Ch 19 - Electric Potential Energy - Electric Potential (569-598)

C H A P T E R 19

ELECTRIC POTENTIAL ENERGY

AND THE ELECTRIC POTENTIAL

Berkeley, California, inventors

Tom Burchill, Jonah Most, and Daniel

Holtmann-Rice (with a portable solar

panel on his back) show their “Solar

Board,” a three-person skateboard

powered by the sun. Sunlight

penetrates the solar cells of the

panel and provides the energy that

separates positive and negative

charges in the materials from which

the cells are made. Thus, each solar

cell develops positive and negative

terminals, much like the terminals of a

battery and, in effect, converts solar

energy into electric energy that

powers the skateboard. Electric

potential energy and the related

concept of electric potential are

the subjects of this chapter.

(© Lonny Shavelson/Zuma Press)

569

In Chapter 18 we discussed the electrostatic force that two point charges exert on

each other, the magnitude of which is F � k �q1��q2�/r 2. The form of this equation is simi-

lar to that for the gravitational force that two particles exert on each other, which is F �Gm1m2/r 2, according to Newton’s law of universal gravitation (see Section 4.7). Both of

these forces are conservative and, as Section 6.4 explains, a potential energy can be asso-

ciated with a conservative force. Thus, an electric potential energy exists that is analogous

to the gravitational potential energy. To set the stage for a discussion of the electric poten-

tial energy, let’s review some of the important aspects of the gravitational counterpart.

Figure 19.1, which is essentially Figure 6.9, shows a basketball of mass m falling from

point A to point B. The gravitational force, m , is the only force acting on the ball, where gis the magnitude of the acceleration due to gravity. As Section 6.3 discusses, the work WAB

done by the gravitational force when the ball falls from a height of hA to a height of hB is

WAB � mghA � mghB � GPEA � GPEB (6.4)

potential energy, potential energy,

GPEA GPEB

Recall that the quantity mgh is the gravitational potential energy* of the ball, GPE � mgh(Equation 6.5), and represents the energy that the ball has by virtue of its position relative

to the surface of the earth. Thus, the work done by the gravitational force equals the initial

gravitational potential energy minus the final gravitational potential energy.

Final

123

gravitational

Initial

123

gravitational

gB

POTENTIAL ENERGY

19.1

hA

F = mg

F = mghB

A

B

B

B

Figure 19.1 Gravity exerts a force,

� m , on the basketball of mass m.

Work is done by the gravitational force

as the ball falls from A to B.

gBFB

*The gravitational potential energy is now being denoted by GPE to distinguish it from the electric potential

energy EPE.

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Figure 19.2 clarifies the analogy between electric and gravitational potential ener-

gies. In this drawing a positive test charge �q0 is situated at point A between two oppo-

sitely charged plates. Because of the charges on the plates, an electric field exists in the

region between the plates. Consequently, the test charge experiences an electric force,

� q0 (Equation 18.2), that is directed downward, toward the lower plate. (The gravi-

tational force is being neglected here.) As the charge moves from A to B, work is done by

this force, in a fashion analogous to the work done by the gravitational force in Figure

19.1. The work WAB done by the electric force equals the difference between the electric

potential energy EPE at A and the electric potential energy at B:

WAB � EPEA � EPEB (19.1)

This expression is similar to Equation 6.4. The path along which the test charge moves

from A to B is of no consequence because the electric force is a conservative force, and so

the work WAB is the same for all paths (see Section 6.4).

EB

FB

EB

570 CHAPTER 19 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL

q0

F = q0E

q0

F = q0E

A

B

− − − − − − − − − − − − − −

+ + + + + + + + + + + + + +

B

B

Figure 19.2 Because of the electric

field , an electric force, � q0 , is

exerted on a positive test charge �q0.

Work is done by the force as the charge

moves from A to B.

EB

FB

EB

Since the electric force is � q0 , the work that it does as the charge moves from

A to B in Figure 19.2 depends on the charge q0. It is useful, therefore, to express this work

on a per-unit-charge basis, by dividing both sides of Equation 19.1 by the charge:

(19.2)

Notice that the right-hand side of this equation is the difference between two terms, each

of which is an electric potential energy divided by the test charge, EPE/q0. The quantity

EPE/q0 is the electric potential energy per unit charge and is an important concept in elec-

tricity. It is called the electric potential or, simply, the potential and is referred to with the

symbol V, as in Equation 19.3.

WAB

q0

�EPEA

q0

�EPEB

q0

EB

FB

THE ELECTRIC POTENTIAL DIFFERENCE

19.2

DEFINITION OF ELECTRIC POTENTIAL

The electric potential V at a given point is the electric potential energy EPE of a small

test charge q0 situated at that point divided by the charge itself:

(19.3)

SI Unit of Electric Potential: joule/coulomb � volt (V)

V �EPE

q0

The SI unit of electric potential is a joule per coulomb, a quantity known as a volt. The

name honors Alessandro Volta (1745–1827), who invented the voltaic pile, the forerunner

of the battery. In spite of the similarity in names, the electric potential energy EPE and the

electric potential V are not the same. The electric potential energy, as its name implies, is

an energy and, therefore, is measured in joules. In contrast, the electric potential is an energyper unit charge and is measured in joules per coulomb, or volts.

We can now relate the work WAB done by the electric force when a charge q0 moves

from A to B to the potential difference VB � VA between the points. Combining Equations

19.2 and 19.3, we have:

(19.4)

Often, the “delta” notation is used to express the difference (final value minus initial value)

in potentials and in potential energies: �V � VB � VA and �(EPE) � EPEB � EPEA. In

terms of this notation, Equation 19.4 takes the following more compact form:

(19.4)�V ��(EPE)

q0

��WAB

q0

VB � VA �EPEB

q0

�EPEA

q0

��WAB

q0

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19.2 THE ELECTRIC POTENTIAL DIFFERENCE 571

Neither the potential V nor the potential energy EPE can be determined in an absolute

sense, because only the differences �V and �(EPE) are measurable in terms of the work

WAB. The gravitational potential energy has this same characteristic, since only the value

at one height relative to the value at some reference height has any significance. Example 1

emphasizes the relative nature of the electric potential.

In Figure 19.2, the work done by the electric force as the test charge (q0 � �2.0 � C)

moves from A to B is WAB � �5.0 � J. (a) Find the value of the difference,

�(EPE) � EPEB � EPEA, in the electric potential energies of the charge between these points.

(b) Determine the potential difference, �V � VB � VA, between the points.

Reasoning The work done by the electric force when the charge travels from A to B is WAB �EPEA � EPEB, according to Equation 19.1. Therefore, the difference in the electric potential

energies (final value minus initial value) is �(EPE) � EPEB � EPEA � �WAB. The potential

difference, �V � VB � VA, is the difference in the electric potential energies divided by the

charge q0, according to Equation 19.4.

Solution (a) The difference in the electric potential energies of the charge between the points

A and B is

(19.1)

The fact that EPEB � EPEA is negative means that the charge has a higher electric potential

energy at A than at B.

(b) The potential difference �V between A and B is

(19.4)

The fact that VB � VA is negative means that the electric potential is higher at A than at B.

� �V14243

�25 VVB � VA �EPEB � EPEA

q0

��5.0 � 10�5 J

2.0 � 10�6 C�

� �(EPE)

144442444443�5.0 � 10�5 JEPEB � EPEA � �WAB �

10�5

10�6

Example 1 Work, Electric Potential Energy, and Electric Potential

�

This electric car stores energy from the

charging station in its battery pack. The

voltage of the battery pack and the

magnitude of the charge sent through it

by the charging station determine how

much energy is stored. (© AP/Wide

World Photos)

The potential difference between two points is measured in volts and, therefore, is of-

ten referred to as a “voltage.” Everyone has heard of “voltage” because, as we will see in

Chapter 20, it is frequently used in connection with everyday devices. For example, your

TV requires a “voltage” of 120 V (which is applied between the two prongs of the plug on

the power cord when it is inserted into an electrical wall outlet), and your cell phone and

laptop computer use batteries that provide, for example, “voltages” of 1.5 V or 9 V (which

exist between the two battery terminals).

In Figure 19.1 the speed of the basketball increases as it falls from A to B. Since point

A has a greater gravitational potential energy than point B, we see that an object of mass

m accelerates when it moves from a region of higher potential energy toward a region of

lower potential energy. Likewise, the positive charge in Figure 19.2 accelerates as it moves

from A to B because of the electric repulsion from the upper plate and the attraction to the

lower plate. Since point A has a higher electric potential than point B, we conclude that apositive charge accelerates from a region of higher electric potential toward a region oflower electric potential. On the other hand, a negative charge placed between the plates in

Figure 19.2 behaves in the opposite fashion, since the electric force acting on the negative

charge is directed opposite to the electric force acting on the positive charge. A negativecharge accelerates from a region of lower potential toward a region of higher potential.The next example illustrates the way positive and negative charges behave.

Problem-solving insight


Three points, A, B, and C, are located along a horizontal line, as Figure 19.3 illustrates. A posi-

tive test charge is released from rest at A and accelerates toward B. Upon reaching B, the test

charge continues to accelerate toward C. Assuming that only motion along the line is possible,

what will a negative test charge do when it is released from rest at B? A negative test charge will

(a) accelerate toward C, (b) remain stationary, (c) accelerate toward A.

Conceptual Example 2 How Positive and Negative Charges Accelerate�

�

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As a familiar application of electric potential energy and electric potential, Figure 19.4

shows a 12-V automobile battery with a headlight connected between its terminals. The

positive terminal, point A, has a potential that is 12 V higher than the potential at the neg-

ative terminal, point B; in other words, VA � VB � 12 V. Positive charges would be repelled

from the positive terminal and would travel through the wires and headlight toward the

negative terminal.* As the charges pass through the headlight, virtually all their potential

energy is converted into heat, which causes the filament to glow “white hot” and emit light.

When the charges reach the negative terminal, they no longer have any potential energy.

The battery then gives the charges an additional “shot” of potential energy by moving them

to the higher-potential positive terminal, and the cycle is repeated. In raising the potential

energy of the charges, the battery does work on them and draws from its reserve of chem-

ical energy to do so. Example 3 illustrates the concepts of electric potential energy and

electric potential as applied to a battery.


*Historically, it was believed that positive charges flow in the wires of an electric circuit. Today, it is known

that negative charges flow in wires from the negative toward the positive terminal. Here, however, we follow

the customary practice of describing the flow of negative charges by specifying the opposite but equivalent

flow of positive charges. This hypothetical flow of positive charges is called the “conventional electric current,”

as we will see in Section 20.1.

Reasoning A positive charge accelerates from a region of higher potential toward a region of

lower potential. A negative charge behaves in an opposite manner, because it accelerates from

a region of lower potential toward a region of higher potential.

Answers (a) and (b) are incorrect. The positive charge accelerates from A to B and then from

B to C. A negative charge placed at B also accelerates, but in a direction opposite to that of the

positive charge. Therefore, a negative charge placed at B will not remain stationary, nor will it

accelerate toward C.

Answer (c) is correct. Since the positive charge accelerates from A to B, the potential at A must

exceed the potential at B. And since the positive test charge accelerates from B to C, the potential

at B must exceed the potential at C. The potential at point B, then, must lie between the potential

at points A and C, as Figure 19.3 illustrates. When the negative test charge is released from rest

at B, it will accelerate toward the region of higher potential, so it will begin moving toward A.

Higherpotential

Lowerpotential

A B C

Figure 19.3 The electric potentials at

points A, B, and C are different. Under

the influence of these potentials,

positive and negative charges accelerate

in opposite directions.

Figure 19.4 A headlight connected to a

12-V battery.

�

Example 3 Operating a Headlight

The wattage of the headlight in Figure 19.4 is 60.0 W. Determine the number of particles, each carrying a charge of

1.60 � 10�19 C (the magnitude of the charge on an electron), that pass between the terminals of the 12-V car battery when

the headlight burns for one hour.

Reasoning The number of particles is the total charge that passes between the battery terminals in one hour divided by the

magnitude of the charge on each particle. The total charge is the amount of charge needed to convey the energy used by the

headlight in one hour. This energy is related to the wattage of the headlight, which specifies the power or rate at which energy

is used, and the time the light is on.

Knowns and Unknowns The following table summarizes the data provided:

Description Symbol Value Comment

Wattage of headlight P 60.0 W

Charge magnitude per particle e 1.60 � 10�19 C

Electric potential difference between battery terminals VA � VB 12 V See Figure 19.4.

Time headlight is on t 3600 s One hour.

Unknown VariableNumber of particles n ?

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

A

B

12-V battery

+−

Filament

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STEP 1

Modeling the Problem

(1)n �q0

eThe Number of Particles The number n of particles is the total charge q0

that passes between the battery terminals in one hour divided by the magnitude e of the

charge on each particle, as expressed by Equation 1 at the right. The value of e is given.

To evaluate q0, we proceed to Step 2.

The Total Charge Provided by the Battery The battery must supply the

total energy used by the headlight in one hour. The battery does this by supplying the

charge q0 to convey this energy. The energy is the difference between the electric

potential energy EPEA at terminal A and the electric potential energy EPEB at terminal

B of the battery (see Figure 19.4). According to Equation 19.4, this total energy is

EPEA � EPEB � q0(VA � VB), where VA � VB is the electric potential difference

between the battery terminals. Solving this expression for q0 gives

which can be substituted into Equation 1, as shown at the right. As the data table shows,

a value is given for VA � VB. In Step 3, we determine a value for EPEA � EPEB.

