electrical field and electrical potential chapter 19 electrical field and electrical potential we...

CHAPTER NINE TEEN

Electrical Field and Electrical Potential

In Chapter 18, we explained a range of observations bydeveloping a model in which charged carriers exert electro-

static forces on one another. In this chapter, we will formulatethe concepts of electrostatic force and electric charge quantita-tively. As we did for gravitational forces moving bodies overdistances, we will develop related ideas of work and potentialenergy. But suppose we want to know, say, how a batteryaffects the charge passing through it. We don’t know whatthe battery will be connected to or for how long. In casessuch as this, we must think about what happens to each unit

of charge. We will introduce quantities that tell how much forceis exerted on each unit of charge and how much potential energy

it gains or loses.

“. . . suppose wewant to know, say,

how a batteryaffects the charge

passing through it.”

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19-1 Making Electrostatic Force and Charge Concepts Quantitative ◆ 567

19-1 Making Electrostatic Force and Charge Concepts Quantitative

French physicist Charles Augustin de Coulomb (1736–1806)put the concepts of electrostatic force and electric charge ona quantitative basis. (The historical record seems to indicatethat Cavendish did this earlier, but Coulomb’s results werethe first to be widely disseminated in print.) Coulombreported a series of measurements on charged bodies usinga torsion balance apparatus much like the one Cavendishused (Figure 8-16) to determine the universal gravitationalconstant G. In Coulomb’s apparatus (Figure 19-1), if the fiberis twisted, it exerts a restoring torque proportional to theangle of twist. The restoring torque it has to exert to keepthe “barbell” in equilibrium is in turn proportional to the electrostatic force thatsphere 1 exerts on sphere 2 when both are charged. It had previously beensuspected, by analogy with gravitational forces, that the electrostatic forcebetween charged point objects varied inversely as By observing the angle oftwist with the spheres at different separations r, Coulomb showed that this wasindeed the case.

Coulomb could also reduce the charge on either sphere by half to see howthe force varied with charge. He did this by touching the sphere in question witha conducting sphere identical to it but uncharged. He reasoned that the excesscharges on the charged sphere would by mutual repulsion spread out equallyover both spheres. Then when the introduced sphere is removed, it should takehalf the excess charge with it. In this way he found the electrostatic force to beproportional both to the charge on sphere 1 and the charge on sphere 2. Theserelationships are summarized by Coulomb’s law.

Coulomb’s law: The magnitude of the electrostaticforce that each of two point objects with charges and exerts on the otherat a separation is given by

(19-1)

Note: The magnitude gives the strength of the interaction.

Direction: The forces on the two bodies are directed along the line connectingthem. The force on either body is toward the other body if the two have oppo-site charges and away if they have like charges.

Note that the signs of the charges are used only in determining the force’s direc-tion, not to find the magnitude of the force.

✦UNITS In SI, the unit of charge, not surprisingly, is called the coulomb (C).(Its definition follows from ideas that we will treat later on.) The value of theproportionality constant k could in principle be found by measuring the force that1 C charges exert on each other at a separation of 1 m. The resulting force wouldbe (verify for yourself that this is about a million tons!). This tells usthat one coulomb is an extremely large amount of charge. It then follows fromEquation 19-1 that

or (19-2)k � 9 � 109 N � m2

C2

k � F

r 122

�q1� �q2�

� 19 � 109 N2 11 m2211 C2 11 C2

9 � 109 N

F � k

�q1� �q2�

r 122

r12

q2q1

F � 0FSon 1 by 2 0 � 0FSon 2 by 1 0

r2.

Knob that can twistfiber to vary r

r

Fiber

Longdistance

Uncharged sphere equalin weight to sphere 2keeps “barbell” balanced

1 2

Figure 19-1 Coulomb’s torsionbalance. Based on drawing inCoulomb’s paper in Histoire etMémoires de l’Académie Royal desSciences, 1785.

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568 ◆ Chapter 19 Electrical Field and Electrical Potential

We cannot physically do the measurement using 1 C charges, but because k is aconstant, we could get the same value from a much smaller measured forcebetween much smaller charges.

