﻿ ﻿Electrostatics - Review the Knowledge You Need to Score High - 5 Steps to a 5: AP Physics C (2016)

## 5 Steps to a 5: AP Physics C (2016)

### Electrostatics

IN THIS CHAPTER

Summary: An electric field provides a force on a charged particle. Electric potential, also called voltage, provides energy to a charged particle. Once you know the force or energy experienced by a charged particle, Newtonian mechanics (i.e., kinematics, conservation of energy, etc.) can be applied to predict the particle”s motion.

Key Ideas

The electric force on a charged particle is qE , regardless of what produces the electric field. The electric potential energy of a charged particle is qV .

Positive charges are forced in the direction of an electric field; negative charges, opposite the field.

Positive charges are forced from high to low potential; negative charges, low to high.

Point charges produce non-uniform electric fields. Parallel plates produce a uniform electric field between them.

Electric field is a vector, and electric potential is a scalar.

Relevant Equations

Electric force on a charge in an electric field:

F = qE

Electric field produced by a point charge 1 :

Electric field produced by parallel plates:

Electric potential energy in terms of voltage:

PE = qV

Voltage produced by a point charge:

Charge stored on a capacitor:

Q = CV

Capacitance of a parallel plate capacitor:

Electricity literally holds the world together. Sure, gravity is pretty important, too, but the primary reason that the molecules in your body stick together is because of electric forces. A world without electrostatics would be no world at all.

This chapter introduces a lot of the vocabulary needed to discuss electricity, and it focuses on how to deal with electric charges that aren”t moving: hence the name, electrostatics . We”ll look at moving charges in the next chapter, when we discuss circuits.

Electric Charge

All matter is made up of three types of particles: protons, neutrons, and electrons. Protons have an intrinsic property called “positive charge.” Neutrons don”t contain any charge, and electrons have a property called “negative charge.”

The unit of charge is the coulomb, abbreviated C. One proton has a charge of 1.6 × 10−19 coulombs.

Most objects that we encounter in our daily lives are electrically neutral—things like couches, for instance, or trees, or bison. These objects contain as many positive charges as negative charges. In other words, they contain as many protons as electrons.

When an object has more protons than electrons, though, it is described as “positively charged”; and when it has more electrons than protons, it is described as “negatively charged.” The reason that big objects like couches and trees and bison don”t behave like charged particles is because they contain so many bazillions of protons and electrons that an extra few here or there won”t really make much of a difference. So even though they might have a slight electric charge, that charge would be much too small, relatively speaking, to detect.

Tiny objects, like atoms, more commonly carry a measurable electric charge, because they have so few protons and electrons that an extra electron, for example, would make a big difference. Of course, you can have very large charged objects. When you walk across a carpeted floor in the winter, you pick up lots of extra charges and become a charged object yourself … until you touch a doorknob, at which point all the excess charge in your body travels through your finger and into the doorknob, causing you to feel a mild electric shock.

Electric charges follow a simple rule: Like charges repel; opposite charges attract . Two positively charged particles will try to get as far away from each other as possible, while a positively charged particle and a negatively charged particle will try to get as close as possible.

You can also have something called “induced charge.” An induced charge occurs when an electrically neutral object becomes polarized—when negative charges pile up in one part of the object and positive charges pile up in another part of the object. The drawing in Figure 18.1 illustrates how you can create an induced charge in an object.

Figure 18.1 Creation of an induced charge.

Electric Fields

Before we talk about electric fields, we”ll first define what a field, in general, is.

Field: A property of a region of space that can apply a force to objects found in that region of space

A gravitational field is a property of the space that surrounds any massive object. There is a gravitational field that you are creating and which surrounds you, and this field extends infinitely into space. It is a weak field, though, which means that it doesn”t affect other objects very much—you”d be surprised if everyday objects started flying toward each other because of gravitational attraction. The Earth, on the other hand, creates a strong gravitational field. Objects are continually being pulled toward the Earth”s surface due to gravitational attraction. However, the farther you get from the center of the Earth, the weaker the gravitational field, and, correspondingly, the weaker the gravitational attraction you would feel.

An electric field is a bit more specific than a gravitational field: it only affects charged particles.

Electric Field: A property of a region of space that applies a force to charged objects in that region of space. A charged particle in an electric field will experience an electric force.