The Energy Used by the Headlight The rate at which the headlight uses en-

ergy is the power P or wattage of the headlight. According to Equation 6.10b, the power is

the total energy EPEA � EPEB divided by the time t, so that P � (EPEA � EPEB)/t. Solving

for the total energy gives

Since P and t are given, we substitute this result into Equation 2, as indicated at the right.

Solution Combining the results of each step algebraically, we find that

The number of particles that pass between the battery terminals in one hour is

Related Homework: Problems 4, 5

1.1 � 1023n �Pt

e(VA � VB)�

(60.0 W)(3600 s)

(1.60 � 10�19 C)(12 V)�

n � q0

e � (EPEA � EPEB)/(VA � VB)

e�

Pt/(VA � VB)

e

EPEA � EPEB � Pt

q0 �EPEA � EPEB

VA � VB

STEP 2

STEP 3

?

(1)n �q0

e

(2)q0 �EPEA � EPEB

VA � VB

?

(1)n �q0

e

(2)q0 �EPEA � EPEB

VA � VB

EPEA � EPEB � Pt

STEP 1 STEP 2 STEP 3

As used in connection with batteries, the volt is a familiar unit for measuring electric po-

tential difference. The word “volt” also appears in another context, as part of a unit that is used

to measure energy, particularly the energy of an atomic particle, such as an electron or a pro-

ton. This energy unit is called the electron volt (eV). One electron volt is the magnitude ofthe amount by which the potential energy of an electron changes when the electron movesthrough a potential difference of one volt. Since the magnitude of the change in potential

energy is �q0�V � � �(�1.60 � 10�19 C) � (1.00 V)� � 1.60 � 10�19 J, it follows that

1 eV � 1.60 � J

One million (10�6) electron volts of energy is referred to as one MeV, and one billion

(10�9) electron volts of energy is one GeV, where the “G” stands for the prefix “giga”

(pronounced “jig�a”).

10�19


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� CONCEPTS AT A GLANCE The Concepts-at-a-Glance chart in Figure 19.5 emphasizes

that the total energy of an object, which is the sum of its kinetic and potential energies, is

an important concept. Its significance lies in the fact that the total energy remains the same

(is conserved) during the object’s motion, provided that nonconservative forces, such as fric-

tion, are either absent or do no net work. While the sum of the energies at each instant remains

constant, energy may be converted from one form to another; for example, gravitational po-

tential energy is converted into kinetic energy as a ball falls. As Figure 19.5 illustrates, we now

include the electric potential energy EPE as part of the total energy that an object can have:

E � � � mgh � � EPE

energy energy energy

If the total energy is conserved as the object moves, then its final energy Ef is equal to its

initial energy E0, or Ef � E0. Example 4 illustrates how the conservation of energy is ap-

plied to a charge moving in an electric field. �

Electric

14243

potential

Elastic

1442443

potential

Gravitational

14243

potential

Rotational

1442443

kinetic energy

Translational

1442443

kinetic energy

Total

14243

energy

1

2kx21

2I 21

2mv2


Wnc = 0 J

Translational KineticEnergy, mv2

(Section 6.2)

–12

–12

–12

–12

–12

–12

Rotational KineticEnergy, Iw2

(Section 9.5)

Gravitational PotentialEnergy, mgh

(Section 6.3)

Elastic PotentialEnergy, kx2

(Section 10.3)

Electric PotentialEnergy, EPE

Conservation of EnergyEf = E0

Total Energy

E = mv2 + Iw2 + mgh + kx2 + EPE

Totalenergy

Translationalkinetic energy

Rotationalkinetic energy

Gravitationalpotentialenergy

Elasticpotentialenergy

Electricpotentialenergy

CONCEPTS AT A GLANCE

Figure 19.5 CONCEPTS AT A GLANCE

Electric potential energy is another

type of energy. The patient in the

photograph is undergoing an X-ray

examination. To produce X-rays,

electrons are accelerated from rest

and allowed to collide with a target

material within the X-ray machine. As

each electron is accelerated, electric

potential energy is converted into

kinetic energy, and the total energy is

conserved during the process.

(Jonathan Nourok/Stone/Getty)

Example 4 The Conservation of Energy

A particle has a mass of 1.8 � 10�5 kg and a charge of

�3.0 � 10�5 C. It is released from rest at point A and ac-

celerates until it reaches point B, as Figure 19.6a shows.

The particle moves on a horizontal straight line and does

not rotate. The only forces acting on the particle are the

gravitational force and an electrostatic force (neither is

shown in the drawing). The electric potential at A is 25 V

greater than that at B; in other words, VA � VB � 25 V.

What is the translational speed of the particle at point B?


+

BA B

A = 0 m/s

(a)

BA A

–

B = 0 m/s

(b)

Figure 19.6 (a) A positive

charge starts from rest at

point A and accelerates toward

point B. (b) A negative charge

starts from rest at B and

accelerates toward A.

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STEP 1

Reasoning The translational speed of the particle is related to the particle’s translational kinetic energy, which forms one part

of the total mechanical energy that the particle has. The total mechanical energy is conserved, because only the gravitational

force and an electrostatic force, both of which are conservative forces, act on the particle (see Section 6.5). Thus, we will deter-

mine the speed at point B by utilizing the principle of conservation of mechanical energy.

Knowns and Unknowns We have the following data:

Description Symbol Value Comment

Explicit DataMass of particle m 1.8 � 10�5 kg

Electric charge of particle q0 �3.0 � 10�5 C

Electric potential difference VA � VB 25 V See Figure 19.6a.between points A and B

Implicit DataSpeed at point A vA 0 m/s Particle released from rest.

Vertical height above ground h Remains constant Particle travels horizontally.

Angular speed � 0 rad/s Particle does not rotate during motion.

Elastic force Felastic 0 N No elastic force acts on particle.

Unknown VariableSpeed at point B vB ?


(1)vB �Bv 2A �

2(EPEA � EPEB)

m

STEP 2

Conservation of Total Mechanical Energy The particle’s total mechanical

energy E is

E � � � mgh � � EPE

energy energy energy

Since the particle does not rotate, the angular speed � is always zero (see the data table)

and since there is no elastic force (see the data table), we may omit the terms and

from this expression. With this in mind, we express the fact that EB � EA (energy is

conserved) as follows:

This equation can be simplified further, since the particle travels horizontally, so that

hB � hA (see the data table), with the result that

Solving for vB gives Equation 1 at the right. Values for vA and m are available, and we turn

to Step 2 in order to evaluate EPEA � EPEB.

The Electric Potential Difference According to Equation 19.4, the differ-

ence in electric potential energies EPEA � EPEB is related to the electric potential

difference VA � VB:

EPEA � EPEB � q0(VA � VB)

The terms q0 and VA � VB are known, so we substitute this expression into Equation 1, as

shown at the right.

1

2mvB

2 � EPEB � 1

2mvA

2 � EPEA

1

2mv 2

B � mghB � EPEB � 1

2mv 2

A � mghA � EPEA

1

2 kx2

1

2 I�2

Electric

1442443

potential

Elastic

1442443

potential

Gravitational

14243

potential

Rotational

1442443

kinetic energy

Translational

1442443

kinetic energy

1

2kx21

2I� 21

2mv2

?

(1)vB �Bv 2A �

2(EPEA � EPEB)

m

Continued

EPEA � EPEB � q0(VA � VB)

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The speed of the particle at point B is

Note that if the particle had a negative charge of and were released

from rest at point B, it would arrive at point A with the same speed of 9.1 m/s (see

Figure 19.6b). This result can be obtained by returning to Modeling Step 1 and solving

for vA instead of vB.

Related Homework: Problems 2, 7, 8

�3.0 � 10�5 C

9.1 m/svB � Bv 2A �

2q0(VA � VB)

m� B(0 m/s)2 �

2(�3.0 � 10�5 C)(25 V)

1.8 � 10�5 kg�

vB � Bv 2A �

2(EPEA � EPEB)

m� Bv 2

A �2q0(VA � VB)

m


A positive charge accelerates from a region of higher potential toward a region of lower potential. In contrast, a negative charge accelerates from a region of lower potential toward a region of higher potential.

(The answers are given at the end of the book.)

1. An ion, starting from rest, accelerates from point A to point B due to a potential differ-ence between the two points. Does the electric potential energy of the ion at point B de-pend on (a) the magnitude of its charge and (b) its mass? Does the speed of the ion atB depend on (c) the magnitude of its charge and (d) its mass?

2. The drawing shows three possibilities for the potentials at two points, A and B. In eachcase, the same positive chargeis moved from A to B. In whichcase, if any, is the most workdone on the positive charge bythe electric force?

3. A proton and an electron arereleased from rest at the midpoint between the plates of a charged parallel plate capacitor(see Chapter 18). Except for these particles, nothing else is between the plates. Ignore theattraction between the proton and the electron, and decide which particle strikes a capaci-tor plate first.

C H E C K Y O U R U N D E R S TA N D I N G�

A positive point charge �q creates an electric potential in a way that Figure 19.7 helps

explain. This picture shows two locations A and B, at distances rA and rB from the charge. At

any position between A and B an electrostatic force of repulsion acts on a positive testFB

THE ELECTRIC POTENTIAL DIFFERENCE CREATED BY POINT CHARGES

19.3

150 V

Case 1

100 V

A B

25 V

Case 2

–25 V

A B

–10 V

Case 3

–60 V

A B

Figure 19.7 The positive test charge

�q0 experiences a repulsive force

due to the positive point charge �q.

As a result, work is done by this force

when the test charge moves from A to

B. Consequently, the electric potential

is higher (uphill) at A and lower

(downhill) at B.

FB

charge �q0. The magnitude of the force is given by Coulomb’s law as F � kq0q/r2, where we

assume for convenience that q0 and q are positive, so that �q0� � q0 and �q� � q. When the test

charge moves from A to B, work is done by this force. Since r varies between rA and rB, the

force F also varies, and the work is not the product of the force and the distance between the

points. (Recall from Section 6.1 that work is force times distance only if the force is constant.)

However, the work WAB can be found with the methods of integral calculus. The result is

This result is valid whether q is positive or negative, and whether q0 is positive or nega-

tive. The potential difference, VB � VA, between A and B can now be determined by sub-

stituting this expression for WAB into Equation 19.4:

(19.5)

As point B is located farther and farther from the charge q, rB becomes larger and

larger. In the limit that rB is infinitely large, the term kq/rB becomes zero, and it is

VB � VA ��WAB

q0

�kq

rB

�kq

rA

WAB �kqq0

rA

�kqq0

rB

STEP 1 STEP 2

Lower (downhill)potential

Higher (uphill)potential

A F B

rArB

+q0+q

B

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19.3 THE ELECTRIC POTENTIAL DIFFERENCE CREATED BY POINT CHARGES 577

customary to set VB equal to zero also. In this limit, Equation 19.5 becomes VA �kq/rA, and it is standard convention to omit the subscripts and write the potential in the

following form:

Potential of a point charge (19.6)

The symbol V in this equation does not refer to the potential in any absolute sense.

Rather, V � kq/r stands for the amount by which the potential at a distance r from a point

charge differs from the potential at an infinite distance away. In other words, V refers to a

potential difference with the arbitrary assumption that the potential at infinity is zero.

With the aid of Equation 19.6, we can describe the effect that a point charge q has on

the surrounding space. When q is positive, the value of V � kq/r is also positive, indicat-

ing that the positive charge has everywhere raised the potential above the zero reference

value. Conversely, when q is negative, the potential V is also negative, indicating that the

negative charge has everywhere decreased the potential below the zero reference value.

The next example deals with these effects quantitatively.

V �kq

r

Using a zero reference potential at infinity, determine the amount by which a point charge of

alters the electric potential at a spot 1.2 m away when the charge is (a) positive

and (b) negative.

Reasoning A point charge q alters the potential at every location in the surrounding space. In

the expression V � kq/r, the effect of the charge in increasing or decreasing the potential is

conveyed by the algebraic sign for the value of q.

Solution (a) Figure 19.8a shows the potential when the charge is positive:

(19.6)

(b) Part b of the drawing illustrates the results when the charge is negative. A calculation

similar to the one in part (a) shows that the potential is now negative: .�300 V

�300 VV �kq

r�

(8.99 � 109 N m2/C2)(�4.0 � 10�8 C)

1.2 m�

4.0 � 10�8 C

Example 5 The Potential of a Point Charge�

�

1.2 m

−300 V

V = 0at r = ∞

−4.0 × 10–8 C

(b)

+300 V

+4.0 × 10–8 C

1.2 m(a)

V = 0at r = ∞

Figure 19.8 A point charge of

alters the potential

at a spot 1.2 m away. The potential

is (a) increased by 300 V when the

charge is positive and (b) decreased

by 300 V when the charge is negative,

relative to a zero reference potential

at infinity.

10�8 C4.0 �

A single point charge raises or lowers the potential at a given location, depending on

whether the charge is positive or negative. When two or more charges are present, thepotential due to all the charges is obtained by adding together the individual potentials,as the next two examples show.


At locations A and B in Figure 19.9, find the total electric potential due to the two point charges.

Reasoning At each location, each charge contributes to the total electric potential. We obtain the

individual contributions by using V � kq/r and find the total potential by adding the individual

contributions algebraically. The two charges have the same magnitude, but different signs. Thus,

at A the total potential is positive because this spot is closer to the positive charge, whose effect

dominates over that of the more distant negative charge. At B, midway between the charges, the

total potential is zero, since the potential of one charge exactly offsets that of the other.