With the humongous coulomb as our unit of charge, the size of nature’s basicunit of charge, that of a single electron, is just a tiny fraction of a coulomb:

(19-3)

(We’ll generally use the more rounded-off value.) The symbol e represents themagnitude only: The charge of a proton is that of an electron is Elec-trostatic forces now join the inventory of forces that we add as vectors when weconsider in Newton’s second law.©F

S

�e.�e;

e � 1.602 � 10�19 C � 1.6 � 10�19 C

Example 19-1 Vector Addition of Electrostatic Forces

For a guided interactive solution, go to Web Example 19-1 atwww.wiley.com/college/touger

The charges and masses of the three point objects shown below are

a. If bodies B and C are fixed in their positions, find the total force on Awhen it is at the origin.

b. Find the acceleration of A at the instant when it is at this position.

Brief SolutionChoice of approach. a. We must (1) identify the individual forces on A andtheir directions in a free-body diagram (see figure), (2) determine the magni-tude of each of these forces using Coulomb’s law, and then (3) find the x andy components and use them to find the resultant vector. In (1), we use thesigns on the charges to determine whether each force is attractive or repulsiveand thus establish its direction along its line of action.

b. To find the acceleration, we use Newton’s second law.

What we know/what we don’t.

To find To find Constants

Second law applied to A

The mathematical solution. a. We find magnitudes by Equation 19-1:

� 5.5 � 10�3 N

Fon A by C � k

�qA� �qC�

rAC2 � 9 � 109

N � m2

C2

12.4 � 10�6 C2 12.0 � 10�6 C212.8 m22

� 7.0 � 10�3 N

Fon A by B � k

�qA� �qB�

rAB2 � 9 � 109

N � m2

C2

12.4 � 10�6 C2 11.3 � 10�6 C212.0 m22

a � ?uF � ?total Fon A � ?

mA � 0.001 kg

rAC � 212.0 m22 � 12.0 m22 � 2.8 mrAB � 2.0 m

0qC 0 � 2.0 � 10�6 C0qB 0 � 1.3 � 10�6 C

k � 9 � 109

N � m2

C20qA 0 � 2.4 � 10�6 C0qA 0 � 2.4 � 10�6 C

Fon A by CFon A by B

mA � 0.001 kg mB � 0.002 kg mC � 0.002 kg

qA � 2.4 � 10�6 C qB � 1.3 � 10�6 C qC � �2.0 � 10�6 C

A+

B

P

C at(2.0m, 2.0m)

+ –

Free bodydiagram of A

Fon A by C

(attractive)

line of actionof Fon A by B

line of actionof Fon A by C

Fon A by B

(repulsive)

y

xq

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19-2 Electric Fields ◆ 569

19-2 Electric FieldsFor an important new perspective on the calculation we did in Example 19-1,notice that we multiplied by the charge in finding each of the forces and We used these in turn to find the components of these forces andthen to find the total components and and the magnitude of the resultant force. So was built in as a multiplier in finding all of these.Thus, each of the electrostatic forces on the charge and each of their com-ponents, can be viewed as the product of and “everything else” multiplying it. For example we can write the total force on A very informally as

It then follows that the magnitude of the total electrostatic force is likewisethe product of and “everything else.” If is at a point P, the “everythingelse” multiplying it involves the other charges, their distances from P, and becausewe consider directions when we find components, the directions from P to thosecharges. In short we can say that the “other stuff” involves the electrical envi-ronment of the point P at which was placed. To see this reasoning presentedin graphic step-by-step detail, go to WebLink 19-1.

Recall from Chapter 4 that a force is one side of an interaction. A body andits environment exert equal and opposite electrostatic forces on each other. Wemay consider the magnitude of either force to be a measure of the strength ofthe interaction. The strength of the interaction is thus the product of a propertyof the object itself (its charge) and a term involving the electrical properties(charges and their configuration) of its environment. This latter term is what we

qA

qA0qA 0Fon A

1Fon A2y � 0qA 0 aeverythingelse

by

1Fon A2x � 0qA 0 aeverythingelse

bx

1FSon A20qA 0

qA,0qA 0

Fon A1Fon A2y1Fon A2xFon A by C.