Unlike a gravitational field, an electric field can either push or pull a charged particle, depending on the charge of the particle. Electric field is a vector; so, electric fields are always drawn as arrows.

Every point in an electric field has a certain value called, surprisingly enough, the “electric field value,” or E , and this value tells you how strongly the electric field at that point would affect a charge. The units of E are newtons/coulomb, abbreviated N/C.

Force of an Electric Field

The force felt by a charged particle in an electric field is described by a simple equation:

F = qE

In other words, the force felt by a charged particle in an electric field is equal to the charge of the particle, q , multiplied by the electric field value, E .

An electron, a proton, and a neutron are each placed in a uniform electric field of magnitude 60 N/C, directed to the right. What is the magnitude and direction of the force exerted on each particle?

The direction of the force on a positive charge is in the same direction as the electric field; the direction of the force on a negative charge is opposite the electric field.

Let”s try this equation on for size. Here”s a sample problem:

The solution here is nothing more than plug-and-chug into F = qE . Notice that we”re dealing with a uniform electric field—the field lines are evenly spaced throughout the whole region. This means that, no matter where a particle is within the electric field, it always experiences an electric field of exactly 60 N/C.

Also note our problem-solving technique. To find the magnitude of the force, we plug in just the magnitude of the charge and the electric field—no negative signs allowed! To find the direction of the force, use the reasoning in the box above (positive charges are forced in the direction of the E field, negative charges opposite the E field).

Let”s start with the electron, which has a charge of 1.6 × 10−19 C (no need to memorize, you can look this up on the constant sheet):

Now the proton:

And finally the neutron:

F = (0 C)(60 N/C) = 0 N

Notice that the proton feels a force in the direction of the electric field, but the electron feels the same force in the opposite direction.

Don”t state a force with a negative sign. Signs just indicate the direction of a force, anyway. So, just plug in the values for q and E , then state the direction of the force in words.

Electric Potential

When you hold an object up over your head, that object has gravitational potential energy. If you were to let it go, it would fall to the ground.

Similarly, a charged particle in an electric field can have electrical potential energy. For example, if you held a proton in your right hand and an electron in your left hand, those two particles would want to get to each other. Keeping them apart is like holding that object over your head; once you let the particles go, they”ll travel toward each other just like the object would fall to the ground.

In addition to talking about electrical potential energy, we also talk about a concept called electric potential.

Electric Potential: Potential energy provided by an electric field per unit charge; also called voltage

Electric potential is a scalar quantity. The units of electric potential are volts. 1 volt = 1 J/C.

Just as we use the term “zero of potential” in talking about gravitational potential, we can also use that term to talk about voltage. We cannot solve a problem that involves voltage unless we know where the zero of potential is. Often, the zero of electric potential is called “ground.”

Unless it is otherwise specified, the zero of electric potential is assumed to be far, far away. This means that if you have two charged particles and you move them farther and farther from each another, ultimately, once they”re infinitely far away from each other, they won”t be able to feel each other”s presence.

The electrical potential energy of a charged particle is given by this equation:

Here, q is the charge on the particle, and V is the voltage.

It is extremely important to note that electric potential and electric field are not the same thing. This example should clear things up:

Three points, labeled A, B, and C, are found in a uniform electric field. At which point will a positron (a positively charged version of an electron) have the greatest electrical potential energy?

Electric field lines point in the direction that a positive charge will be forced, which means that our positron, when placed in this field, will be pushed from left to right. So, just as an object in Earth”s gravitational field has greater potential energy when it is higher off the ground (think “mgh ”), our positron will have the greatest electrical potential energy when it is farthest from where it wants to get to. The answer is A.

We hope you noticed that, even though the electric field was the same at all three points, the electric potential was different at each point.

A positron is given an initial velocity of 6 × 106 m/s to the right. It travels into a uniform electric field, directed to the left. As the positron enters the field, its electric potential is zero. What will be the electric potential at the point where the positron has a speed of 1 × 106 m/s?

This is a rather simple conservation of energy problem, but it”s dressed up to look like a really complicated electricity problem.

As with all conservation of energy problems, we”ll start by writing our statement of conservation of energy.

Ki + Ui = Kf + Uf

Next, we”ll fill in each term with the appropriate equations. Here the potential energy is not due to gravity (mgh ), nor due to a spring (1/2 kx 2 ). The potential energy is electric, so it should be written as qV .