Solution

Example 6 The Total Electric Potential�

Location Contribution from � Charge Contribution from � Charge Total Potential

A �

B � 0 V(8.99 � 109 N m2/C2)(�8.0 � 10�9 C)

0.40 m�

(8.99 � 109 N m2/C2)(�8.0 � 10�9 C)

0.40 m

�240 V(8.99 � 109 N m2/C2)(�8.0 � 10�9 C)

0.60 m�

(8.99 � 109 N m2/C2)(�8.0 � 10�9 C)

0.20 m

�

+8.0 × 10–9 C −8.0 × 10–9 C

0.20 m 0.20 m 0.40 m

A B

Figure 19.9 Both the positive and

negative charges affect the electric

potential at locations A and B.

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 577

In Example 6 we determined the total potential at a spot due to two point charges. In

Example 8 we now extend this technique to find the total potential energy of three charges.


Two point charges are fixed in place, as in Figure 19.10. The positive charge is and has

twice the magnitude of the negative charge, which is On the line that passes through the

charges, three spots are identified, and At which of these spots could the potential be

equal to zero? (a) and (b) and (c) and

Reasoning The total potential is the algebraic sum of the individual potentials created by each

charge. It will be zero if the potential due to the positive charge is exactly offset by the poten-

tial due to the negative charge. The potential of a point charge is directly proportional to the

charge and inversely proportional to the distance from the charge.

Answers (b) and (c) are incorrect. The total potential at cannot be zero. The positive

charge has the larger magnitude and is closer to than is the negative charge. As a result, the

potential of the positive charge at dominates over the potential of the negative charge, so the

total potential cannot be zero.

Answer (a) is correct. Between the charges there is a location at which the individual poten-

tials cancel each other. We saw a similar situation in Example 6, where the canceling occurred

at the midpoint between the two charges that had equal magnitudes. Now the charges have un-

equal magnitudes, so the cancellation point does not occur at the midpoint. Instead, it occurs at

the location which is closer to the charge with the smaller magnitude—namely, the nega-

tive charge. At since the potential of a point charge is inversely proportional to the distance

from the charge, the effect of the smaller charge will be able to offset the effect of the more dis-

tant larger charge.

To the right of the negative charge there is also a location at which the individual poten-

tials exactly cancel each other. All places on this section of the line are closer to the negative

charge than to the positive charge. Therefore, there is a location in this region at which the

potential of the smaller negative charge exactly cancels the potential of the more distant and

larger positive charge.

Related Homework: Problem 66

P3

P2,

P2,

P1

P1

P1

P2P1P3P1P3P2

P3.P1, P2,

�q.

�2q

Conceptual Example 7 Where Is the Potential Zero?�


At any location, the total electric potential isthe algebraic sum of the individual potentialscreated by each point charge that is present.

−q+2qP1 P2 P3

Figure 19.10 Two point charges, one

positive and one negative. The positive

charge, �2q, has twice the magnitude

of the negative charge, �q.

�

Three point charges initially are infinitely far apart. Then, as Figure 19.11 shows, they are

brought together and placed at the corners of an equilateral triangle. Each side of the triangle

has a length of 0.50 m. Determine the electric potential energy of the triangular group. In other

words, determine the amount by which the electric potential energy of the group differs from

that of the three charges in their initial, infinitely separated, locations.

Reasoning We will proceed in steps by adding charges to the triangle one at a time, and then

determining the electric potential energy at each step. According to Equation 19.3, EPE � q0V,

the electric potential energy is the product of the charge and the electric potential at the spot

where the charge is placed. The total electric potential energy of the triangular group is the sum

of the energies of each step in assembling the group.

Solution The order in which the charges are put on the triangle does not matter; we begin with

the charge of �5.0 �C. When this charge is placed at a corner of the triangle, it has no electric

potential energy, according to EPE � q0V. This is because the total potential V produced by the

other two charges is zero at this corner, since they are infinitely far away. Once the charge is in

place, the potential it creates at either empty corner (r � 0.50 m) is

(19.6)

Therefore, when the �6.0 �C charge is placed at the second corner of the triangle, its electric

potential energy is

(19.3)EPE � qV � (�6.0 � 10�6 C)(�9.0 � 104 V) � �0.54 J

V �kq

r�

(8.99 � 109 N m2/C2)(�5.0 � 10�6 C)

0.50 m� �9.0 � 104 V

Example 8 The Potential Energy of a Group of Charges�+5.0 C μ

+6.0 C μ −2.0 C μ

Figure 19.11 Three point charges are

placed on the corners of an equilateral

triangle. Example 8 illustrates how to

determine the total electric potential

energy of this group of charges.

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 578

19.3 THE ELECTRIC POTENTIAL DIFFERENCE CREATED BY POINT CHARGES 579

The electric potential at the remaining empty corner is the sum of the potentials due to the two

charges that are already in place:

When the third charge, �2.0 �C, is placed at the remaining empty corner, its electric potential

energy is

EPE � qV � (�2.0 � 10�6 C)(�2.0 � 105 V) � �0.40 J (19.3)

The total potential energy of the triangular group differs from that of the widely separated

charges by an amount that is the sum of the potential energies calculated above:

Total potential energy � 0 J � 0.54 J � 0.40 J �

This energy originates in the work done to bring the charges together.

�0.14 J

�

(8.99 � 109 N m2/C2)(�6.0 � 10�6 C)

0.50 m� �2.0 � 105 V

V �(8.99 � 109 N m2/C2)(�5.0 � 10�6 C)

0.50 m

�


Be careful to distinguish between the concepts of potential V and electric potential energy EPE. Potential is electric potential energy per unit charge: V � EPE/q.


4. The drawing shows four arrangements (A–D) of two point charges. In each arrange-ment consider the total electric potential that the charges produce at location P. Rank thearrangements (largest to smallest) according to the total potential. (a) B, C, A and D (a tie)(b) D, C, A, B (c) A and C (a tie), B, D (d) C, D, A, B

5. A positive point charge and a negative point charge have equal magnitudes. One chargeis fixed to one corner of a square, and the other is fixed to another corner. On which cornersshould the charges be placed, so that the same potential exists at the empty corners? Thecharges should be placed at (a) adjacent corners, (b) diagonally opposite corners.

6. Three point charges have identical magnitudes, but two of the charges are positive and one is negative. These charges are fixed to the corners of a square, one to a corner. No matter how the charges are arranged, the potential at the empty corner is always(a) zero, (b) negative, (c) positive.

7. Consider a spot that is located midway between two identical point charges. Which oneof the following statements concerning the electric field and the electric potential at thisspot is true? (a) The electric field is zero, but the electric potential is not zero. (b) Theelectric field is not zero, but the electric potential is zero. (c) Both the electric field and theelectric potential are zero. (d) Neither the electric field nor the electric potential is zero.

8. Four point charges have thesame magnitude (but they mayhave different signs) and areplaced at the corners of a square,as the drawing shows. Whatmust be the sign (� or �) of eachcharge so that both the electricfield and the electric potential arezero at the center of the square? Assume that the potential has a zero value at infinity.

9. An electric potential energy exists when two protons are separated by a certain dis-tance. Does the electric potential energy increase, decrease, or remain the same (a) whenboth protons are replaced by electrons, and (b) when only one of the protons is replacedby an electron?

10. A proton is fixed in place. An electron is released from rest and allowed to collide with theproton. Then the roles of the proton and electron are reversed, and the same experiment is re-peated. Which, if either, is traveling faster when the collision occurs, the proton or the electron?

+2q +2q +2q +2q

2r0 2r0 2r0 r0

2r0 2r0

r0

r0

−q

−q −q

−q

A

P P P P

B C D


q1 q2

q4 q3

q1 q2 q3 q4

(a) � � � �(b) � � � �(c) � � � �(d) � � � �

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 579


An equipotential surface is a surface on which the electric potential is the same

everywhere. The easiest equipotential surfaces to visualize are those that surround an iso-

lated point charge. According to Equation 19.6, the potential at a distance r from a point

charge q is V � kq/r. Thus, wherever r is the same, the potential is the same, and the

equipotential surfaces are spherical surfaces centered on the charge. There are an infinite

number of such surfaces, one for every value of r, and Figure 19.12 illustrates two of them.

The larger the distance r, the smaller is the potential of the equipotential surface.

The net electric force does no work as a charge moves on an equipotential surface.This important characteristic arises because when an electric force does work WAB as a

charge moves from A to B, the potential changes according to VB � VA � �WAB/q0

(Equation 19.4). Since the potential remains the same everywhere on an equipotential sur-

face, VA � VB, and we see that WAB � 0 J. In Figure 19.12, for instance, the electric force

does no work as a test charge moves along the circular arc ABC, which lies on an equipo-

tential surface. In contrast, the electric force does work when a charge moves betweenequipotential surfaces, as from A to D in the picture.

The spherical equipotential surfaces that surround an isolated point charge illustrate

another characteristic of all equipotential surfaces. Figure 19.13 shows two of the surfaces

around a positive point charge, along with some electric field lines. The electric field lines

give the direction of the electric field, and for a positive point charge the electric field is

directed radially outward. Therefore, at each location on an equipotential sphere the elec-

tric field is perpendicular to the surface and points outward in the direction of decreasing

potential, as the drawing emphasizes. This perpendicular relation is valid whether or not

the equipotential surfaces result from a positive charge or have a spherical shape. The elec-tric field created by any charge or group of charges is everywhere perpendicular to theassociated equipotential surfaces and points in the direction of decreasing potential. For

example, Figure 19.14 shows the electric field lines (in red) around an electric dipole,

along with some equipotential surfaces (in blue), shown in cross section. Since the field

lines are not simply radial, the equipotential surfaces are no longer spherical but, instead,

have the shape necessary to be everywhere perpendicular to the field lines.

To see why an equipotential surface must be perpendicular to the electric field, con-

sider Figure 19.15, which shows a hypothetical situation in which the perpendicular rela-

tion does not hold. If were not perpendicular to the equipotential surface, there would

be a component of parallel to the surface. This field component would exert an electric

force on a test charge placed on the surface. As the charge moved along the surface, work

would be done by this component of the electric force. The work, according to Equation

19.4, would cause the potential to change, and, thus, the surface could not be an equipo-

tential surface as assumed. The only way out of the dilemma is for the electric field to be

perpendicular to the surface, so there is no component of the field parallel to the surface.

We have already encountered one equipotential surface. In Section 18.8, we found that

the direction of the electric field just outside an electrical conductor is perpendicular to the

EB

EB

EQUIPOTENTIAL SURFACES AND THEIR RELATION TO THE ELECTRIC FIELD

19.4

+qD

B

A

C

Equipotentialsurfaces

+q

Higher potential

Lower potential90°

90° 90°Electric fieldline

Figure 19.12 The equipotential surfaces

that surround the point charge �q are

spherical. The electric force does no

work as a charge moves on a path that

lies on an equipotential surface, such as

the path ABC. However, work is done

by the electric force when a charge

moves between two equipotential

surfaces, as along the path AD.

Figure 19.13 The radially directed

electric field of a point charge is

perpendicular to the spherical

equipotential surfaces that surround

the charge. The electric field points in

the direction of decreasing potential.


+ −

Figure 19.14 A cross-sectional view

of the equipotential surfaces (in blue)

of an electric dipole. The surfaces are

drawn to show that at every point they

are perpendicular to the electric field

lines (in red) of the dipole.

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 580

19.4 EQUIPOTENTIAL SURFACES AND THEIR RELATION TO THE ELECTRIC FIELD 581

conductor’s surface, when the conductor is at equilibrium under electrostatic conditions.

Thus, the surface of any conductor is an equipotential surface under such conditions. In

fact, since the electric field is zero everywhere inside a conductor whose charges are in

equilibrium, the entire conductor can be regarded as an equipotential volume.

There is a quantitative relation between the electric field and the equipotential sur-

faces. One example that illustrates this relation is the parallel plate capacitor in Figure

19.16. As Section 18.6 discusses, the electric field between the metal plates is perpen-

dicular to them and is the same everywhere, ignoring fringe fields at the edges. To be per-

pendicular to the electric field, the equipotential surfaces must be planes that are parallel to

the capacitor plates, which themselves are equipotential surfaces. The potential difference

between the plates is given by Equation 19.4 as �V � VB � VA � �WAB/q0, where A is a

point on the positive plate and B is a point on the negative plate. The work done by the

electric force as a positive test charge q0 moves from A to B is WAB � F�s, where F refers

to the electric force and �s to the displacement along a line perpendicular to the plates.

The force equals the product of the charge and the electric field E (F � q0E), so the work

becomes WAB � F�s � q0E�s. Therefore, the potential difference between the capacitor

plates can be written in terms of the electric field as �V � �WAB/q0 � �q0E�s/q0, or

(19.7a)

The quantity �V/�s is referred to as the potential gradient and has units of volts per

meter. In general, the relation E � ��V/�s gives only the component of the electric

field along the displacement �s; it does not give the perpendicular component.

The next example deals further with the equipotential surfaces between the plates of a

capacitor.

E � �

�V

�s

EB

Equipotentialsurface

E

Component of Eparallel to the

equipotential surface

B

B

Figure 19.15 In this hypothetical

situation, the electric field is not

perpendicular to the equipotential

surface. As a result, there is a component

of parallel to the surface.EB

EB

Equipotential surfaces

Δs

A B

EB

Figure 19.16 The metal plates of a

parallel plate capacitor are equipotential

surfaces. Two additional equipotential

surfaces are shown between the plates.

These two equipotential surfaces

are parallel to the plates and are

perpendicular to the electric field

between the plates.

EB

The plates of the capacitor in Figure 19.16 are separated by a distance of 0.032 m, and the

potential difference between them is �V � VB � VA � �64 V. Between the two labeled equipo-

tential surfaces there is a potential difference of �3.0 V. Find the spacing between the two

labeled surfaces.