Fon A by BqA

In finding the x and y components of these forces, we note that is inthe negative y direction, and from the geometry of triangle APC, Thus,

Now we add components to get the components of the total force on A:

Finally, we convert to a magnitude and direction description:

so

b. By Newton’s second law,

◆ Related homework: Problems 19-7, 19-8, 19-9, and 19-11.

a �Fon A

mA�

5 � 10�3 N

1 � 10�3 kg� 5 m�s2

uF � �38� 138° below the �x axis2. tan uF �

1Fon A2y1Fon A2x �

�3.1 � 10�3 N

3.9 � 10�3 N� �0.79

� 5.0 � 10�3 N

Fon A � 21Fon A2x2 � 1Fon A2y2 � 213.9 � 10�3 N22 � 1�3.1 � 10�3 N22

� �3.1 � 10�3 N

1Fon A2y � 1Fon A by B2y � 1Fon A by C2y � �7.0 � 10�3 N � 3.9 � 10�3 N

1Fon A2x � 1Fon A by B2x � 1Fon A by C2x � 0 � 3.9 � 10�3 N � 3.9 � 10�3 N

1Fon A by C2y � 1Fon A by C2 sin 45° � 15.5 � 10�3 N2 10.7072 � 3.9 � 10�3 N

1Fon A by C2x � 1Fon A by C2 cos 45° � 15.5 � 10�3 N2 10.7072 � 3.9 � 10�3 N

1Fon A by B2y � �7.0 � 10�3 N1Fon A by B2x � 0

u � 45°.Fon A by B

For WebLink 19-1:The Electric Field—

The Charge Carrier’sEnvironment, go towww.wiley.com/college/touger

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call the electrical field, and we denote its magnitude by E:

or (19-4)strength of property properties of A’selectrical of A environment

interactionbetween A and its

environment

(We use the absolute value of because F and E are magnitudes, so they are always positive.)

You can associate the symbol E with electrical environment, or with “everything else.”Notice that if were moved to a different position, its distance and direc-

tion from each other charge would change, that is, the “other stuff” contributingto the field E would change. In essence, experiences a different environment—its surroundings appear different—when it is in a different position. The mathe-matical way of saying this is: The electric field is a function of position. Changingposition affects each component of the force on A, so in two dimensions we canwrite an equation like 19-4 for each direction

(19-5x, 19-5y)

Here we use the charge rather than its absolute value because the x and y com-ponents of a force can be either positive or negative. We summarize the twocomponent equations by the vector equation

(19-5)

STOP&Think If after doing Example 19-1, you had to find the force on a differentcharge (call it ) placed at the origin, would you have had to do the entire cal-culation over again using the value of ? ◆

The above question suggests there is an advantage to factoring the force asin Equation 19-5. If we did the calculation in Example 19-1 with the factor left out, we could as easily have multiplied by or at the end. In otherwords, we could first have calculated the field at a particular point, and then mul-tiplied by the charge to find the force on any charge carrier placed at that point.

From Equation 19-4, the magnitude of the electric field is so it ismeasured in N/C. This is the number of newtons on each coulomb of charge.

The electric field at a point is the force that will be exerted on each coulomb of chargeplaced at that point.

In Equation 19-5, the force vector is the product of the scalar charge and thevector field. Multiplying a vector by a positive scalar preserves its direction; mul-tiplying it by a negative scalar reverses its direction:

If the charge is positive, the electrostatic force on it is in the same direction as the field;if the charge is negative, the force is opposite in direction to the field.

Another important point follows from this:

The direction of the electric field at any point is the direction of the total electrostaticforce that a positive charge placed at that point would experience.

✦ELECTRIC FIELD DUE TO A POINT CHARGE For two isolated chargesand Equation 19-4 becomes

(19-6)

where is the field at the point—let’s call it P—where is placed. Thefield is due to the environment of charges surrounding P; in this case the onlycharge in the environment is Comparing Equation 19-6 with Coulomb’s law q1.

q2Edue to 1

Fon 2 by 1 � 0q2 0Edue to 1

q2,q1

E �Fon A0qA 0 ,

qDqA

qA

qD

qD

FS

on A � qAES

Fy on A � qAEyFx on A � qAEx

qA

qA

qA

Fon A � 0qA 0 E

Fon A � 0qA 0 a“everythingelse”

b

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19-2 Electric Fields ◆ 571

(Equation 19-1), we see that Because the field is the force on eachcoulomb (C) placed at P, and because 1 C is a positive amount of charge, the fieldat P is in the direction of the force that any positive charge placed there wouldexperience. Simplifying our notation, we can now summarize.

Electric field at a position a distance r from a point charge q

(19-7)

Direction: Away from q if q positive,toward q if q negative.

