a glass of ice-cold water. After some time goes by, the mercury thermometer can be read to determine the temperature of the water 0ºC. Several minutes later, the reading on the thermometer has not changed. What has happened to the molecules of mercury in the thermometer?
(A) Over time, all the molecules slow down, but the rate of slowing decreases as temperature decreases so it is not noticeable between the two readings.
(B) The mercury molecules stopped moving upon reaching a temperature of 0ºC and remained stopped for the second reading.
(C) The mercury molecules reached thermal equilibrium before the first reading, so all their molecules had the same non-zero speed when both readings were taken.
(D) The mercury molecules reached thermal equilibrium before the first reading, but between the readings some of the molecules sped up and others slowed down, resulting in the same reading.
31. A current carrying wire and coordinate system is shown above. Initially, the wire carries a current I towards the top of the page. The amount of current is steadily decreased until it is 0 A, then steadily increased until it reaches a value of I in the downward direction. This change in current takes time t. The magnitude of the magnetic field versus time at the observation point P from t = 0 until t is graphed. Which statement is true concerning the graph?
(A) The graph has a constant slope.
(B) The graph is piecewise linear with a negative slope for the first half and a positive slope for the second half.
(C) The graph is piecewise linear with a positive slope for the first half and a negative slope for the second half.
(D) The graph is a curve showing an inverse relationship.
32. Charges are distributed as shown above. At point A is a charge of +3Q. At B is a charge of +1Q. At C is a charge of −1Q. What is the direction of the force on a proton located at point P?
(A) Up and to the left.
(B) Down and to the left.
(C) Up and to the right.
(D) Down and to the right.
33. All of the following observations of electric field diagrams are correct EXCEPT:
(A) Longer arrows correspond to a greater field magnitude.
(B) Field lines will be larger at positions close to sources or sinks.
(C) Arrow lengths will decrease linearly with distance from sources or sinks.
(D) Arrows point in the direction in which a positive charge would experience a force.
34. A neutral sphere of metal is placed in a region of uniform electric field. The electric field points towards the top of the page. Which of the following diagrams shows the arrangement of the charges on the surface of the sphere once electrostatic equilibrium is reached?
(A)
(B)
(C)
(D)
35. An electron enters a region of magnetic field created by a long current carrying wire. Along which path could the electron travel without being deflected?
(A) The electron is traveling parallel to the wire and in the same direction that the current is flowing.
(B) The electron is traveling parallel to the wire and in the opposite direction from the direction that the current is flowing.
(C) The electron is traveling straight away from the wire.
(D) The electron is traveling in a circle centered at the wire.
36. Which process will result in charging an uncharged sphere using a charged rod?
(A) Bringing the rod near the sphere and touching the two objects together, then removing the rod
(B) Grounding the sphere first, then bringing a charged object near the sphere and touching the two objects together, then removing the rod
(C) Bringing the rod near the sphere, but not touching them together, then removing the rod
(D) Grounding the sphere, moving the rod near the sphere, bringing the rod near the sphere but not touching them together, then removing the rod
Questions 37 and 38 refer to the following diagram.
37. The current drawn from the battery is most nearly equal to what value?
(A) 0.042 A
(B) 0.19 A
(C) 0.30 A
(D) 0.68 A
38. A fifth resistor is placed in the circuit. It is parallel beside the 5 Ω resistor. The voltage drop across this resistor is found to be V = 0.50 V. What is the resistance of the additional resistor?
(A) 5 Ω
(B) 10 Ω
(C) 15 Ω
(D) 30 Ω
39. Three cylinders of the same metal act as resistors arranged in series, as shown above. Which of the following correctly ranks the voltage drops across the three resistors?
(A) V1 = V2 = V3
(B) V3 > V1 > V2
(C) V2 > V3 > V1
(D) V1 > V3 > V2
40. An electron experiences both electric and magnetic forces. Those forces are balanced. What must be true about the motion of the particle and the alignment of the two fields?