Reasoning The electric field is E � ��V/�s. To find the spacing between the two labeled

equipotential surfaces, we solve this equation for �s, with �V � �3.0 V and E equal to the

electric field between the plates of the capacitor. A value for E can be obtained by using the val-

ues given for the distance and potential difference between the plates.

Solution The spacing between the labeled equipotential surfaces is given by

(1)

The electric field between the capacitor plates is

(19.7a)

Substituting Equation 19.7a into Equation 1 gives

1.5 � 10�3 m�s � ��V

E� �

�3.0 V

2.0 � 103 V/m�

E � ��V

�s� �

�64 V

0.032 m� 2.0 � 103 V/m

�s � ��V

E

Example 9 The Electric Field and Potential Are Related�

�Equation 19.7a gives the relationship between the electric field and the electric poten-

tial. It gives the component of the electric field along the displacement �s in a region of

space where the electric potential changes from place to place and applies to a wide vari-

ety of situations. When applied strictly to a parallel plate capacitor, however, this expres-

sion is often used in a slightly different form. In Figure 19.16, the metal plates of the

capacitor are marked A (higher potential) and B (lower potential). Traditionally, in discus-

sions of such a capacitor, the potential difference between the plates is referred to by us-

ing the symbol V to denote the amount by which the higher potential exceeds the lower

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 581

potential (V � VA � VB). In this tradition, the symbol V is often referred to as simply the

“voltage.” For example, if the potential difference between the plates of a capacitor is

5 volts, it is common to say that the “voltage” of the capacitor is 5 volts. In addition, the

displacement from plate A to plate B is expressed in terms of the separation d between the

plates (d � sB � sA). With this nomenclature, Equation 19.7a becomes

(parallel plate capacitor) (19.7b)E � ��V

�s� �

VB � VA

sB � sA

�VA � VB

sB � sA

�V

d



11. The drawing shows a cross-sectional view of twospherical equipotential surfaces and two electric fieldlines that are perpendicular to these surfaces. When anelectron moves from point A to point B (against theelectric field), the electric force does �3.2 � 10�19 J of work. What are the electric potential differences(a) VB � VA, (b) VC � VB, and (c) VC � VA?

12. The electric potential is constant throughout agiven region of space. Is the electric field zero ornonzero in this region?

13. In a region of space where the electric field is con-stant everywhere, as it is inside a parallel plate capaci-tor, is the potential constant everywhere? (a) Yes. (b) No, the potential is greatest at thepositive plate. (c) No, the potential is greatest at the negative plate.

14. A positive test charge is placed in an electric field. In what direction should the chargebe moved relative to the field, so that the charge experiences a constant electric potential?The charge should be moved (a) perpendicular to the electric field, (b) in the same direction as the electric field, (c) opposite to the direction of the electric field.

15. The location marked P in the drawing lies midwaybetween the point charges �q and �q. The blue lines labeled A, B, and C are edge-on views of three planes.Which of the planes is an equipotential surface? (a) Aand C (b) A, B, and C (c) Only B (d) None of theplanes is an equipotential surface.

16. Imagine that you are moving a positive test chargealong the line between two identical point charges. Withregard to the electric potential, is the midpoint on theline analogous to the top of a mountain or the bottom ofa valley when the two point charges are (a) positiveand (b) negative?


A

B

C

Equipotential surfaces(Cross-sectional view)

Electricfield lines

+q –qP

CA

B

Question 15

Question 11

THE CAPACITANCE OF A CAPACITOR

In Section 18.6 we saw that a parallel plate capacitor consists of two parallel metal

plates placed near one another but not touching. This type of capacitor is only one among

many. In general, a capacitor consists of two conductors of any shape placed near one an-

other without touching. For a reason that will become clear later on, it is common practice

to fill the region between the conductors or plates with an electrically insulating material

called a dielectric, as Figure 19.17 illustrates.

A capacitor stores electric charge. Each capacitor plate carries a charge of the samemagnitude, one positive and the other negative. Because of the charges, the electric poten-

tial of the positive plate exceeds that of the negative plate by an amount V, as Figure 19.17

indicates. Experiment shows that when the magnitude q of the charge on each plate is dou-

bled, the magnitude V of the electric potential difference is also doubled, so q is propor-

tional to V: q � V. Equation 19.8 expresses this proportionality with the aid of a

proportionality constant C, which is the capacitance of the capacitor.

CAPACITORS AND DIELECTRICS

19.5

Voltmeter

Dielectric

Metal plate(area = A)

Plateseparation

= d

Electric potentialdifference = V

Figure 19.17 A parallel plate capacitor

consists of two metal plates, one

carrying a charge �q and the other a

charge �q. The potential of the positive

plate exceeds that of the negative plate

by an amount V. The region between

the plates is filled with a dielectric.

2762T_ch19_569-598.qxd 10/18/08 8:24 PM Page 582

2nd

19.5 CAPACITORS AND DIELECTRICS 583

Equation 19.8 shows that the SI unit of capacitance is the coulomb per volt (C/V). This

unit is called the farad (F), named after the English scientist Michael Faraday

(1791–1867). One farad is an enormous capacitance. Usually smaller amounts, such as a

microfarad (1 �F � 10�6 F) or a picofarad (1 pF � 10�12 F), are used in electric circuits.

The capacitance reflects the ability of the capacitor to store charge, in the sense that a

larger capacitance C allows more charge q to be put onto the plates for a given value of

the potential difference V.

The physics of random-access memory (RAM) chips. The ability of a capacitor to store

charge lies at the heart of the random-access memory (RAM) chips used in computers,

where information is stored in the form of the “ones” and “zeros” that comprise binary

numbers. Figure 19.18 illustrates the role of a capacitor in a RAM chip. The capacitor is

connected to a transistor switch, to which two lines are connected, an address line and a

data line. A single RAM chip often contains millions of such transistor–capacitor units.

The address line is used by the computer to locate a particular transistor–capacitor combi-

nation, and the data line carries the data to be stored. A pulse on the address line turns on

the transistor switch. With the switch turned on, a pulse coming in on the data line can

cause the capacitor to charge. A charged capacitor means that a “one” has been stored,

whereas an uncharged capacitor means that a “zero” has been stored.

THE DIELECTRIC CONSTANT

If a dielectric is inserted between the plates of a capacitor, the capacitance can in-

crease markedly because of the way in which the dielectric alters the electric field between

the plates. Figure 19.19 shows how this effect comes about. In part a, the region between

the charged plates is empty. The field lines point from the positive toward the negative

plate. In part b, a dielectric has been inserted between the plates. Since the capacitor is not

connected to anything, the charge on the plates remains constant as the dielectric is in-

serted. In many materials (e.g., water) the molecules possess permanent dipole moments,

even though the molecules are electrically neutral. The dipole moment exists because one

end of a molecule has a slight excess of negative charge while the other end has a slight

excess of positive charge. When such molecules are placed between the charged plates of

the capacitor, the negative ends are attracted to the positive plate and the positive ends are

attracted to the negative plate. As a result, the dipolar molecules tend to orient themselves

end to end, as in part b. Whether or not a molecule has a permanent dipole moment, the

electric field can cause the electrons to shift position within a molecule, making one end

slightly negative and the opposite end slightly positive. Because of the end-to-end orienta-

tion, the left surface of the dielectric becomes positively charged, and the right surface be-

comes negatively charged. The surface charges are shown in red in the picture.

Because of the surface charges on the dielectric, not all the electric field lines generated

by the charges on the plates pass through the dielectric. As Figure 19.19c shows, some of the

field lines end on the negative surface charges and begin again on the positive surface

charges. Thus, the electric field inside the dielectric is less strong than the electric field in-

side the empty capacitor, assuming the charge on the plates remains constant. This reduction

in the electric field is described by the dielectric constant , which is the ratio of the field

magnitude E0 without the dielectric to the field magnitude E inside the dielectric:

(19.9) �E0

E

THE RELATION BETWEEN CHARGE AND POTENTIAL DIFFERENCE FOR A CAPACITOR

The magnitude q of the charge on each plate of a capacitor is directly proportional to

the magnitude V of the potential difference between the plates:

q � CV (19.8)

where C is the capacitance.

SI Unit of Capacitance: coulomb/volt � farad (F)

Data line

Capacitor

Addressline

Transistorswitch

Figure 19.18 A transistor–capacitor

combination is part of a RAM chip

used in computer memories.

–q +q

(a)

Positivesurface chargeon dielectric

Negativesurface chargeon dielectric

Molecule(b)

(c)

–q +q

–q +q

Figure 19.19 (a) The electric field

lines inside an empty capacitor. (b) The

electric field produced by the charges

on the plates aligns the molecular

dipoles within the dielectric end to

end. The space between the dielectric

and the plates is added for clarity. In

reality, the dielectric fills the region

between the plates. (c) The surface

charges on the dielectric reduce the

electric field inside the dielectric.

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 583

Being a ratio of two field strengths, the dielectric constant is a number without units.

Moreover, since the field 0 without the dielectric is greater than the field inside the di-EB

EB


Table 19.1 Dielectric Constants of Some Common Substancesa

Dielectric

Substance Constant,

Vacuum 1

Air 1.000 54

Teflon 2.1

Benzene 2.28

Paper (royal gray) 3.3

Ruby mica 5.4

Neoprene rubber 6.7

Methyl alcohol 33.6

Water 80.4

aNear room temperature.

Example 10 Storing Electric Charge

The capacitance of an empty capacitor is 1.2 �F. The capacitor is connected to a 12-V battery and charged up. With the capaci-

tor connected to the battery, a slab of dielectric material is inserted between the plates. As a result, 2.6 � 10�5 C of additionalcharge flows from one plate, through the battery, and onto the other plate. What is the dielectric constant of the material?

Reasoning The charge stored by a capacitor is q � CV, according to Equation 19.8. The battery maintains a constant potential dif-

ference of V � 12 V between the plates of the capacitor while the dielectric is inserted. Inserting the dielectric causes the capacitance

C to increase, so that with V held constant, the charge q must increase. Thus, additional charge flows onto the plates. To find the

dielectric constant, we will apply Equation 19.8 to the capacitor filled with the dielectric material and then to the empty capacitor.

Knowns and Unknowns The following table summarizes the given data:

Description Symbol Value

Capacitance of empty capacitor C0 1.2 �F

Potential difference between capacitor plates V 12 V

Additional charge that flows when dielectric is inserted �q 2.6 � 10�5 C

Unknown VariableDielectric constant ?


electric, the dielectric constant is greater than unity. The value of depends on the nature

of the dielectric material, as Table 19.1 indicates.

THE CAPACITANCE OF A PARALLEL PLATE CAPACITOR

The capacitance of a capacitor is affected by the geometry of the plates and the di-

electric constant of the material between them. For example, Figure 19.17 shows a paral-

lel plate capacitor in which the area of each plate is A and the separation between the plates

is d. The magnitude of the electric field inside the dielectric is given by Equation 19.7b as

E � V/d, where V is the magnitude of the potential difference between the plates. If the

charge on each plate is kept fixed, the electric field inside the dielectric is related to the

electric field in the absence of the dielectric via Equation 19.9. Therefore,

Since the electric field within an empty capacitor is E0 � q/(�0A) (see Equation 18.4), it

follows that q/( �0A) � V/d, which can be solved for q to give

A comparison of this expression with q � CV (Equation 19.8) reveals that the capaci-

tance C is

(19.10)

Notice that only the geometry of the plates (A and d) and the dielectric constant affect the

capacitance. With C0 representing the capacitance of the empty capacitor ( � 1), Equation

19.10 shows that C � C0. In other words, the capacitance with the dielectric present is in-

creased by a factor of over the capacitance without the dielectric. It can be shown that the

relation C � C0 applies to any capacitor, not just to a parallel plate capacitor. One reason,

then, that capacitors are filled with dielectric materials is to increase the capacitance. Example

10 illustrates the effect that increasing the capacitance has on the charge stored by a capacitor.

C � �0A

dParallel plate capacitorfilled with a dielectric

q � � �0A

d �V

E �E0

�

V

d

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 584

19.5 CAPACITORS AND DIELECTRICS 585

STEP 1 Dielectric Constant According to Equations 19.8 and 19.10, the charge

stored by the dielectric-filled capacitor is q � �C0V, where � is the dielectric constant,

�C0 is the capacitance of the capacitor with the dielectric inserted, C0 is the capacitance

of the empty capacitor, and V is the voltage provided by the battery. Solving for the

dielectric constant gives Equation 1 at the right. In this expression, C0 and V are known,

and we will deal with the unknown quantity q in Step 2.

The Charge Stored by the Dielectric-Filled Capacitor The charge qstored by the dielectric-filled capacitor is the charge q0 stored by the empty capacitor plus

the additional charge �q that arises when the dielectric is inserted. Therefore, we have

q � q0 � �q

This result can be substituted into Equation 1, as indicated at the right. �q is given, and a

value for q0 will be obtained in Step 3.

The Charge Stored by the Empty Capacitor According to Equation 19.8,

the charge stored by the empty capacitor is

q0 � C0V

The substitution of this expression into Equation 2 is shown at the right.


The dielectric constant can now be determined:


2.8� �C0V � �q

C0V� 1 �

�q

C0V� 1 �

2.6 � 10�5 C

(1.2 � 10�6 F)(12 V)�

� � q

C0V �

q0 � �q

C0V �

C0V � �q

C0V


(1)� �q

C0V

(1)� �q

C0V

STEP 2

(1)� �q

C0VSTEP 3

STEP 1 STEP 2 STEP 3

An empty capacitor is connected to a battery and charged up. The capacitor is then discon-

nected from the battery, and a slab of dielectric material is inserted between the plates. Does

the potential difference across the plates (a) increase, (b) remain the same, or (c) decrease?