If the environment of a point P is made up of many point charges, the totalfield at P is the vector sum of the fields due to the individual charges. An impor-tant first step is to draw a diagram showing the individual fields contributing tothe sum. In the diagram in Figure 19-2a,

1. and are directed away from the positive charges on B and C (see rulesfor direction under Equation 19-7) but is toward the negative charge on A;

2. although A and B are the same distance from point P, because themagnitude of the charge on A is greater; and

3. although B and C are equally charged, because B is further from P(so r in Equation 19-7 is larger).

The diagram is much like a free-body diagram but inventories fields rather thanforces. Another important difference is this: There need not be any object at P. Theenvironment of point P is present whether or not anything is at P to interact withit. In contrast to the two-tone arrows representing forces in Figures 19-2b and c, thefield arrows are a single color to remind you that fields represent only one contri-bution to the interaction. If a charged body is then placed at P, it provides the othercontribution, resulting in a force. We then get the free-body diagrams in Figures19-2b and c, showing the forces on a positively and a negatively charged body at P.Note that for the negatively charged body, the forces are opposite to the fields.

EB � EC

EA � EB

ES

A

ES

CES

B

Magnitude: E � k

0q 0r2

Edue to 1 � k

0q1 0r 12

2 .

A

B

CEA

Fon by B

EB

EC

++

– –

–

– ––

+

+

+

Fon by A+

Fon by C+ Fon by A–

Fon by C–

Fon by B–

+

–

(a) (b) (c)

PROCEDURE 19-1

Finding the Electric Field at a Point PDue to an Array of Point Charges

1. Draw a field inventory diagram (such as Figure 19-2a) showing the fieldsat P due to each individual charge. The directions are away from positivecharges and toward negative charges.

2. Find the distance from P to each charge.3. Use Equation 19-7 to calculate the magnitude of each field.4. Find the x and y components of each field and do vector addition to find

the resultant field vector (Procedure 3-4).

Figure 19-2 Field inventory andfree-body diagrams. (a) Fieldinventory diagram showing fields atpoint P due to surrounding charges.(b) Free-body diagram of pointobject with charge placed at P.(c) Free-body diagram of pointobject with charge placed at P.�1 C

�1 C

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Example 19-2 A Field Approach to Example 19-1


a. Redo Example 19-1a by finding the field first.b. Repeat for the case when point object A is an electron.

Brief SolutionChoice of approach. In following Procedure 19-1, we repeat the calculationwe did for Example 19-1a but leaving out the factor until the end.STOP&Think How does the above field inventory diagram compare with thefree-body diagram in Example 19-1? ◆

What we know/what we don’t. See Example 19-1.

The mathematical solution. a. (Compare with Example 19-1) Apply Equation19-7 at the origin:

As before, we add components to get the component description of thetotal field at the origin (the position where A is or will be placed).

and convert to a magnitude and direction description:

E is in the same direction as the force in Example 19-1 (except for error due torounding off). Also agreeing with Example 19-1, the magnitude of the force is

b. Because we calculated the field first, we need only multiply by the differ-ent charge ( for an electron) placed at the origin:

The direction of the force on a negative charge is reversed (changed by see part b of figure) from that of the field, so in this case


uF � �39° � 180° � 141�

180°;

Fon e� � e E � 11.6 � 10�19 C2 12.1 � 103 N/C2 � 3.4 � 10�16 N

0qA 0 � e

Fon A � 0qA 0E � 12.4 � 10�6 C2 12.1 � 103 N/C2 � 5.0 � 10�3 N

tan uE �Ey

Ex� �0.81 so u � �39�E � 2E x

2� E y

2� 2.1 � 103 N/C

Ey � 1Edue to B2y � 1Edue to C2y � �1.3 � 103 N/C

Ex � 1Edue to B2x � 1Edue to C2x � 1.6 � 103 N/C

1Edue to C2y � 1Edue to C2 sin 45° � 1.6 � 103 N/C

1Edue to C2x � 1Edue to C2 cos 45° � 1.6 � 103 N/C

1Edue to B2x � 0 1Edue to B2y � �2.9 � 103 N/C

Edue to C � k

0qC 0r2AC

� 9 � 109 N � m2

C2

12.0 � 10�6 C2

12.8 m22 � 2.3 � 103 N/C

Edue to B � k

0qB 0r2AB

� 9 � 109

N � m2

C2

11.3 � 10�6C212.0 m22 � 2.9 � 103 N/C

0qA 0

19-3 Fields Due to Continuous Charge DistributionsWhen the source of the field is not a point object, we may have to considerfields due to charges spread out more or less continuously over part or all of thebody. We will use qualitative reasoning to draw conclusions about some com-mon distributions of charge.

