ACKNOWLEDGMENTS

The Princeton Review would like to extend special thanks to Peter Vaughan for his contributions to the 2018 edition of this book. We are also, as always, very appreciative of the time and attention given to each page by Harmony Quiroz, Liz Rutzel, and Gabriel Berlin.

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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.

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.

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.

HOW TO BEGIN

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.

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.

Questions 7 and 8 refer to the following diagram.

Questions 16-18 refer to the following diagram.

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.

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.

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.

STOP

END OF EXAM

Practice Test 1: Answers and Explanations

PRACTICE TEST 1 ANSWER KEY

SECTION I

SECTION II

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:

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:

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:

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:

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:

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 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:

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:

Part V

Content Review for the AP Physics 2 Exam

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 F_net = 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, P_atm, is about 101,300 Pa; this is 1 atmosphere.

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

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

F_g = 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, P_liquid = 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, P_atm—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

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

P_total = P_atm + 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 F_buoy. We calculate the magnitude of the buoyant force using Archimedes’ principle; in words, Archimedes’ principle says

When an object is partially or completely submerged in a fluid, the volume of the object submerged, which we call V_sub, 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:

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 ρ_object and its (total) volume is V, its weight will be mg = ρVg. The buoyant force it feels is ρ_fluidV_sub g. Setting these equal to each other, we find that

So, if ρ_object < ρ_fluid, 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 V_sub 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.

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.

Solution.

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 F_buoy + F_N = F_g, so F_N = F_g – F_buoy = 30 N – 15 N = 15 N. So the apparent weight of the brick, that is, the normal force it feels, is 15 N.

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, V_sub = V, and the buoyant force is

The weight of the balloon is

Because F_buoy > w, the net force on the balloon is upward and has magnitude

F_net = F_buoy – F_g = 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:

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.

Solution.

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.

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.

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 p_atm , Bernoulli’s Equation becomes

P₁ + ρv = P_atm + ρ gy₂ + ρv

Now, P₁ is the total pressure at point 1. Recall that gauge pressure is P_tot – P_atm, so P_gauge = P₁ – P_atm.

Therefore,

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

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

Section II: Free Response

Summary

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

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 = Nk_BT , where k_B is Boltzmann’s constant (k_B = 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 = NK_avg. Comparing this to the Ideal Gas Law we see that NK_avg = nRT. We can rewrite this equation in the form N_AK_avg = RT, since, by definition, N = nN_A. The ratio R/N_A is a fundamental constant of nature called Boltzmann’s constant (k_B = 1.38 × 10^–23 J/K), so our equation becomes

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 K_avg = m(v²)_avg, the equation above becomes m(v²)_avg = k_BT, 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, v_rms , so

v_rms =

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 v_rms. 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.

Solution. Temperature is a measure of the average kinetic energy. The velocity is determined from the following equation:

v_rms =

Since v_rms is proportional to the square root of T, the temperature must quadruple, again, assuming the temperature is given in kelvins.

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

F_net = (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:

The First Law of Thermodynamics

Simply put, the First Law of Thermodynamics is a statement of the Conservation of Energy that includes heat.

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

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.

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

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.

Solution.

Solution. Note that the initial and final states of the gas are the same as in the preceding example, but the path is different.

Q = ∆U ‒W = 4,500 J ‒ (‒1,000 J) = 5,500 J

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.

Solution. A process such as this, which begins and ends at the same state, is said to be cyclical.

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 ∆U_ad = 0, it must be true that Q_ad = –W_ad. Since W_ad equals –1,650 J, Q_ad 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:

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.

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, Q_net = –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 Q_H (H for hot), and the heat that is discharged into the low-temperature reservoir is denoted Q_C (C for cold). Because heat coming in is positive and heat going out is negative, Q_H is positive and Q_C is negative, and the net heat absorbed is Q_H + Q_C. Instead of writing Q_net in this way, it’s customary to write it as Q_H – |Q_C|, to show explicitly that Q_net is less than Q_H.

This is one of the forms of

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| = Q_H ‒|Q_C| = 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

Section II: Free Response

W =________________
Q =________________
U =________________

Summary

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

This is Coulomb’s Law. We interpret a negative F_E as an attraction between the charges and a positive F_E 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 =

Solution. The electric force between the proton and the electron is given by Coulomb’s Law:

The fact that F_E 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 F_1-on-2, the electric force on q₂ due to q₁, and F_3-on-2, the electric force on q₂ due to q₃:

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.

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 F_1y = F₁ sin 45° and F_2y = F₂ sin 45°. Also, the magnitude of F₁ equals that of F₂. So the net electric force on the negative charge is F_1y + F_2y = 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.