Reasoning Our reasoning is guided by the following fact: Once the capacitor is disconnected

from the battery, the charge on its plates remains constant, for there is no longer any way for

charge to be added or removed. According to Equation 19.8, the magnitude q of the charge

stored by the capacitor is where C is its capacitance and V is the magnitude of the po-

tential difference between the plates.

Answers (a) and (b) are incorrect. Placing a dielectric between the plates of a capacitor re-

duces the electric field in that region (see Figure 19.19c). The magnitude V of the potential dif-

ference is related to the magnitude E of the electric field by where d is the distanceV � Ed,

q � CV,

Conceptual Example 11

The Effect of a Dielectric When a Capacitor Has a Constant Charge�

?

In Example 10 the capacitor remains connected to the battery while the dielectric is

inserted between the plates. The next example discusses what happens if the capacitor is

disconnected from the battery before the dielectric is inserted.

q � q0 � �q (2)

q � q0 � �q (2)

?

q0 � C0V

2762T_ch19_569-598.qxd 7/7/08 9:16 PM Page 585


`

esc

1 2 3 4 5

F1 F2 F3 F4 F5 F6

6 7~ ! @ # $ % ^ &

Q

control option

alt

W E R T Y U

A S D F G H J

Zshift X C V B N

tab

capslock

option

alt

Key

Plunger

DielectricFixed metal plate

Movable metal plate

between the capacitor plates. Since E decreases when the dielectric is inserted and d is un-

changed, the potential difference does not increase or remain the same.

Answer (c) is correct. Inserting the dielectric causes the capacitance C to increase. Since

and q is fixed, the potential difference V across the plates must decrease in order for qto remain unchanged. The amount by which the potential difference decreases from the value

initially established by the battery depends on the dielectric constant of the slab.


q � CV

One kind of computer keyboard is based on the idea of capacitance. Each key is mounted on

one end of a plunger, and the other end is attached to a movable metal plate (see Figure 19.20).

The movable plate is separated from a fixed plate, the two plates forming a capacitor. When the

key is pressed, the movable plate is pushed closer to the fixed plate, and the capacitance in-

creases. Electronic circuitry enables the computer to detect the change in capacitance, thereby

recognizing which key has been pressed. The separation of the plates is normally 5.00 � 10�3 m

but decreases to 0.150 � 10�3 m when a key is pressed. The plate area is 9.50 � 10�5 m2, and

the capacitor is filled with a material whose dielectric constant is 3.50. Determine the change

in capacitance that is detected by the computer.

Reasoning We can use Equation 19.10 directly to find the capacitance of the key, since the

dielectric constant , the plate area A, and the plate separation d are known. We will use this

relation twice, once to find the capacitance when the key is pressed and once when it is not

pressed. The change in capacitance will be the difference between these two values.

Solution When the key is pressed, the capacitance is

(19.10)

A calculation similar to the one above reveals that when the key is not pressed, the capacitance

has a value of 0.589 � 10�12 F (0.589 pF). The change in capacitance is an increase of

. The change in the capacitance is greater with the dielectric present,

which makes it easier for the circuitry within the computer to detect it.

19.0 � 10�12 F (19.0 pF)

� 19.6 � 10�12 F (19.6 pF)

C � �0 A

d�

(3.50)[8.85 � 10�12 C2/(N m2)](9.50 � 10�5 m2)

0.150 � 10�3 m

Example 12 A Computer Keyboard�The physics ofa computer keyboard.

Figure 19.20 In one kind of computer

keyboard, each key, when pressed,

changes the separation between the

plates of a capacitor.

�

ENERGY STORAGE IN A CAPACITOR

When a capacitor stores charge, it also stores energy. In charging up a capacitor, for ex-

ample, a battery does work in transferring an increment of charge from one plate of the capac-

itor to the other plate. The work done is equal to the product of the charge increment and the

potential difference between the plates. However, as each increment of charge is moved, the

potential difference increases slightly, and a larger amount of work is needed to move the next

increment. The total work W done in completely charging the capacitor is the product of the

total charge q transferred and the average potential difference . Since the average

potential difference is one-half the final potential V, or the total work done by the bat-

tery is This work does not disappear but is stored as electric potential energy in the

capacitor, so that Energy � . Equation 19.8 indicates that q � CV or, equivalently, that

V � q/C. We can see, then, that our expression for the energy can be cast into two additional

equivalent forms by substituting for q or for V. Equations 19.11a–c summarize these results:

(19.11a)

(19.11b)

(19.11c)Energy � 1

2q � q

C � �q2

2C

Energy � 1

2(CV )V � 1

2CV 2

Energy � 1

2qV

1

2 qV

W � 1

2 qV.

V � 1

2 V,

V; W � qV

�Capacitors are used often in electronic devices, and Example 12 deals with one familiar

application.

2762T_ch19_569-598.qxd 7/4/08 2:31 PM Page 586

19.6 BIOMEDICAL APPLICATIONS OF ELECTRIC POTENTIAL DIFFERENCES 587

It is also possible to regard the energy as being stored in the electric field between the

plates. The relation between energy and field strength can be obtained for a parallel plate

capacitor by substituting V � Ed (Equation 19.7b) and C � �0A/d (Equation 19.10) into

Equation 19.11b:

Since the area A times the separation d is the volume between the plates, the energy per

unit volume or energy density is

(19.12)

It can be shown that this expression is valid for any electric field strength, not just that be-

tween the plates of a capacitor.

The physics of an electronic flash attachment for a camera. The energy-storing capabil-

ity of a capacitor is often put to good use in electronic circuits. For example, in an elec-

tronic flash attachment for a camera, energy from the battery pack is stored in a capacitor.

The capacitor is then discharged between the electrodes of the flash tube, which converts

the energy into light. Flash duration times range from 1/200 to 1/1 000 000 second or less,

with the shortest flashes being used in high-speed photography (see Figure 19.21). Some

flash attachments automatically control the flash duration by monitoring the light reflected

from the photographic subject and quickly stopping or quenching the capacitor discharge

when the reflected light reaches a predetermined level.

Energy density �Energy

Volume� 1

2 �0E 2

Energy � 1

2 CV 2 � 1

2 � �0 A

d �(Ed )2

Figure 19.21 This time-lapse photo of

a gymnast was obtained using a camera

with an electronic flash attachment.

The energy for each flash of light

comes from the electrical energy stored

in a capacitor. (© Bongarts/Getty Images)

Figure 19.22 A portable defibrillator is

being used in an attempt to revive this

heart attack victim. A defibrillator uses

the electrical energy stored in a

capacitor to deliver a controlled electric

current that can restore normal heart

rhythm. (Adam Hart-Davis/SPL/Photo

Researchers, Inc.)

The physics of a defibrillator. During a heart attack, the heart produces a rapid,

unregulated pattern of beats, a condition known as cardiac fibrillation. Cardiac

fibrillation can often be stopped by sending a very fast discharge of electrical

energy through the heart. For this purpose, emergency medical personnel use defibrillators,

such as the one shown in Figure 19.22. A paddle is connected to each plate of a large ca-

pacitor, and the paddles are placed on the chest near the heart. The capacitor is charged to

a potential difference of about a thousand volts. The capacitor is then discharged in a few

thousandths of a second; the discharge current passes through a paddle, the heart, and the

other paddle. Within a few seconds, the heart often returns to its normal beating pattern.


17. An empty parallel plate capacitor is connected to a battery that maintains a constant po-tential difference between the plates. With the battery connected, a dielectric is then insertedbetween the plates. Do the following quantities decrease, remain the same, or increase whenthe dielectric is inserted? (a) The electric field between the plates (b) The capacitance(c) The charge on the plates (d) The energy stored by the capacitor

18. A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. The plates are then pulled apart. Do the following quan-tities decrease, remain the same, or increase as the distance between the plates increases?(a) The capacitance of the capacitor (b) The potential difference between the plates (c) Theelectric field between the plates (d) The electric potential energy stored by the capacitor


CONDUCTION OF ELECTRICAL SIGNALS IN NEURONS

The human nervous system is remarkable for its ability to transmit information in

the form of electrical signals. These signals are carried by the nerves, and the concept of

electric potential difference plays an important role in the process. For example, sensory

information from our eyes and ears is carried to the brain by the optic nerves and auditory

nerves, respectively. Other nerves transmit signals from the brain or spinal column to mus-

cles, causing them to contract. Still other nerves carry signals within the brain.

*BIOMEDICAL APPLICATIONS OF ELECTRIC POTENTIAL DIFFERENCES

19.6

2762T_ch19_569-598.qxd 7/4/08 2:32 PM Page 587

The physics of an action potential. A “resting” neuron is one that is not conduct-

ing an electrical signal. The change in the resting membrane potential is the key

factor in the initiation and conduction of a signal. When a sufficiently strong

stimulus is applied to a given point on the neuron, “gates” in the membrane open and

sodium ions flood into the cell, as Figure 19.25 illustrates. The sodium ions are driven into

the cell by attraction to the negative ions on the inner side of the membrane as well as by

the relatively high concentration of sodium ions outside the cell. The large influx of

ions first neutralizes the negative ions on the interior of the membrane and then causes it

to become positively charged. As a result, the membrane potential in this localized region

goes from �70 mV, the resting potential, to about �30 mV in a very short time (see Figure

19.26). The sodium gates then close, and the cell membrane quickly returns to its normal

resting potential. This change in potential, from �70 mV to �30 mV and back to �70 mV,

is known as the action potential. The action potential lasts for a few milliseconds, and it is

the electrical signal that propagates down the axon, typically at a speed of about 50 m/s,

to the next neuron or to a muscle cell.

Na�

A nerve consists of a bundle of axons, and each axon is one part of a nerve cell, or

neuron. As Figure 19.23 illustrates, a typical neuron consists of a cell body with numerous

extensions, called dendrites, and a single axon. The dendrites convert stimuli, such as pres-

sure or heat, into electrical signals that travel through the neuron. The axon sends the sig-

nal to the nerve endings, which transmit the signal across a gap (called a synapse) to the

next neuron or to a muscle.

The fluid inside a cell, the intracellular fluid, is quite different from that outside the cell,

the extracellular fluid. Both fluids contain concentrations of positive and negative ions.

However, the extracellular fluid is rich in sodium and chlorine ions, whereas

the intracellular fluid is rich in potassium ions and negatively charged proteins. These

concentration differences between the fluids are extremely important to the life of the cell.

If the cell membrane were freely permeable, the ions would diffuse across it until the con-

centrations on both sides were equal. (See Section 14.4 for a review of diffusion.) This does

not happen, because a living cell has a selectively permeable membrane. Ions can enter or

leave the cell only through membrane channels, and the permeability of the channels varies

markedly from one ion to another. For example, it is much easier for ions to diffuse out

of the cell than it is for to enter the cell. As a result of selective membrane permeabil-

ity, there is a small buildup of negative charges just on the inner side of the membrane and

an equal amount of positive charges on the outer side (see Figure 19.24). The buildup of

charge occurs very close to the membrane, so the membrane acts like a capacitor (see

Problems 44 and 57). Elsewhere in the intracellular and extracellular fluids, there are equal

numbers of positive and negative ions, so the fluids are overall electrically neutral. Such a

separation of positive and negative charges gives rise to an electric potential difference

across the membrane, called the resting membrane potential. In neurons, the resting mem-

brane potential ranges from �40 to �90 mV, with a typical value of �70 mV. The minus

sign indicates that the inner side of the membrane is negative relative to the outer side.

Na�

K�

(K�)

(Cl�)(Na�)


Dendrites

Cell body

Axon

Synapse

Another neuronor a muscle

Nerve endings

Figure 19.23 The anatomy of a typical

neuron.

–

–

–

–

+

+

– – – –

–

––

–

+ +

+ +

+

+

+ +

+

+

Positivecharge layer

Extracellularfluid

(outside)

Intracellularfluid

(inside)

Cellmembrane

Negativecharge layer

– – – –Intracellularfluid

(inside)

+

+

+

++

Na+

Figure 19.24 Positive and negative charge layers

form on the outside and inside surfaces of a membrane

during its resting state.

Figure 19.25 When a stimulus is applied to the

cell, positive sodium ions (Na�) rush into the cell,

causing the interior surface of the membrane to

become momentarily positive.

Action potential

Na+ influx

0 1 2 3 4

+30

Time, ms

Mem

bran

e po

tent

ial,

mV

0

–70

Membranerestingpotential

Figure 19.26 The action potential is

caused by the rush of positive sodium

ions into the cell and a subsequent

return of the cell to its resting potential.

2762T_ch19_569-598.qxd 7/4/08 2:32 PM Page 588

19.6 BIOMEDICAL APPLICATIONS OF ELECTRIC POTENTIAL DIFFERENCES 589

MEDICAL DIAGNOSTIC TECHNIQUES

Several important medical diagnostic techniques depend on the fact that the surface

of the human body is not an equipotential surface. Between various points on the body

there are small potential differences (approximately 30–500 �V), which provide the basis

for electrocardiography, electroencephalography, and electroretinography. The potential

differences can be traced to the electrical characteristics of muscle cells and nerve cells. In

carrying out their biological functions, these cells utilize positively charged sodium and

potassium ions and negatively charged chlorine ions that exist within the cells and in the

extracellular fluid. As a result of such charged particles, electric fields are generated that

extend to the surface of the body and lead to the small potential differences.