–

Eat origin due to B

Ftotal onnegative charge

Eat origin due to C

Etotal

qE

qF

qE = field makes with +x axis

(b) Directions of resultant field and force vectors

(a) Field inventory diagram

qF = force makes with +x axis

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19-3 Fields Due to Continuous Charge Distributions ◆ 573

✦CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM In Example 18-2, wededuced that if a body is a conductor in electrostatic equilibrium, all excess chargemust be on the body’s surface. If there is an electric field in the interior of such abody, positive charge or conventional current will flow in the direction of the field.In electrostatic equilibrium, there is no net motion of charge, so we must conclude:

There is no electric field anywhere in the interior of a conductor in electrostaticequilibrium.

What about at the surface? Suppose a field at a point P on the surface (Fig-ure 19-3) has components parallel and perpendicular to the surface. By Equation19-5, the electrostatic force on any excess charge at that point would also haveparallel and perpendicular components. But although would be opposedby the constraining forces that keep the charges from leaving the body, would be unopposed, so charge would move along the surface in the directionof But then the conductor would not be in electrostatic equilibrium. Weavoid this contradiction only if and therefore is zero.

The electric field must be perpendicular to the surface at all points on the surface of aconductor in electrostatic equilibrium.

Applying Procedure 19-1 in detail to a continuous charge distribution wouldrequire calculus. But we can apply Procedure 19-1 in part, together with symmetryarguments, to determine the direction of the field due to some simple charge config-urations. Fortunately, these are the configurations commonly found in electric circuits.

E//,1Fe2//,1Fe2//.

1Fe2//1Fe2�

Fe

(Fe)⊥(Fe)

+

+E

Fconstraining

Fe on

E⊥E

P P

Figure 19-3 Fields and forces atthe surface of a conductor.

Example 19-3 The Electric Field along the Axis of a Uniformly Charged Circular Ring

Find the direction of the electric field along the axis of a circular ring with auniformly distributed positive charge (Figure 19-4a).

SolutionChoice of approach. Mentally we can break up the ring of charge into piecesso tiny that they are nearly point charges. Guided by symmetry considerations,we can select representative pieces. We will draw a field inventory diagramfor a point P on the axis (Step 1 of Procedure 19-1), and then sketch the vec-tor addition to determine the direction of the total field.

The reasoning. We can choose pairs of pieces (e.g., 1 and 2, 3 and 4 in Fig-ure 19-4b) that are diametrically across from each other. The field due to each

E1 + E2

E2 E1

E2y E1y

E1xE2x

1

3

4

2

(b)

E1 + E2

E3 + E4

etc...

(c)

+

+ +

+

+

++

+

+

+

+

+

++++

++

+ +

++

++

++

P

(a)

Figure 19-4 The electric fieldalong the axis of a uniformlycharged circular ring. SeeExample 19-3.

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✦CONDUCTORS NOT IN ELECTROSTATIC EQUILIBRIUM By maintaininga difference in the surface charge density at its two poles, a battery in a simplecircuit prevents the conducting path from reaching electrostatic equilibrium, sothere continues to be a net motion of charge, that is, a current. Because elec-trostatic equilibrium is not reached, the interior field does not become zero. InExample 19-4, we find its direction by extending the approach of Example 19-3.

piece is directed away from that piece. Figure 19-4b shows that due to thesymmetry in our choice, the horizontal components of the fields and (due to charges 1 and 2) are equal and opposite. Then the resultant of thesetwo fields is along the axis. The same reasoning applies to each such pair ofcharges, so the vector sum of the contributions of all pairs (Figure 19-4c)—thefield due to the entire ring—is likewise along the axis. By considering the sur-face of a conducting wire to be made up of many such rings, we can buildon this result to find the field along the axis of the wire—see Example 9-4.

◆ Related homework: Problem 19-22.

ES

2ES

1

EXAMPLE 19-3 continued

Figure 19-5 Electric field in astraight wire. (Summary of mainfield ideas in Example 19-4).

Example 19-4 Electric Field in a Wire Connected to a Battery


In Chapter 18 we showed that when a conducting wire is connected betweenthe terminals of a battery, the battery causes a surface charge density gradientalong the wire.a. Find the direction of the electric field in the interior of the wire. In your

reasoning, consider a long, straight segment of the wire (Figure 19-5a).b. Find the direction of the electrostatic force on an electron moving through

the wire, and on a positive charge carrier (such as we consider in a con-ventional current) moving through the wire.