½ mvi 2 + qVi = ½ mvf 2 + qVf

Finally, we”ll plug in the corresponding values. The mass of a positron is exactly the same as the mass of an electron, and the charge of a positron has the same magnitude as the charge of an electron, except a positron”s charge is positive. Both the mass and the charge of an electron are given to you on the “constants sheet.” Also, the problem told us that the positron”s initial potential V i was zero.

½ (9.1 × 10−31 kg)(6 × 106 m/s)2 + (1.6 × 10−19 C)(0) =
½ (9.1 × 10−31 kg)(1 × 106 m/s)2 + (1.6 × 10−19 C)(Vf )

Solving for Vf , we find that Vf is about 100 V.

For forces , a negative sign simply indicates direction. For potentials, though, a negative sign is important. −300 V is less than −200 V, so a proton will seek out a −300 V position in preference to a −200 V position. So, be careful to use proper + and − signs when dealing with potential.

Just as you can draw electric field lines, you can also draw equipotential lines.

Equipotential Lines: Lines that illustrate every point at which a charged particle would experience a given potential

Figure 18.2 shows a few examples of equipotential lines (shown with solid lines) and their relationship to electric field lines (shown with dotted lines):

Figure 18.2 Two examples of equipotential lines (in bold) and electric field lines (dotted).

On the left in Figure 18.2 , the electric field points away from the positive charge. At any particular distance away from the positive charge, you would find an equipotential line that circles the charge—we”ve drawn two, but there are an infinite number of equipotential lines around the charge. If the potential of the outermost equipotential line that we drew was, say, 10 V, then a charged particle placed anywhere on that equipotential line would experience a potential of 10 V.

On the right in Figure 18.2 , we have a uniform electric field. Notice how the equipotential lines are drawn perpendicular to the electric field lines. In fact, equipotential lines are always drawn perpendicular to electric field lines, but when the field lines aren”t parallel (as in the drawing on the left), this fact is harder to see.

Moving a charge from one equipotential line to another takes energy. Just imagine that you had an electron and you placed it on the innermost equipotential line in the drawing on the left. If you then wanted to move it to the outer equipotential line, you”d have to push pretty hard, because your electron would be trying to move toward, and not away from, the positive charge in the middle.

In the diagram above, point A and point B are separated by a distance of 30 cm. How much work must be done by an external force to move a proton from point A to point B?

The potential at point B is higher than at point A ; so moving the positively charged proton from A to B requires work to change the proton”s potential energy. The question here really is asking how much more potential energy the proton has at point B .

Well, potential energy is equal to qV ; here, q is 1.6 × 10−19 C, the charge of a proton. The potential energy at point A is (1.6 × 10−19 C)(50 V) = 8.0 × 10−18 J; the potential energy at point B is (1.6 × 10−19 C)(60 V) = 9.6 × 10−18J. Thus, the proton”s potential is 1.6 × 10−18 J higher at point B , so it takes 1.6 × 10−18 J of work to move the proton there.

Um, didn”t the problem say that points A and B were 30 cm apart? Yes, but that”s irrelevant. Since we can see the equipotential lines, we know the potential energy of the proton at each point; the distance separating the lines is irrelevant.

Special Geometries for Electrostatics

There are two situations involving electric fields that are particularly nice because they can be described with some relatively easy formulas. Let”s take a look:

Parallel Plates

If you take two metal plates, charge one positive and one negative, and then put them parallel to each other, you create a uniform electric field in the middle, as shown in Figure 18.3 :

Figure 18.3 Electric field between charged, parallel plates.

The electric field between the plates has a magnitude of

V is the voltage difference between the plates, and d is the distance between the plates. Remember, this equation only works for parallel plates.

Charged parallel plates can be used to make a capacitor , which is a charge-storage device. When a capacitor is made from charged parallel plates, it is called, logically enough, a “parallel-plate capacitor.” A schematic of this type of capacitor is shown in Figure 18.4 .

Figure 18.4 Basic parallel-plate capacitor.

The battery in Figure 18.4 provides a voltage across the plates; once you”ve charged the capacitor, you disconnect the battery. The space between the plates prevents any charges from jumping from one plate to the other while the capacitor is charged. When you want to discharge the capacitor, you just connect the two plates with a wire.