Figure 19.27 shows electrodes placed on the body to measure potential differences in

electrocardiography. The potential difference between two locations changes as the heart

beats and forms a repetitive pattern. The recorded pattern of potential difference versus

time is called an electrocardiogram (ECG or EKG), and its shape depends on which pair

of points in the picture (A and B, B and C, etc.) is used to locate the electrodes. The figure

also shows some EKGs and indicates the regions (P, Q, R, S, and T ) associated with spe-

cific parts of the heart’s beating cycle. The differences between the EKGs of normal and

abnormal hearts provide physicians with a valuable diagnostic tool.

In electroencephalography the electrodes are placed at specific locations on the head,

as Figure 19.28 indicates, and they record the potential differences that characterize brain

behavior. The graph of potential difference versus time is known as an electroencephalo-

gram (EEG). The various parts of the patterns in an EEG are often referred to as “waves”

or “rhythms.” The drawing shows an example of the main resting rhythm of the brain, the

so-called alpha rhythm, and also illustrates the distinct differences that are found between

the EEGs generated by healthy (normal) and diseased (abnormal) tissue.

The electrical characteristics of the retina of the eye lead to the potential differences

measured in electroretinography. Figure 19.29 shows a typical electrode placement used to

Strip-chartrecorder

ElectrodesED

C

A BEKG (Normal) EKG (Abnormal)

Time

Potentialdifference

Time

Potentialdifference

PP

Q

R

STQ

R

S

T

Electrodesconnectedto detectionand recordingdevice

Time Time

Potentialdifference

Potentialdifference

EEG (Abnormal)EEG (Normal alpha rhythm)

Figure 19.27 The potential differences

generated by heart muscle activity

provide the basis for electrocardiography.

The normal and abnormal EKG patterns

correspond to one heartbeat.

Figure 19.28 In electroencephalography

the potential differences created by the

electrical activity of the brain are used

for diagnosing abnormal behavior.

The physics ofelectrocardiography.

The physics ofelectroencephalography.

The physics ofelectroretinography.

Electrodesconnectedto detectionand recordingdevice

Electronic flash Time Time

Potentialdifference

Potentialdifference

ERG (Normal) ERG (Abnormal)B

A

B

A Figure 19.29 The electrical activity

of the retina of the eye generates

the potential differences used

in electroretinography.

2762T_ch19_569-598.qxd 7/7/08 9:16 PM Page 589


The conservation of energy (Chapter 6) and the conservation of linear momentum

(Chapter 7) are two of the most broadly applicable principles in all of science. In this chap-

ter, we have seen that electrically charged particles obey the conservation-of-energy prin-

ciple, provided that the electric potential energy is taken into account. The behavior of

electrically charged particles, however, must also be consistent with the conservation-of-

momentum principle, as the first example in this section emphasizes.

CONCEPTS & CALCULATIONS

19.7

Particle 1 has a mass of while particle 2 has a mass of m2 � m1 � 3.6 � 10�6 kg,

Concepts & Calculations Example 13

Conservation Principles�

q

m1

q

m2

q

m1

q

m2

1f 2f

(b) Final

(a) Initial (at rest)

Figure 19.30 (a) Two particles have

different masses, but the same electric

charge q. They are initially held at rest.

(b) At an instant following the release

of the particles, they are flying apart

due to the mutual force of electric

repulsion.

record the pattern of potential difference versus time that occurs when the eye is stimulated

by a flash of light. One electrode is mounted on a contact lens, while the other is often

placed on the forehead. The recorded pattern is called an electroretinogram (ERG), and

parts of the pattern are referred to as the “A wave” and the “B wave.” As the graphs show,

the ERGs of normal and diseased (abnormal) eyes can differ markedly.

6.2 � Each has the same electric charge. These particles are initially held at rest,

and the two-particle system has an initial electric potential energy of 0.150 J. Suddenly, the

particles are released and fly apart because of the repulsive electric force that acts on each

one (see Figure 19.30). The effects of the gravitational force are negligible, and no other

forces act on the particles. At one instant following the release, the speed of particle 1 is

measured to be What is the electric potential energy of the two-particle system

at this instant?

Concept Questions and Answers What types of energy does the two-particle system have

initially?

Answer Initially, the particles are at rest, so they have no kinetic energy. However, they

do have electric potential energy. They also have gravitational potential energy, but it is

negligible.

What types of energy does the two-particle system have at the instant illustrated in part b of the

drawing?

Answer Since each particle is moving, each has kinetic energy. The two-particle system also

has electric potential energy at this instant. Gravitational potential energy remains negligible.

Does the principle of conservation of energy apply?

Answer Yes. Since the gravitational force is negligible, the only force acting here is the

conservative electric force. Nonconservative forces are absent. Thus, the principle of con-

servation of energy applies.

Does the principle of conservation of linear momentum apply to the two particles as they fly apart?

Answer Yes. This principle states that the total linear momentum of an isolated system

remains constant. An isolated system is one for which the vector sum of the external forces

acting on the system is zero. There are no external forces acting here. The only apprecia-

ble force acting on each of the two particles is the force of electric repulsion, which is an

internal force.

Solution The conservation-of-energy principle indicates that the total energy of the two-particle

system is the same at the later instant as it was initially. Considering that only kinetic energy and

electric potential energy are present, the principle can be stated as follows:

We have used the fact that KE0 is zero, since the particles are initially at rest. The final kinetic

energy of the system is In this expression, we have values for bothKEf � 1

2m1v 2

f1 � 1

2m2v 2

f2.

Initial total energy

144424443Final total energy

144424443KEf � EPEf � KE0 � EPE0 or EPEf � EPE0 � KEf

v1 � 170 m/s.

10�6 kg.

2762T_ch19_569-598.qxd 7/4/08 2:32 PM Page 590

19.7 CONCEPTS & CALCULATIONS 591

Chapter 18 introduces the electric field, and the present chapter introduces the electric

potential. These two concepts are central to the study of electricity, and it is important to dis-

tinguish between them, as the next example emphasizes.

masses and the speed vf1. We can find the speed vf2 by using the principle of conservation of

momentum:

We have again used the fact that the particles are initially at rest, so that v01 � v02 � 0 m/s. Using

this result for vf2, we can now obtain the final electric potential energy from the conservation-

of-energy equation:

� 0.068 J

� 0.150 J � 1

2(3.6 � 10�6 kg) �1 �

3.6 � 10�6 kg

6.2 � 10�6 kg � (170 m/s)2

� EPE0 � 1

2m1 �1 �

m1

m2� vf1

2

EPEf � EPE0 � �1

2 m1v f1

2 � 1

2 m2 ��

m1

m2

vf1�2

�

Initial total

144424443

momentum

Final total

144424443

momentum

m1vf1 � m2vf2 � m1v01 � m2v02 or vf2 � �m1

m2

vf1

�

Two identical point charges (�2.4 � 10�9 C) are fixed in place, separated by 0.50 m (see

Figure 19.31). Find the electric field and the electric potential at the midpoint of the line

between the charges qA and qB.

Concept Questions and Answers The electric field is a vector and has a direction. At the mid-

point, what are the directions of the individual electric-field contributions from qA and qB?

Answer Since both charges are positive, the individual electric-field contributions from qA

and qB point away from each charge. Thus, at the midpoint they point in opposite directions.

Is the magnitude of the net electric field at the midpoint greater than, less than, or equal to zero?

Answer The magnitude of the net electric field is zero, because the electric field from qA can-

cels the electric field from qB. These charges have the same magnitude �q �, and the midpoint

is the same distance r from each of them. According to E � k �q �/r2 (Equation 18.3), then,

each charge produces a field of the same strength. Since the individual electric-field contribu-

tions from qA and qB point in opposite directions at the midpoint, their vector sum is zero.

Is the total electric potential at the midpoint positive, negative, or zero?

Answer The total electric potential at the midpoint is the algebraic sum of the individual

contributions from each charge. According to V � kq/r (Equation 19.6), each contribution

is positive, since each charge is positive. Thus, the total electric potential is also positive.

Does the total electric potential have a direction associated with it?

Answer No. The electric potential is a scalar quantity, not a vector quantity. Therefore, it

has no direction associated with it.

Solution As discussed in the second concept question, the total electric field at the midpoint

is . Using V � kq/r (Equation 19.6), we find that the total potential at the midpoint is

Since and we have

170 VV �2kqA

rA

�2(8.99 � 109 N m2/C2)(�2.4 � 10�9 C)

0.25 m�

rA � rB � 0.25 m,qA � qB � �2.4 � 10�9 C

VTotal �kqA

rA

�kqB

rB

� � 0 N/C

Concepts & Calculations Example 14

Electric Field and Electric Potential�qA qB

rA rB

Midpoint

Figure 19.31 Example 14 determines

the electric field and the electric

potential at the midpoint (rA � rB)

between the identical charges

(qA � qB).

�

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If you need more help with a concept, use the Learning Aids noted next to the discussion or equation. Examples (Ex.) are in the text

of this chapter. Go to www.wiley.com/college/cutnell for the following Learning Aids:

Topic Discussion Learning Aids

Interactive LearningWare (ILW) — Additional examples solved in a five-step interactive format.

Concept Simulations (CS) — Animated text figures or animations of important concepts.

Interactive Solutions (IS) — Models for certain types of problems in the chapter homework. The calculations are carried out interactively.

CONCEPT SUMMARY

Work and electric potential energy

Path independence

Electric potential

Electric potential difference

Acceleration of positive and negative charges

Electron volt

Total energy

Conservation of energy

Electric potential of a point charge

Total electric potential

Total potential energy of a group of charges

Equipotential surface

19.1 POTENTIAL ENERGY When a positive test charge �q0 moves from point A to point B in an

electric field, work WAB is done by the electric force. The work equals the electric potential energy

(EPE) at A minus that at B:

WAB � EPEA � EPEB (19.1)

The electric force is a conservative force, so the path along which the test charge moves from A toB is of no consequence, for the work WAB is the same for all paths.

19.2 THE ELECTRIC POTENTIAL DIFFERENCE The electric potential V at a given point is the elec-

tric potential energy of a small test charge q0 situated at that point divided by the charge itself:

(19.3)

The SI unit of electric potential is the joule per coulomb (J/C), or volt (V).

The electric potential difference between two points A and B is

VB � VA � (19.4)

The potential difference between two points (or between two equipotential surfaces) is often

called the “voltage.”

A positive charge accelerates from a region of higher potential toward a region of lower potential.

Conversely, a negative charge accelerates from a region of lower potential toward a region of higher

potential.

An electron volt (eV) is a unit of energy. The relationship between electron volts and joules is

1 eV � 1.60 � 10�19 J.

The total energy E of a system is the sum of its translational ( mv2) and rotational ( I�2) kinetic

energies, gravitational potential energy (mgh), elastic potential energy ( kx2), and electric poten-

tial energy (EPE):

If external nonconservative forces like friction do no net work, the total energy of the system is

conserved. That is, the final total energy Ef is equal to the initial total energy E0; Ef � E0.

19.3 THE ELECTRIC POTENTIAL DIFFERENCE CREATED BY POINT CHARGES The electric poten-

tial V at a distance r from a point charge q is

(19.6)

where k � 8.99 � 109 N m2/C2. This expression for V assumes that the electric potential is zero at

an infinite distance away from the charge.

The total electric potential at a given location due to two or more charges is the algebraic sum of

the potentials due to each charge.

The total potential energy of a group of charges is the amount by which the electric potential en-

ergy of the group differs from its initial value when the charges are infinitely far apart and far

away. It is also equal to the work required to assemble the group, one charge at a time, starting

with the charges infinitely far apart and far away.

19.4 EQUIPOTENTIAL SURFACES AND THEIR RELATION TO THE ELECTRIC FIELD An equipotential

surface is a surface on which the electric potential is the same everywhere. The electric force does

no work as a charge moves on an equipotential surface, because the force is always perpendicular

to the displacement of the charge.

V �kq

r

E � 1

2mv2 � 1

2I�2 � mgh � 1

2kx2 � EPE

1

2

1

2

1

2

EPEB

q0

�EPEA

q0

��WAB

q0

V �EPE

q0

Ex. 1, 3

ILW 19.1

IS 19.5

Ex. 2

Ex. 4, 13

Ex. 5

CS 19.1

IS 19.25

Ex. 6, 7, 14

Ex. 8

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FOCUS ON CONCEPTS 593

Relation between the electric field and the potential gradient

A capacitor

Relation between charge and potential difference

Dielectric constant

Capacitance of a parallel plate capacitor

Energy stored in a capacitor

Energy density

The electric field created by any group of charges is everywhere perpendicular to the associated

equipotential surfaces and points in the direction of decreasing potential.

The electric field is related to two equipotential surfaces by

(19.7a)

where �V is the potential difference between the surfaces and �s is the displacement. The term

�V/�s is called the potential gradient.

19.5 CAPACITORS AND DIELECTRICS A capacitor is a device that stores charge and energy. It

consists of two conductors or plates that are near one another, but not touching. The magnitude qof the charge on each plate is given by

(19.8)

where V is the magnitude of the potential difference between the plates and C is the capacitance.

The SI unit for capacitance is the coulomb per volt (C/V), or farad (F).

The insulating material included between the plates of a capacitor is called a dielectric. The di-

electric constant of the material is defined as

(19.9)

where E0 and E are, respectively, the magnitudes of the electric fields between the plates without

and with a dielectric, assuming the charge on the plates is kept fixed.

The capacitance of a parallel plate capacitor filled with a dielectric is

(19.10)

where is the permittivity of free space, A is the area of each plate,

and d is the distance between the plates.