Brief SolutionChoice of approach. We can find the direction of the field by the same approachas the previous example. Because a straight wire is cylindrical, we can thinkof the charge on its surface as being made up of a stack of loops of charge,each like the one in Example 19-3.

vector sumE1 + E2 downward

regionblown upin (b)

regionblown upin (c)

+

+

E2E1

(b) (c)(a)

P

–

ring 1

ring 2–

+

+

–

+

+

+

+

++

–

–

–

–––

+

+

+

+

++

–

–

–

–––

–

vector sumE3 + E4 downward

+

+

E3

E4

P

ring 3

ring 4

+++++

+

++

++

1

4

3

2

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Figure 19-6 Electric fields due to point charges. (a) Force per unit charge (field) on atest charge at several different positions. (b) A more detailed mapping of field vectors atdifferent positions. (c) Electric field lines for the mapping in b. (d) Electric field lines fora greater point charge. (e) Electric field lines for a negative point charge.

19-4 Picturing Electric Fields ◆ 575

19-4 Picturing Electric FieldsIt is often useful to get an overall picture of the electric field, which we can doas follows. We can sample the effects of the field by placing a point positivecharge—a “test charge”—at different positions and observing how the magnitudeand direction of the force on it vary from point to point. This is shown in Fig-ure 19-6. The test charge in Figure 19-6a is sampling the field due to the pointcharge in the center; the figure shows the force per unit charge on it (the field)at several different points. Figure 19-6b shows a more detailed mapping, but it isstill representative; if we made the map complete, the arrows would totally blackenthe region and make the map unreadable. So we have to imagine the arrows atthe in-between points; in reality, the field is continuous, changing gradually inmagnitude and direction from one point to the next except right at the pointcharge(s) giving rise to the field. The procedure and some of the implications ofthe mapping that we get are shown in step-by-step detail in WebLink 19-2.

The mappings are sometimes summarized by drawing a continuous set oflines, called electric field lines (Figure 19-6c). The field vectors in Figure 19-6bwould be directed along these lines. For fields due to more complicated arrange-ments of charge, the field lines might be curved. The field vector at each pointwould then be directed tangent to the field line.

Because the field lines in Figure 19-6c don’t show the lengths of the fieldvectors, they must communicate the field’s strength in another way. Figure 19-6dsuggests how this happens. There, because the field is due to a greater point

The details.a. First consider the field at a point P on the axis. From Example 19-3, we

know the field contribution from each ring is along the axis at point P.Consider pairs of rings at equal distances above and below P. If the ringabove is positive and the ring below is negative, as in Figure 19-5b, thecontribution due to each ring is in the positive-to-negative direction, as istheir sum. If both rings are positive, as in Figure 19-5c, the contributionsare in opposite directions, but because the lower ring is less positive, itsfield contribution is weaker, so the sum of these two contributions is like-wise in the positive-to-negative direction. We can similarly show that thefield due to each pair of rings chosen in this way, and therefore the totalfield, is in the direction of decreasing surface charge density.

b. Electrostatic forces are in the same direction as the field on positive chargesand opposite on negative charges. Thus, conventional current flows in thedirection of decreasing surface charge density, but electrons flow oppositely.


+ + + ++ –

(b) (c) (d ) (e)(a)

For WebLink 19-2:Mapping an Electric

Field, go towww.wiley.com/college/touger

0540T_c19_566-597.qxd 9/24/04 15:54 Page 575 EQA


charge, the lines are more densely packed. The density of field lines visually com-municates the strength of the field. Notice, too, that in both Figures 19-6c and19-6d the lines fan out (become less dense) as they go further out from thecharge. By doing so, they show us that the field gets weaker as the distance rfrom a point charge increases. Figure 19-6e shows that the direction of the fieldis reversed if the point charge is negative.

In summary, the field line “map” communicates information because they fol-low a few basic rules:

• Rule 1: Field lines start and end at charge carriers (or else go to infinity). Theyare directed away from positive charge and toward negative charge.

• Rule 2: The direction of the field line at any point (or of the tangent at that pointif the field line is curved) is simply the direction of the field vector at that point.

• Rule 3: The density of field lines (how closely packed they are) in a smallregion about a point is drawn to be proportional to the magnitude of the fieldvector at that point.