(A) The fields point along the same axis and the velocity of the electron is parallel to that axis.
(B) The fields point along the same axis and the velocity of the electron is perpendicular to that axis.
(C) The fields point along the perpendicular axes and the velocity is parallel to the axis of the electric field.
(D) The fields point along the perpendicular axes and the velocity along the third perpendicular axis.
41. A pair of electrons are held a fixed distance apart. Which of the following correctly describes the electrical and gravitational forces between the electrons?
(A) The electrical force is much greater than the gravitational force.
(B) The gravitational force is much greater than the electrical force.
(C) The forces are equal as both the gravitational force and electrical force are governed by inverse square laws.
(D) The forces cannot be compared without knowing the value of the distance between the electrons.
42. An electron travels down the center of a solenoid, carrying a current I. Which of the following explanations for the motion of the electron is correct?
(A) The electron accelerates due to the electric force from the charges in the current in the wire.
(B) The electron accelerates due to the magnetic field generated by the current in the wires of the solenoid.
(C) The electron travels at a constant speed but changes direction due to the magnetic field generated by the current in the wires.
(D) The electron travels at a constant velocity because there is no net force on the electron.
43. A voltage is sent through a tube containing hydrogen gas. The gas emits light. When the light is sent into a spectrometer, several distinct bright lines are seen. Why does this occur?
(A) The voltage causes the gas to heat up to specific temperatures, which are characterized by the lines.
(B) The voltage causes gas molecules to move at specific speeds, which are characterized by the lines.
(C) The voltage sets up standing waves due to the pressure of the gas in the tube, which can be characterized by the lines.
(D) The voltage excites electrons in the gas into specific energy levels, which can be characterized by the lines.
44. An experiment is conducted to determine the critical angle for light going from glass into air, as shown above. A linear plot is made with a vertical axis of sin (θ1) and a horizontal axis of sin (θ2). How is the critical angle determined from the graph?
(A) The critical angle can be found from the slope of the line.
(B) The critical angle can be found from the y-intercept of the line.
(C) The critical angle can be found from the horizontal axis value which corresponds to the maximum vertical value.
(D) The critical angle can be found from the vertical axis value which corresponds to the maximum horizontal value.
45. The circuit shown above is set up. The switch is closed and a long time passes. What conditions on the two resistors result in the greatest amount of energy stored in the capacitor?
(A) The energy stored in the capacitor will be greatest if R1 > R2.
(B) The energy stored in the capacitor will be greatest if R1 = R2.
(C) The energy stored in the capacitor will be greatest if R1 < R2.
(D) The energy will be the same regardless of the resistor values.
Directions: For questions 46-50 below, two of the suggested answers will be correct. Select the two answers that are best in each case, and then fill in both of the corresponding circles on the answer sheet.
46. A wire is placed vertically and carries a current in the upward direction. A compass is located directly to the north of the wire. Which of the following observations can be made about the compass needle deflection? Select two answers.
(A) The deflection grows larger as the compass is moved farther from the wire.
(B) The deflection is to the west.
(C) The deflection grows larger as the current is increased.
(D) The deflection changes direction as the compass is moved upward.
47. An ideal gas is confined to a leak-proof box. What type of processes could occur to cause the gas to absorb heat, but have no net work done on the gas? Select two answers.
(A) Isothermal expansion to the original volume, then isobaric compression to the original volume
(B) A doubling of pressure at constant volume
(C) A doubling of pressure at constant volume, then doubling the volume at constant pressure, then halving the pressure at constant volume, and finally halving the volume at constant pressure to return to the initial state
(D) A doubling of pressure at constant volume, then finally halving the pressure at constant volume, then doubling the volume at constant pressure, then finally halving the volume at constant pressure to return to the initial state
48. Which of the following results of the photoelectric effect give support for the particle nature of light? Select two answers.
(A) There is a threshold frequency below which no electrons are emitted.