The electric potential energy stored in a capacitor is

(19.11a–c)

The energy density is the energy stored per unit volume and is related to the magnitude E of the

electric field as follows:

(19.12)Energy density � 1

2 �0E 2

Energy � 1

2 qV � 1

2CV 2 � q2/(2C )

�0 � 8.85 � 10�12 C2 / (N m2)

C � �0 A

d

�E0

E

q � CV

E � ��V

�sEx. 9

IS 19.37

Ex. 10, 11, 12

IS 19.53

ILW 19.2

IS 19.45

Topic Discussion Learning Aids

Note to Instructors: The numbering of the questions shown here reflects the fact that they are only a representative subset of the total number that areavailable online. However, all of the questions are available for assignment via an online homework management program such as WileyPLUS or WebAssign.

Section 19.2 The Electric Potential Difference

2. Two different charges, q1 and q2, are placed at two different loca-

tions, one charge at each location. The locations have the same elec-

tric potential V. Do the charges have the same electric potential

energy? (a) Yes. If the electric potentials at the two locations are

the same, the electric potential energies are also the same, regardless

of the type (� or �) and magnitude of the charges placed at these

locations. (b) Yes, because electric potential and electric potential

energy are just different names for the same concept. (c) No, be-

cause the electric potential V at a given location depends on the

charge placed at that location, whereas the electric potential energy

EPE does not. (d) No, because the electric potential energy EPE at

a given location depends on the charge placed at that location as well

as the electric potential V.

4. A proton is released from rest at point A in a constant electric field

and accelerates to point B (see part a of the drawing). An electron is

released from rest at point B and accelerates to point A (see part b of

the drawing). How does the change in the proton’s electric potential

energy compare with the change in the electron’s electric potential

energy? (a) The change in the proton’s electric potential energy is

the same as the change in the electron’s electric potential energy.

(b) The proton experiences a greater change in electric potential

energy, since it has a greater charge magnitude. (c) The proton ex-

periences a smaller change in electric potential energy, since it has a

smaller charge magnitude. (d) The proton experiences a smaller

change in electric potential energy, since it has a smaller speed at Bthan the electron has at A. This is due to the larger mass of the proton.

(e) One cannot compare the change in electric potential energies

because the proton and electron move in opposite directions.

FOCUS ON CONCEPTS

+

Proton

E

AB

(a)

E

E

−

Electron

B

(b)

E

A

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Section 19.3 The Electric Potential Difference Created by Point Charges

6. The drawing shows three arrangements of charged particles, all

the same distance from the origin. Rank the arrangements, largest

to smallest, according to the total electric potential V at the origin.

(a) A, B, C

(b) B, A, C

(c) B, C, A

(d) A, B and C (a tie)

(e) A and C (a tie), B

9. Four pairs of charged particles with identical separations are

shown in the drawing. Rank the pairs according to their electric po-

tential energy EPE, greatest (most positive) first.

(a) A and C (a tie), B and D (a tie)

(b) A, B, C, D

(c) C, B, D, A

(d) B, A, C and D (a tie)

(e) A and B (a tie), C, D

A

−2q +6q

B

+4q

−2q +7q

−5q

C

+5q −3q


(a) A, B, C

(b) A, C, B

(c) C, B, A

(d) C, A, B

(e) B, C, A

12. The drawing shows a plot of the electric potential V versus the

displacement s. The plot consists of four segments. Rank the magni-tude of the electric fields for the four segments, largest to smallest.

(a) D, C, B, A

(b) A and C (a tie), B and D (a tie)

(c) A, B, D, C

(d) B, D, C, A

(e) D, B, A and C (a tie)

Section 19.5 Capacitors and Dielectrics

17. Which two or more of the following actions would increase the

energy stored in a parallel plate capacitor when a constant potential

difference is applied across the plates?

1. Increasing the area of the plates

2. Decreasing the area of the plates

3. Increasing the separation between the plates

4. Decreasing the separation between the plates

5. Inserting a dielectric between the plates

(a) 2, 4 (b) 2, 3, 5 (c) 1, 4, 5 (d) 1, 3

A

+4q +3q

B

−2q +6q

C

−12q −q

D

−4q +3q

A

+50 V +250 V

B

−100 V +150 V

C

−250 V +50 V

Section 19.4 Equipotential Surfaces and Their Relation to the Electric Field

11. The drawing shows edge-on views of three parallel plate capaci-

tors with the same separation between the plates. The potential of

each plate is indicated above it. Rank the capacitors as to the magni-tude of the electric field inside them, largest to smallest.

Note to Instructors: Most of the homework problems in this chapter are available for assignment via an online homework management program suchas WileyPLUS or WebAssign, and those marked with the icon are presented in WileyPLUS using a guided tutorial format that provides enhanced interactivity. See Preface for additional details.

Note: All charges are assumed to be point charges unless specified otherwise.

ssm Solution is in the Student Solutions Manual.This icon represents a biomedical application.

www Solution is available online at www.wiley.com/college/cutnell

PROBLEMS

Section 19.1 Potential Energy,Section 19.2 The Electric Potential Difference

1. During a particular thunderstorm, the electric potential difference

between a cloud and the ground is Vcloud � Vground � 1.3 � 108 V, with

the cloud being at the higher potential. What is the change in an elec-

tron’s electric potential energy when the electron moves from the

ground to the cloud?

2. Multiple-Concept Example 4 provides useful background for this

problem. Point A is at a potential of �250 V, and point B is at a po-

tential of �150 V. An �-particle is a helium nucleus that contains two

protons and two neutrons; the neutrons are electrically neutral. An

�-particle starts from rest at A and accelerates toward B. When the

�-particle arrives at B, what kinetic energy (in electron volts) does it

have?

A

BC

D

Ele

ctri

c po

tent

ial,

V

Displacement, s

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PROBLEMS 595

3. ssm The work done by an electric force in moving a charge from

point A to point B is 2.70 � 10�3 J. The electric potential difference

between the two points is VA � VB � 50.0 V. What is the charge?

4. Refer to Multiple-Concept Example 3 to review the concepts

that are needed here. A cordless electric shaver uses energy at a rate

of 4.0 W from a rechargeable 1.5-V battery. Each of the charged

particles that the battery delivers to the shaver carries a charge that

has a magnitude of 1.6 � 10�19 C. A fully charged battery allows

the shaver to be used for its maximum operation time, during which

3.0 � 1022 of the charged particles pass between the terminals of

the battery as the shaver operates. What is the shaver’s maximum

operation time?

5. Consult Interactive Solution 19.5 at www.wiley.com/college/cutnell to review one method for modeling this problem. Multiple-

Concept Example 3 employs some of the concepts that are needed

here. An electric car accelerates for 7.0 s by drawing energy from its

290-V battery pack. During this time, 1200 C of charge passes through

the battery pack. Find the minimum horsepower rating of the car.

6. An electron and a proton, starting from rest, are accelerated

through an electric potential difference of the same magnitude. In the

process, the electron acquires a speed ve, while the proton acquires a

speed vp. Find the ratio ve/vp.

7. ssm Multiple-Concept Example 4 deals with the concepts that

are important in this problem. As illustrated in Figure 19.6b, a nega-

tively charged particle is released from rest at point B and accelerates

until it reaches point A. The mass and charge of the particle are

4.0 � 10�6 kg and �2.0 � 10�5 C, respectively. Only the gravita-

tional force and the electrostatic force act on the particle, which moves

on a horizontal straight line without rotating. The electric potential at

A is 36 V greater than that at B; in other words, VA � VB � 36 V. What

is the translational speed of the particle at point A?

8. Review Multiple-Concept Example 4 to see the concepts that

are pertinent here. In a television picture tube, electrons strike the

screen after being accelerated from rest through a potential difference

of 25 000 V. The speeds of the electrons are quite large, and for ac-

curate calculations of the speeds, the effects of special relativity must

be taken into account. Ignoring such effects, find the electron speed

just before the electron strikes the screen.

*9. During a lightning flash, there exists a potential difference

reaches the same maximum height h when thrown vertically upward

with a speed of 30.0 m/s. The electric potential at the height h ex-

ceeds the electric potential at ground level. Finally, the particle is

given a negative charge �q. Ignoring air resistance, determine the

speed with which the negatively charged particle must be thrown ver-

tically upward, so that it attains exactly the maximum height h. In all

three situations, be sure to include the effect of gravity.

Section 19.3 The Electric Potential Difference Created by Point Charges

13. Two point charges, �3.40 �C and �6.10 �C, are separated by

1.20 m. What is the electric potential midway between them?

14. Point A is located 0.25 m away from a charge of �2.1 � 10�9 C.

Point B is located 0.50 m away from the charge. What is the electric

potential difference VB � VA between these two points?

15. ssm Two charges A and B are fixed in place, at different dis-

tances from a certain spot. At this spot the potentials due to the two

charges are equal. Charge A is 0.18 m from the spot, while charge B

is 0.43 m from it. Find the ratio qB/qA of the charges.

16. The drawing shows

a square, each side of which

has a length of L � 0.25 m.

On two corners of the square

are fixed different positive

charges, q1 and q2. Find the

electric potential energy of

a third charge q3 � �6.0 �10�9 C placed at corner A and

then at corner B.

17. ssm www Two identical

point charges are fixed to diago-

nally opposite corners of a square

that is 0.500 m on a side. Each

charge is �3.0 � 10�6 C. How

much work is done by the electric

force as one of the charges moves

to an empty corner?

18. The drawing shows four point

charges. The value of q is 2.0 �C,

and the distance d is 0.96 m. Find

the total potential at the location

P. Assume that the potential of a point charge is zero at infinity.

19. The drawing

shows six point

charges arranged in

a rectangle. The

value of q is 9.0 �C,

and the distance d is

0.13 m. Find the

total electric poten-

tial at location P,

which is at the center of the rectangle.

20. Charges of �q and �2q are fixed in place, with a distance

of 2.00 m between them. A dashed line is drawn through the negative

charge, perpendicular to the line between the charges. On the dashed

line, at a distance L from the negative charge, there is at least one spot

where the total potential is zero. Find L.

21. Identical �1.8 �C charges are fixed to adjacent corners of a

square. What charge (magnitude and algebraic sign) should be fixed

to one of the empty corners, so that the total electric potential at the

remaining empty corner is 0 V?

of Vcloud � Vground � 1.2 � 109 V between a cloud and the ground.

As a result, a charge of �25 C is transferred from the ground to the

cloud. (a) How much work Wground–cloud is done on the charge by

the electric force? (b) If the work done by the electric force were

used to accelerate a 1100-kg automobile from rest, what would be

its final speed? (c) If the work done by the electric force were con-

verted into heat, how many kilograms of water at 0 °C could be

heated to 100 °C?

*10. A moving particle encounters an external electric field that de-

creases its kinetic energy from 9520 eV to 7060 eV as the particle

moves from position A to position B. The electric potential at A is

�55.0 V, and the electric potential at B is �27.0 V. Determine the

charge of the particle. Include the algebraic sign (� or �) with your

answer.

*11. ssm www The potential at location A is 452 V. A positively

charged particle is released there from rest and arrives at location Bwith a speed vB. The potential at location C is 791 V, and when re-

leased from rest from this spot, the particle arrives at B with twice the

speed it previously had, or 2vB. Find the potential at B.

**12. A particle is uncharged and is thrown vertically upward from

ground level with a speed of 25.0 m/s. As a result, it attains a maxi-

mum height h. The particle is then given a positive charge �q and

q1 = +1.5 × 10–9 C q2 = +4.0 × 10–9 C

B A

q

q

d

d

d

d

−q

−q

P

−5q −3q +7q

+7q +3q +5q

d d

d

d d

d

P

Problem 18

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22. Three point charges, �5.8 � 10�9 C, �9.0 � 10�9 C, and

�7.3 � 10�9 C, are fixed at different positions on a circle. The total

electric potential at the center of the circle is �2100 V. What is the

radius of the circle?

*23. ssm A charge of �3.00 �C is fixed in place. From a horizontal

distance of 0.0450 m, a particle of mass 7.20 � kg and charge

�8.00 �C is fired with an initial speed of 65.0 m/s directly toward the

fixed charge. How far does the particle travel before its speed is zero?

*24. Two identical point charges (q ��7.20 � 10�6 C) are fixed at di-

agonally opposite corners of a square with sides of length 0.480 m. A

test charge (q0 ��2.40 � 10�8 C), with a mass of 6.60 � 10�8 kg, is

released from rest at one of the empty corners of the square. Determine

the speed of the test charge when it reaches the center of the square.

*25. Refer to Interactive Solution 19.25 at www.wiley.com/college/cutnell to review one way in which this problem can be solved. Two

protons are moving directly toward one another. When they are very

far apart, their initial speeds are 3.00 � 106 m/s. What is the distance

of closest approach?

*26. Four identical charges (�2.0 �C each) are brought from infinity

and fixed to a straight line. The charges are located 0.40 m apart.

Determine the electric potential energy of this group.

*27. ssm Determine the electric

potential energy for the array of

three charges shown in the draw-

ing, relative to its value when the

charges are infinitely far away.

**28. Charges q1 and q2 are fixed

in place, q2 being located at a

distance d to the right of q1. A

third charge q3 is then fixed to

the line joining q1 and q2 at a

distance d to the right of q2. The

third charge is chosen so the potential energy of the group is zero;

that is, the potential energy has the same value as that of the three

charges when they are widely separated. Determine the value for q3,

assuming that (a) q1 � q2 � q and (b) q1 � q and q2 � �q.

Express your answers in terms of q.