Quantitatively, the density of field lines is the number of field lines crossingeach unit of area perpendicular to the lines in a three-dimensional picture (Fig-ure 19-7a). The following line of reasoning demonstrates that with this definition,the depicted field due to a point charge (Figure 19-7b) is proportional to asCoulomb’s law requires it to be. In preparation for our argument, the figure showsimaginary spheres of different radii r centered at the point charge; these providethe surface areas that we picture the field lines crossing.

1�r2,

(b) Same total number of lines crosses each sphere, solfewer lines cross each cm2 of larger spherical surface.

(a)a Density at this surface is 3 lines/cm2

3 linescross onesquare cm

eeeeeeeeeeeeeeeeeeeeeqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqquququququququuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuauauauauauauauauaaaaaaaaaaaaaaaaaaaaaalalalalalalalalalalalalalallllllllllllllllllllaaaaaaaaaaaaararararrrrrrrrrrrrrrrrrrrerererereeeeeeeeeeeeeeeeeeeeeeeeeeaeaeaaaaaaaaaaaaaaaaaaaaaaaaasasasasasasassssssssssssssssssssssss

Argument Statement in Words

1. The same number of field lines crosses all thespheres in Figure 19-7b.

2. The number of field lines crossing each unitarea on a particular sphere of radius r is thetotal number of lines divided by the sphere’ssurface area

3. The number of field lines per unit area thatwe draw is proportional to the field.

4. Then the field due to the point charge isinversely proportional to r2.

4pr2.

Equivalent Mathematical Statement

Number of field (a constant)

Number of

a constantthat we call k

E � a c1

4pc2b

1

r2

c2E �c1

4pr2 1c2 is another constant2

lines � c1

b

Figure 19-7 The density of fieldlines indicates the magnitude ofthe field. (a) The density at thissurface is (b) The sametotal number of lines crosses eachsphere, so fewer lines cross each

of the larger spherical surface.cm2

3 lines�cm2.

lines/unit area �c1

4pr2

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For a more intuitive approach to this dependence, work through Problem 19-1at the end of the chapter.

✦GAUSS’S LAW Doing Problem 6-55 shows that the irradiance due to a pointlight source is proportional to The irradiance is also called the energy fluxdensity. In this terminology there is the same total flow or flux of energy persecond across each sphere, rather than the same “total flow” of field lines. Thenwe can speak of the electric field as the electric flux density. A total electricflux, proportional to the total number of lines, can be obtained for the pointcharge by multiplying the field at any distance r by the spherical surface to whichit is everywhere perpendicular, giving us (a constant times q).Thus, the total flux across any of the imaginary spheres enclosing the charge isproportional to the charge enclosed. By deforming the spheres, we could showthat the total flux across an enclosing surface of any shape is proportional to thetotal amount of charge enclosed, however the charge is configured. This gener-alization is called Gauss’s law, after the great German mathematician KarlFriedrich Gauss (1777–1855). It is useful for finding the elec-tric fields due to various continuous arrangements of charge,but its detailed mathematical treatment is beyond the scopeof this book.

A charged conductor is not a point object. To get a pic-ture of the field due to such an object (Figure 19-8a), weuse the fact that close to a conducting object, the field mustbe perpendicular to the surface (Figure 19-8b). At a great dis-tance, though, the object looks like a point object, so thefield at this distance must look like the field due to a pointobject (Figure 19-8c). Because field lines always start and endat charge carriers or else go to infinity, there can be no breaksin the lines between the near and far regions, so we can fillin the connecting parts (Figure 19-8d). For a more detailedgraphic presentation of this reasoning, go to WebLink 19-3.

In general, we can think of an extended charged objectas though it were built up of point charges. Each point chargehas a field like the fields in Figure 19-6. We now considerhow these fields combine to give us a total field.

✦SUPERPOSITION OF FIELDS The electric field lines dueto an arrangement of charges communicate information inthe same way (Rules 1–3) as the electric field lines due to apoint charge. But now the rules tell us how to read the direc-tion and comparative magnitude of the total electric fieldfrom the field lines. From the field lines for each of two pointcharges, we can find the two fields at each point in spaceand then combine them by vector addition (Figure 19-9a).

14pr22 � 4pkq1kqr

2 2

1r

2.

1r

2

Figure 19-8 Electric field linesdue to a charged conductingplate.