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 F_T must cancel out F_w and the horizontal component of F_T must cancel out F_E:

F_T cos θ = F_w   and   F_T sin θ = F_E

Dividing the second equation by the first, we get tan θ = F_E/F_w. 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 F_{on q}. Divide this force by the test charge q; the resulting vector is the electric field vector, E, at that location:

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:

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: E_total = 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 E_total 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.

Solution. From the definition of the electric field, we have the following equation:

F_{on q} = qE

Therefore, in this case, F_{on q} = qE = (3 × 10^–9 C)(400 N/C) = 1.2 × 10^–6 N.

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

Solution. Since the field E at this location is known, the force felt by q is easy to calculate:

F_{on q} = qE = (–5.0 × 10⁻¹² C)(8.0 × 10⁴ N/C to the right) = 4.0 × 10⁻⁷ N to the left

Solution. Remember that F_{on q} = qE. So if E = 0, then F_{on 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).

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 F_E to balance mg, we must have

Solution.

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.

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

Section II: Free Response

Summary

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 W_E is the work done by the electric force, then the change in the charge’s electrical potential energy is defined by

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 (∆U_G = −W_G).

Solution. Since the field is uniform, the electric force that the charge feels, F_E = qE, is constant. Since q is positive, F_E 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 W_E = F_Er = qEr, so the change in the electrical potential energy is

∆U_E = −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.

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, F_E = (−q)E, points in the direction opposite to the displacement, so the work it does is W_E = −F_Er = −qEr = −qEr. Thus, the change in electrical potential energy is positive: ∆U_E = −W_E = −(−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.

Solution. The electric force felt by the charge q is F_E = qE and this force is parallel to E, because q is positive. In this case, because F_E is not parallel to r (as it was in Example 1), we will use the more general definition of work for a constant force:

W_E = F_E · r = F_E r cos θ = qEr cos θ

But r cos θ = d, so

W_E = qEd and ∆U_E = − W_E = −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 W_E = F_Ed = qEd, just as before.

Electric Potential Energy from a Point Charge

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 r_A to r_B, 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.

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, W_E, 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 W_E = −∆U_E and the one we gave above for ∆U_E, we have

Therefore, the work that an external agent must do to bring q₃ into position is

ELECTRIC POTENTIAL

Let W_E 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, 2W_E. 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: U_E/q = U_E/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 r_A to a distance r_B 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 V_A → 0 as r_A → ∞ (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.

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.

Solution. If the charge always remains on a single equipotential, then, by definition, the potential, V, never changes. Therefore, ∆V = 0, so ∆U_E = 0. Since W_E = –∆U_E, 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.

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.

Solution. ∆U_E = 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: ΔU_{A → B} = qΔV_{A → 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: V_A = –1080 V, so ∆V_{A → B} = V_B – V_A = (–900 V) – (–1080 V) = +180 V. This means that the change in electrical potential energy as q moves from A to B is

∆V_{A → B} = qΔV_{A → 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

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.

Solution. Imagine a positive charge q moving from the positive plate to the negative plate:

Since the work done by the electric field is

W_E,+→– = F_Ed = 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),

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.

Solution.

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.

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

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

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 ∆U_E is equal to charge times voltage, we can write ΔU_E = Q · ΔV = QΔV. At the beginning of the charging process, when there was no charge on the capacitor, we had U_i = 0, so ∆U_E = U_f – U_i = U_f – 0 = U_f; therefore, we have

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:

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.

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 (E_i), 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: E_total = E + E_i, and E_total= E – E_i. 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

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:

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:

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

Section II: Free Response

Summary

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, v_d, would be relatively slow, on the order of a millimeter per second.

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

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:

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.

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.

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

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:

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.

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.

Solution.

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 = V_B – Ir = 20 V – 4 V = 16 V.

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/R_P = (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 R_S = 30 Ω + 30 Ω + 30 Ω = 90 Ω, and the current drawn is only I = V/R_S = (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.

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.

Solution. R₁ and R₂ are in parallel; their equivalent resistance is R_P, where

R_P is in series with R₃, so the equivalent resistance is:

R = R_P + 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:

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, C_P, 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, C_S, 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.

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 C_2-3) is in parallel with C₁. We can represent this as follows:

So, the first step is to find C_2-3:

Now this is in parallel with C₁, so the overall equivalent capacitance (C_1-2-3) 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.

Solution.

Solution.

The 6 Ω and 12 Ω resistors are in parallel and can be thought of as 4 Ω (from 1/R_eq = 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/R_eq = 1/R₁ + 1/R₂). The equivalent resistance of these two branches, which are in series with each other, is 6.67 Ω (from R_eq = 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).

The 6 Ω and 4 Ω resistors are in series and can be thought of as a 10 Ω resistor (from R_eq = R₁ + R₂). Similarly, the 12 Ω and 8 Ω resistors can be thought of as 20 Ω (again, from R_eq = 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.