**29. Two particles each have a mass of 6.0 � kg. One has a

charge of �5.0 � C, and the other has a charge of �5.0 � C.

They are initially held at rest at a distance of 0.80 m apart. Both are

then released and accelerate toward each other. How fast is each

particle moving when the separation between them is one-third its

initial value?

**30. A positive charge of �q1 is located 3.00 m to the left of a negative

charge �q2. The charges have different magnitudes. On the line through

the charges, the net electric field is zero at a spot 1.00 m to the right of

the negative charge. On this line there are also two spots where the po-

tential is zero. Locate these two spots relative to the negative charge.

Section 19.4 Equipotential Surfaces and Their Relation to the Electric Field

31. Two equipotential surfaces surround a �1.50 � 10�8-C point

charge. How far is the 190-V surface from the 75.0-V surface?

32. An equipotential surface that surrounds a point charge q has a

potential of 490 V and an area of 1.1 m2. Determine q.

33. ssm A spark plug in an automobile engine consists of two metal

conductors that are separated by a distance of 0.75 mm. When an

electric spark jumps between them, the magnitude of the electric field

is 4.7 � 107 V/m. What is the magnitude of the potential difference

�V between the conductors?

10�610�6

10�3

10�3


34. A positive point charge (q � �7.2 � 10�8 C) is surrounded

+8.00 Cμ

+20.0 Cμ–15.0 Cμ

4.00 m

3.00 m

90.0°

Problem 27

by an equipotential surface A, which has a radius of rA � 1.8 m. A pos-

itive test charge (q0 ��4.5 � 10�11 C) moves from surface A to an-

other equipotential surface B, which has a radius rB. The work done as

the test charge moves from surface A to surface B is WAB � �8.1 �10�9 J. Find rB.

35. ssm www The inner and outer surfaces of a cell membrane

carry a negative and a positive charge, respectively. Because

of these charges, a potential difference of about 0.070 V exists across

the membrane. The thickness of the cell membrane is 8.0 � 10�9 m.

What is the magnitude of the electric field in the membrane?

36. The drawing shows

a graph of a set of

equipotential surfaces

seen in cross section.

Each is labeled according

to its electric potential.

A �2.8 � 10�7 C point

charge is placed at posi-

tion A. Find the work

that is done on the point charge by the electric force when it is moved

(a) from A to B, and (b) from A to C.

*37. Refer to Interactive Solution 19.37 at www.wiley.com/college/cutnell to review a method by which this problem can be solved. An

electric field has a constant value of 4.0 � 103 V/m and is directed

downward. The field is the same everywhere. The potential at a point

P within this region is 155 V. Find the potential at the following

points: (a) 6.0 � m directly above P, (b) 3.0 � m

directly below P, (c) 8.0 � m directly to the right of P.

*38. An electron is released

from rest at the negative plate of a

parallel plate capacitor and accel-

erates to the positive plate (see the

drawing). The plates are separated

by a distance of 1.2 cm, and the

electric field within the capacitor

has a magnitude of 2.1 � 106 V/m.

What is the kinetic energy of the

electron just as it reaches the positive plate?

*39. ssm The drawing shows the potential at

five points on a set of axes. Each of the four

outer points is 6.0 � m from the point at

the origin. From the data shown, find the mag-

nitude and direction of the electric field in the

vicinity of the origin.

*40. Equipotential surface A has a potential of 5650 V, while

equipotential surface B has a po-

tential of 7850 V. A particle has

a mass of 5.00 � kg and a

charge of �4.00 � C. The

particle has a speed of 2.00 m/s

on surface A. A nonconservative

outside force is applied to the

particle, and it moves to surface

B, arriving there with a speed of

3.00 m/s. How much work is

done by the outside force in

moving the particle from A to B?

*41. The drawing shows a uni-

form electric field that points in

the negative y direction; the

10�5

10�2

10�3

10�3

10�310�3

A

B

C

D+150.0 V

+250.0 V

+350.0 V

+650.0 V

+550.0 V

+450.0 V

2.0 cm

++++++++++

−−−−−−−−−−

−

Electric field

Electron

FB

505 V

505 V

505 V

515 V495 V

8.0 cm10.0 cm

6.0 cmA

E E

B

C

+y

+x

Problem 41

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ADDITIONAL PROBLEMS 597

magnitude of the field is 3600 N/C. Determine the electric potential

difference (a) VB � VA between points A and B, (b) VC � VB be-

tween points B and C, and (c) VA � VC between points C and A.

Section 19.5 Capacitors and Dielectrics

42. What voltage is required to store 7.2 � C of charge on the

plates of a 6.0-�F capacitor?

43. ssm The electric potential energy stored in the capacitor of

a defibrillator is 73 J, and the capacitance is 120 �F. What

is the potential difference that exists across the capacitor plates?

44. An axon is the relatively long tail-like part of a neuron, or

nerve cell. The outer surface of the axon membrane (dielec-

tric constant � 5, thickness � 1 � m) is charged positively, and

the inner portion is charged negatively. Thus, the membrane is a kind

of capacitor. Assuming that the membrane acts like a parallel plate ca-

pacitor with a plate area of 5 � m2, what is its capacitance?

45. Refer to Interactive Solution 19.45 at www.wiley.com/college/cutnell for one approach to this problem. The electronic flash attach-

ment for a camera contains a capacitor for storing the energy used to

produce the flash. In one such unit, the potential difference between

the plates of an 850-�F capacitor is 280 V. (a) Determine the energy

that is used to produce the flash in this unit. (b) Assuming that the

flash lasts for 3.9 � s, find the effective power or “wattage” of

the flash.

46. Two capacitors have the same plate separation, but one has

square plates and the other has circular plates. The square plates are a

length L on each side, and the diameter of the circular plate is L. The

capacitors have the same capacitance because they contain different

dielectric materials. The dielectric constant of the material between

the square plates has a value of square � 3.00. What is the dielectric

constant circle of the material between the circular plates?

47. ssm A parallel plate capacitor has a capacitance of 7.0 �F when

filled with a dielectric. The area of each plate is 1.5 m2 and the sepa-

ration between the plates is 1.0 � m. What is the dielectric con-

stant of the dielectric?

48. Two capacitors are identical, except that one is empty and

10�5

10�3

10�6

10�8

10�5

away the same amount of water at 100 °C. The capacitance of capac-

itor A is 9.3 �F. What is the capacitance of capacitor B?

*51. ssm Interactive LearningWare 19.2 at www.wiley.com/college/cutnell reviews the concepts pertinent to this problem. What is the

potential difference between the plates of a 3.3-F capacitor that stores

sufficient energy to operate a 75-W light bulb for one minute?

*52. A capacitor is constructed of two concentric conducting cylindri-

cal shells. The radius of the inner cylindrical shell is 2.35 � 10�3 m,

and that of the outer shell is 2.50 � 10�3 m. When the cylinders carry

equal and opposite charges of magnitude 1.7 � 10�10 C, the electric

field between the plates has an average magnitude of 4.2 � 104 V/m

and is directed radially outward from the inner shell to the outer shell.

Determine (a) the magnitude of the potential difference between

the cylindrical shells and (b) the capacitance of this capacitor.

*53. Refer to Interactive Solution 19.53 at www.wiley.com/college/cutnell and Multiple-Concept Example 10 for help in solving this

problem. An empty capacitor has a capacitance of 3.2 �F and is con-

nected to a 12-V battery. A dielectric material ( � 4.5) is inserted

between the plates of this capacitor. What is the magnitude of the sur-

face charge on the dielectric that is adjacent to either plate of the

capacitor? (Hint: The surface charge is equal to the difference in thecharge on the plates with and without the dielectric.)

*54. An empty parallel plate capacitor is connected between the ter-

minals of a 9.0-V battery and charged up. The capacitor is then dis-

connected from the battery, and the spacing between the capacitor

plates is doubled. As a result of this change, what is the new voltage

between the plates of the capacitor?

**55. ssm The drawing shows a parallel plate capac-

itor. One-half of the region between the plates is

filled with a material that has a dielectric constant

1. The other half is filled with a material that has

a dielectric constant 2. The area of each plate is A,

and the plate separation is d. The potential differ-

ence across the plates is V. Note especially that the charge stored by

the capacitor is q1 � q2 � CV, where q1 and q2 are the charges on

the area of the plates in contact with materials 1 and 2, respectively.

Show that C � �0 A( 1 � 2)/(2d ).

**56. If the electric field inside a capacitor exceeds the dielectricstrength of the dielectric between its plates, the dielectric will break

down, discharging and ruining the capacitor. Thus, the dielectric

strength is the maximum magnitude that the electric field can have

without breakdown occurring. The dielectric strength of air is 3.0 �106 V/m, and that of neoprene rubber is 1.2 � 107 V/m. A certain air-

gap, parallel plate capacitor can store no more than 0.075 J of elec-

trical energy before breaking down. How much energy can this

capacitor store without breaking down after the gap between its

plates is filled with neoprene rubber?

the other is filled with a dielectric ( � 4.50). The empty capacitor is

connected to a 12.0-V battery. What must be the potential difference

across the plates of the capacitor filled with a dielectric so that it

stores the same amount of electrical energy as the empty capacitor?

49. A capacitor has a capacitance of 2.5 � F. In the charging

process, electrons are removed from one plate and placed on the other

plate. When the potential difference between the plates is 450 V, how

many electrons have been transferred?

*50. Capacitor A and capacitor B both have the same voltage

across their plates. However, the energy of capacitor A can melt mkilograms of ice at 0 °C, while the energy of capacitor B can boil

10�8

k1 k2

ADDITIONAL PROBLEMS

57. ssm The membrane that surrounds a certain type of living

cell has a surface area of 5.0 � m2 and a thickness of

1.0 � m. Assume that the membrane behaves like a parallel

plate capacitor and has a dielectric constant of 5.0. (a) The poten-

tial on the outer surface of the membrane is �60.0 mV greater than

that on the inside surface. How much charge resides on the outer sur-

10�8

10�9

face? (b) If the charge in part (a) is due to K� ions (charge �e),

how many such ions are present on the outer surface?

58. A particle has a charge of �1.5 �C and moves from point A to

point B, a distance of 0.20 m. The particle experiences a constant

electric force, and its motion is along the line of action of the force.

2762T_ch19_569-598.qxd 7/4/08 2:32 PM Page 597




The difference between the particle’s electric potential energy at Aand at B is EPEA � EPEB � �9.0 � 10�4 J. (a) Find the magnitude

and direction of the electric force that acts on the particle. (b) Find

the magnitude and direction of the electric field that the particle

experiences.

59. An electron and a proton are initially very far apart (effectively

an infinite distance apart). They are then brought together to form a

hydrogen atom, in which the electron orbits the proton at an average

distance of 5.29 � m. What is EPEfinal � EPEinitial, which is the

change in the electric potential energy?

60. The drawing that accompanies Problem 36 shows a graph of a set

of equipotential surfaces in cross section. The grid lines are 2.0 cm

apart. Determine the magnitude and direction of the electric field at

position D. Specify whether the electric field points toward the top or

the bottom of the drawing.

61. ssm Suppose that the electric potential outside a living cell

is higher than that inside the cell by 0.070 V. How much

work is done by the electric force when a sodium ion (charge � �e)

moves from the outside to the inside?

62. Location A is 3.00 m to the right of a point charge q. Location Blies on the same line and is 4.00 m to the right of the charge. The po-

tential difference between the two locations is VB � VA � 45.0 V.

What are the magnitude and sign of the charge?

63. A capacitor stores 5.3 � C of charge when connected to a

6.0-V battery. How much charge does the capacitor store when con-

nected to a 9.0-V battery?

*64. Review Conceptual Example 11 before attempting this problem.

An empty capacitor is connected to a 12.0-V battery and charged up.

The capacitor is then disconnected from the battery, and a slab of di-

electric material ( � 2.8) is inserted between the plates. Find the

amount by which the potential difference across the plates changes.

Specify whether the change is an increase or a decrease.

*65. The drawing shows the electric potential as a function of dis-

tance along the x axis. Determine the magnitude of the electric field

in the region (a) A to B, (b) B to C, and (c) C to D.

10�5

10�11


**66. Review Conceptual Example 7 as background for this problem.

A positive charge �q1 is located to the left of a negative charge �q2.

On a line passing through the two charges, there are two places where

the total potential is zero. The first place is between the charges and is

4.00 cm to the left of the negative charge. The second place is 7.00 cm

to the right of the negative charge. (a) What is the distance between

the charges? (b) Find �q1�/�q2�, the ratio of the magnitudes of the

charges.

**67. ssm www The potential difference between the plates of a ca-

pacitor is 175 V. Midway between the plates, a proton and an electron

are released. The electron is released from rest. The proton is pro-

jected perpendicularly toward the negative plate with an initial speed.

The proton strikes the negative plate at the same instant that the elec-

tron strikes the positive plate. Ignore the attraction between the two

particles, and find the initial speed of the proton.

**68. One particle has a mass of 3.00 � kg and a charge of

�8.00 �C. A second particle has a mass of 6.00 � kg and the

same charge. The two particles are initially held in place and then re-

leased. The particles fly apart, and when the separation between

them is 0.100 m, the speed of the 3.00 � -kg particle is 125 m/s.

Find the initial separation between the particles.

10�3

10�3

10�3

A B

C

D

0

5.0

4.0

3.0

2.0

1.0

00.20

Pot

enti

al,

V

0.40

x, m

0.60 0.80

Problem 65

2762T_ch19_569-598.qxd 7/4/08 2:32 PM Page 598

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Ch 19 - Electric Potential Energy - Electric Potential (569-598)