The amount of charge that each plate can hold is described by the following equation:

Q is the charge on each plate, C is called the “capacitance,” and V is the voltage across the plates. The capacitance is a property of the capacitor you are working with, and it is determined primarily by the size of the plates and the distance between the plates, as well as by the material that fills the space between the plates. The units of capacitance are farads, abbreviated F; 1 coulomb/volt = 1 farad.

The only really interesting thing to know about parallel-plate capacitors is that their capacitance can be easily calculated. The equation is:

In this equation, A is the area of each plate (in m2 ), and d is the distance between the plates (in m). The term ε 0 (pronounced “epsilon-naught”) is called the “permittivity of free space.” This term will show up again soon, when we introduce the constant k . The value of ε 0 is 8.84 × 10−12 C/V·m, which is listed on the constants sheet.

Capacitors become important when we work with circuits. So we”ll see them again in Chapter 19 .

Point Charges

As much as the writers of the AP exam like parallel plates, they love point charges. So you”ll probably be using these next equations quite a lot on the test.

But, please don”t go nuts… . The formulas for force on a charge in an electric field (F = qE ) and a charge”s electrical potential energy (PE = qV ) are your first recourse, your fundamental tools of electrostatics. On the AP exam, most electric fields are NOT produced by point charges! Only use the equations in this section when you have convinced yourself that a point charge is creating the electric field or the voltage in question.

First, the value of the electric field at some distance away from a point charge:

Q is the charge of your point charge, ε 0 is the permittivity of free space (on the table of information), and r is the distance away from the point charge. 2 The field produced by a positive charge points away from the charge; the field produced by a negative charge points toward the charge. When finding an electric field with this equation, do NOT plug in the sign of the charge or use negative signs at all .

Second, the electric potential at some distance away from a point charge:

When using this equation, you must include a + or − sign on the charge creating the potential. (See Figure 18.5 .)

Figure 18.5 Electric field produced by point charges.

And third, the force that one point charge exerts on another point charge:

In this equation, Q 1 is the charge of one of the point charges, and Q 2 is the charge on the other one. This equation is known as Coulomb”s Law.

To get comfortable with these three equations, we”ll provide you with a rather comprehensive problem.

Two point charges, labeled “A” and “B”, are located on the x -axis. “A” has a charge of —3 μC, and “B” has a charge of +3 μC. Initially, there is no charge at point P , which is located on the y -axis as shown in the diagram.

(a) What is the electric field at point P due to charges “A” and “B”?

(b) If an electron were placed at point P , what would be the magnitude and direction of the force exerted on the electron?

(c) What is the electric potential at point P due to charges “A” and “B”?

Yikes! This is a monster problem. But if we take it one part at a time, you”ll see that it”s really not too bad.

Part 1—Electric Field

Electric field is a vector quantity. So we”ll first find the electric field at point P due to charge “A,” then we”ll find the electric field due to charge “B,” and then we”ll add these two vector quantities. One note before we get started: to find r , the distance between points P and “A” or between P and “B,” we”ll have to use the Pythagorean theorem. We won”t show you our work for that calculation, but you should if you were solving this on the AP exam.

Note that we didn”t plug in any negative signs! Rather, we calculated the magnitude of the electric field produced by each charge, and showed the direction on the diagram.

Now, to find the net electric field at point P , we must add the electric field vectors. This is made considerably simpler by the recognition that the y -components of the electric fields cancel … both of these vectors are pointed at the same angle, and both have the same magnitude. So, let”s find just the x -component of one of the electric field vectors:

E x = E cos θ , where θ is measured from the horizontal.

Some quick trigonometry will find cos θ … since cos θ is defined as , inspection of the diagram shows that . So, the horizontal electric field Ex = (510 m) … this gives 140 N/C.

And now finally, there are TWO of these horizontal electric fields adding together to the left—one due to charge “A” and one due to charge “B”. The total electric field at point P , then, is

280 N/C, to the left.

Part 2—Force

The work that we put into Part 1 makes this part easy. Once we have an electric field, it doesn”t matter what caused the E field—just use the basic equation F = qE to solve for the force on the electron, where q is the charge of the electron. So,

F = (1.6 × 10−19 C) 280 N/C = 4.5 × 10−17 N.

The direction of this force must be OPPOSITE the E field because the electron carries a negative charge; so, to the right .

Part 3—Potential

The nice thing about electric potential is that it is a scalar quantity, so we don”t have to concern ourselves with vector components and other such headaches.

The potential at point P is just the sum of these two quantities. V = zero!

Notice that when finding the electric potential due to point charges, you must include negative signs … negative potentials can cancel out positive potentials, as in this example.

Gauss”s Law

A more thorough understanding of electric fields comes from Gauss”s law. But before looking at Gauss”s law itself, it is necessary to understand the concept of electric flux.

Electric flux: The amount of electric field that penetrates an area

Φ E = E · A

The electric flux, Φ E , equals the electric field multiplied by the surface area through which the field penetrates.

Flux only exists if the electric field lines penetrate straight through a surface. (Or, if the electric field lines have a component that”s perpendicular to a surface.) If an electric field exists parallel to a surface, there is zero flux through that surface. One way to think about this is to imagine that electric field lines are like arrows, and the surface you”re considering is like an archer”s bull”s-eye. There would be flux if the arrows hit the target; but if the archer is standing at a right angle to the target (so that his arrows zoom right on past the target without even nicking it) there”s no flux.

In words, Gauss”s law states that the net electric flux through a closed surface is equal to the charge enclosed divided by ε 0 . This is often written as

How and When to Use Gauss”s Law

Gauss”s law is valid the universe over. However, in most cases Gauss”s law is not in any way useful—no one expects you to be able to evaluate a three-dimensional integral with a dot product! ONLY use Gauss”s law when the problem has spherical, cylindrical, or planar symmetry.

First, identify the symmetry of the problem. Then draw a closed surface, called a “Gaussian surface,” that the electric field is everywhere pointing straight through. A Gaussian surface isn”t anything real … it”s just an imaginary closed surface that you”ll use to solve the problem. The net electric flux is just E times the area of the Gaussian surface you drew.

You should NEVER, ever, try to evaluate the integralE · dA in using Gauss”s law!

Here is an example problem.

Consider a metal sphere of radius R that carries a surface charge density σ . What is the magnitude of the electric field as a function of the distance from the center of the sphere?

There are two possibilities here. One possibility is that the function describing the electric field will be a smooth, continuous function. The other possibility is that the function inside the sphere will be different from the function outside the sphere (after all, they”re different environments—inside the sphere you”re surrounded by charge, and outside the sphere you”re not). So we”ll assume that the function is different in each environment, and we”ll consider the problem in two parts: inside the sphere and outside the sphere. If it turns out that the function is actually smooth and continuous, then we”ll have done some extra work, but we”ll still get the right answer.

Inside the sphere, draw a Gaussian sphere of any radius. No charge is enclosed, because in a conductor, all the charges repel each other until all charge resides on the outer edge. So, by Gauss”s law since the enclosed charge is zero, the term E·A has to be zero as well. A refers to the area of the Gaussian surface you drew, which sure as heck has a surface area.

The electric field inside the conducting sphere must be zero everywhere . This is actually a general result that you should memorize—the electric field inside a conductor is always zero.

Outside the sphere, draw a Gaussian sphere of radius r . This sphere, whatever its radius, always encloses the full charge of the conductor. What is that charge? Well, σ represents the charge per area of the conductor, and the area of the conductor is 4πR 2 . So the charge on the conductor is σ 4πR 2 . Now, the Gaussian surface of radius r has area 4πr 2 . Plug all of this into Gauss”s law:

E ·4πr 2 = σ 4πR 2 /ε 0 .

All the variables are known, so just solve for electric field: E = σR 2 /ε 0 r 2 .

Now we can state our answer, where r is the distance from the center of the charged sphere:

What is interesting about this result? We solved in terms of the charge density σ on the conductor. If we solve instead in terms of Q , the total charge on the conductor, we recover the formula for the electric field of a point charge:

Practice Problems

Multiple Choice:

Questions 1 and 2

Two identical positive charges Q are separated by a distance a , as shown above.

1 . What is the electric field at a point halfway between the two charges?

(A) kQ /a 2

(B) 2kQ /a 2

(C) zero

(D) kQQ /a 2

(E) 2kQ /a

2 . What is the electric potential at a point halfway between the two charges?

(A) kQ /a

(B) 2kQ /a

(C) zero

(D) 4kQ /a

(E) 8kQ /a

Questions 3 and 4

The diagram above shows two parallel metal plates that are separated by distance d . The potential difference between the plates is V . Point A is twice as far from the negative plate as is point B .

3 . Which of the following statements about the electric potential between the plates is correct?

(A) The electric potential is the same at points A and B .

(B) The electric potential is two times larger at A than at B .

(C) The electric potential is two times larger at B than at A .

(D) The electric potential is four times larger at A than at B .

(E) The electric potential is four times larger at B than at A .

4 . Which of the following statements about the electric field between the plates is correct?

(A) The electric field is the same at points A and B .

(B) The electric field is two times larger at A than at B .

(C) The electric field is two times larger at B than at A .

(D) The electric field is four times larger at A than at B .

(E) The electric field is four times larger at B than at A .

5 . A very long cylindrical conductor is surrounded by a very long cylindrical conducting shell, as shown above. A length L of the inner conductor carries positive charge Q . The same length L of the outer shell carries total charge −3Q . How much charge is distributed on a length L of the outside surface of the outer shell?

(A) none

(B) −Q

(C) −2Q

(D) −3Q

(E) −4Q

Free Response:

6 . Two conducting metal spheres of different radii, as shown above, each have charge −Q .

(a) Consider one of the spheres. Is the charge on that sphere likely to clump together or to spread out? Explain briefly.

(b) Is charge more likely to stay inside the metal spheres or on the surface of the metal spheres? Explain briefly.

(c) If the two spheres are connected by a metal wire, will charge flow from the big sphere to the little sphere, or from the little sphere to the big sphere? Explain briefly.

(d) Which of the following two statements is correct? Explain briefly.

1. If the two spheres are connected by a metal wire, charge will stop flowing when the electric field at the surface of each sphere is the same.
2. If the two spheres are connected by a metal wire, charge will stop flowing when the electric potential at the surface of each sphere is the same.

(e) Explain how the correct statement you chose from part (d) is consistent with your answer to (c).

Solutions to Practice Problems

1 . C— Electric field is a vector . Look at the field at the center due to each charge. The field due to the left-hand charge points away from the positive charge; i.e., to the right; the field due to the right-hand charge points to the left. Because the charges are equal and are the same distance from the center point, the fields due to each charge have equal magnitudes. So the electric field vectors cancel! E = 0.

2 . D— Electric potential is a scalar . Look at the potential at the center due to each charge: Each charge is distance a /2 from the center point, so the potential due to each is kQ /(a /2), which works out to 2kQ /a . The potentials due to both charges are positive, so add these potentials to get 4kQ /a .

3 . B— If the potential difference between plates is, say, 100 V, then we could say that one plate is at +100 V and the other is at zero V. So, the potential must change at points in between the plates. The electric field is uniform and equal to V /d (d is the distance between plates). Thus, the potential increases linearly between the plates, and A must have twice the potential as B.

4 . A— The electric field by definition is uniform between parallel plates. This means the field must be the same everywhere inside the plates.

5 . C— We have cylindrical symmetry, so use Gauss”s law. Consider a Gaussian surface drawn within the outer shell. Inside a conducting shell, the electric field must be zero, as discussed in the chapter. By Gauss”s law, this means the Gaussian surface we drew must enclose zero net charge. Because the inner cylinder carries charge +Q , the inside surface of the shell must carry charge −Q to get zero net charge enclosed by the Gaussian surface. What happens to the −2Q that is left over on the conducting shell? It goes to the outer surface.

6 . (a) Like charges repel, so the charges are more likely to spread out from each other as far as possible.

(b) “Conducting spheres” mean that the charges are free to move anywhere within or onto the surface of the spheres. But because the charges try to get as far away from each other as possi-ble, the charge will end up on the surface of the spheres. This is actually a property of conductors—charge will always reside on the surface of the conductor, not inside.

(c) Charge will flow from the smaller sphere to the larger sphere. Following the reasoning from parts (a) and (b), the charges try to get as far away from each other as possible. Because both spheres initially carry the same charge, the charge is more concentrated on the smaller sphere; so the charge will flow to the bigger sphere to spread out. (The explanation that negative charge flows from low to high potential, and that potential is less negative at the surface of the bigger sphere, is also acceptable here.)

(d) The charge will flow until the potential is equal on each sphere. By definition, negative charges flow from low to high potential. So, if the potentials of the spheres are equal, no more charge will flow.

(e) The potential at the surface of each sphere is −kQ /r , where r is the radius of the sphere. Thus, the potential at the surface of the smaller sphere is initially more negative, and the charge will initially flow low-to-high potential to the larger sphere.

Rapid Review

• Matter is made of protons, neutrons, and electrons. Protons are positively charged, neutrons have no charge, and electrons are negatively charged.
• Like charges repel, opposite charges attract.
• An induced charge can be created in an electrically neutral object by placing that object in an electric field.
• Electric field lines are drawn from positive charges toward negative charges. Where an electric field is stronger, the field lines are drawn closer together.
• The electric force on an object depends on both the object”s charge and the electric field it is in.
• Unless stated otherwise, the zero of electric potential is at infinity.
• Equipotential lines show all the points where a charged object would feel the same electric force. They are always drawn perpendicular to electric field lines.
• The electric field between two charged parallel plates is constant. The electric field around a charged particle depends on the distance from the particle.
• Gauss”s law says that the net electric flux through a closed surface is equal to the charge enclosed divided byε 0 . To solve a problem using Gauss”s law, look for planar, cylindrical, or spherical symmetry.

1 The actual equation sheet for the AP exam includes this equation: . To interpret this (and other point charge equations), recognize that this equation for the force between two point charges is an amalgamation of the first two equations above: combine F = qE with and you get .

2 For calculations, it might be easier to recognize that . This value, often labeled as k , shows up repeatedly in point-charge problems.

CHAPTER 18

Electrostatics

1 . A cubic Gaussian surface of side length x sits centered in a uniform electric field (E) between two capacitor plates. The net electric flux through the Gaussian surface is

(A) 0

(B) Ex2

(C) 2Ex2

(D) 4Ex2

(E) 6Ex2

2 . Two spheres with charges +Q and –Q of equal magnitude are placed a vertical distance of d apart on the y -axis, as shown in the figure. A third charge of +q is brought from a distance of x , where x >>d , horizontally toward the midpoint between +Q and –Q. The net force on +q as it is moved to the left along the x -axis

(A) increases and continues in the same direction.

(B) increases and changes direction.

(C) remains the same magnitude and continues in the same direction.

(D) decreases and continues in the same direction.

(E) decreases and changes direction.

3 . An electron of charge –e and mass m is launched with a velocity of v 0 through a small hole in the right plate of a parallel plate capacitor toward the opposite plate a distance of d away. The electric potential of both plates are equal in magnitude but opposite in sign ±V , as shown in the figure. The velocity of the electron as it reaches the left plate is

(A)

(B)

(C)

(D)

(E)

4 . Two metal spheres with different radii but the same charge (+Q) are separated by a distance of d , which is much greater than their combined radii. The two spheres exert a force of F on each other. They are brought into contact and then again placed a distance of d apart. The new force between the spheres is

(A)

(B)

(C)

(D)

(E)

1 . A —The uniform electric field is downward from the positive plate to the negative plate. The top surface has a negative electric flux of –Ex2 because the E-field is passing into the closed Gaussian surface. The bottom surface of the cube will have a positive electric flux of +Ex2 as the E-field is passing out of the closed Gaussian surface. The four sides of the cube have no flux because the E-field does not pass through the Gaussian surface. Thus, the net flux for the entire surface is zero. A more elegant solution method is to use Gauss”s law: . All of the charge is spread across the capacitor plates. Since there is no charge enclosed in the Gaussian surface, the net flux must be zero.

2 . A —In the original location x , the two electrostatic forces on +q add up, as shown in the figure on the left. As the charge +q is moved toward +Q and –Q, the electrostatic forces increase in strength, and their direction shifts as shown in the figure on the right. Thus, the net force increases in strength but continues to point directly downward due to the symmetry of the charge arrangement.

3 . B —Using conservation of energy, the change in electrostatic potential energy plus the initial kinetic energy equals the final kinetic energy:

Note: Please be careful with your signs of the potential and electron charge!

4 . C —The original force between the spheres is

When the spheres are brought into contact, they share a charge until they have the same electric potential:

In addition, we know that Q1 and Q2 must add up to be the total original charge of 2Q:

Q 1 + Q 2 = 2Q

Substituting, we get

Calculating the new electrostatic force, we get the following:

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