(b) Field near surface

(c) Field far from surface (zoom out)

(d ) Lines are continuous between near and far limits

(a) Charge configuration

++++++++

++++++++

+

+

+

(a) The two fields add vectorially at each point.

A

B

C

+

+ +

(b) The field lines are a map of the total field

A

B

C

Figure 19-9 Superposition of fields due to two pointcharges. (a) The two fields add by vector addition at eachpoint. (b) The field lines are a map of the total field.

For WebLink 19-3:Electric Field Due

to a Charged ConductingPlate, go towww.wiley.com/college/touger

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The resulting electric field lines for the combination of charges map informationabout the resultant of the two field vectors at each point (Figure 19-9b). Thispoint-by point vector addition is called a superposition of the two fields. Tosee how we apply this reasoning to obtain the result in Figure 19-10, workthrough Example 19-5.

Example 19-5 Electric Field of a Parallel Plate Capacitor


Draw the electric field lines for a charged parallel plate capacitor, in which thedistance between the plates is very small compared to the dimensions of eitherplate.

Brief SolutionChoice of approach.1. Draw a charge configuration diagram.

2. Consider the superposition of fields in three separate regions: a. betweenthe plates, b. to the left and right of both plates, and c. around the edges.

The details.1. As in Example 18-3, the charge distributes over the inner faces of the two

plates, forming two opposite planes of charge (Figure 19-10a).

2. A typical point A between the plates (away from the edges) is closeenough to each plane of charge so that they both look effectively infinitefrom A (Figure 19-10b). The uniform fields due to the two planes areboth away from positive and toward negative charge, as is the resultantfield.

At a typical point B a short distance to either side of both plates, the planesof charge still look more or less infinite (Figure 19-10c). But now their fields,still away from positive and toward negative and still equal in magnitude, are

+

+

+

+

+

+

–

–

–

–

–

+

+

+

+

+

+A

+

+

+

–

–

+–

–

–

–

+

+

–

–

–

–

–

–

–

+

+

+

+

+

+

–

–

–

–

–

–

E due to

E due to

Total E

–E dueto

E dueto

+

+B

+

+

+

–

+

–

–

–

–Total E ≈ 0

(e) Overall picture

(b) Superposition of fields between plates

(approx. zero) (approx. zero)

(a) Charge configuration

CD

(c) Superposition of fields outside

(d ) Superposition of fields near edges

Figure 19-10 Electric field of acharged parallel plate capacitor.

0540T_c19_566-597.qxd 9/24/04 15:54 Page 578 EQA


opposite in direction, for a total field of zero. At much larger distances fromboth plates, the fields due to the two planes of charge are very nearly (butnot exactly) equal and opposite, so the total field is nearly (but not exactly)zero.

At points near the edges, such as C and D in Figure 19-10d, the fields arenot equal and opposite, and the total field varies in direction from point topoint. This effect is called fringing.

Figure 19-10e shows the overall field lines resulting from these superposi-tions.

◆ Related homework: Problems 19-27, 19-28, and 19-29.

DNA testing. The photo shows theseparation of DNA components bygel electrophoresis.

Terminal Speed, Electrophoresis, and DNA TestingAn object falling through a fluid (such as air) in a gravitational field is subjectto a drag force that increases with speed. When the magnitude of the dragforce becomes as great as the gravitational force, the total force on the objectbecomes zero and the object stops accelerating. The speed it has reached atthis stage is called the terminal speed. If the magnitude of the drag force isproportional to the speed where the proportionality constant or dragcoefficient the Greek letter zeta, depends on the object’s size and shape),then at terminal speed

In the same way, when a uniform electric field is applied to ions in solution(also a fluid), the electric force on the ions is opposed by a drag force. As inair, the drag coefficient will be greater for ions with larger cross-sections, andthose objects will reach smaller terminal speeds. In this case, the electricalforce equals the drag force at terminal speed: Becausethese terminal speeds are typically reached over very small distances, eachtype of ion travels through most of the solution at its terminal speed—adifferent terminal speed for each one. The ions thus separate out. The processof separating out ions in solution by applying an electric field is calledelectrophoresis.

If the ions are permitted to pass through a porous medium, such as a gel,procedures such as staining will show how far the different ions have traveledthrough the gel. In a given time interval, ions of different sizes, with differentterminal speeds, travel different characteristic distances (see Problem 19-72).The resulting separation, and the density of each ion that shows up on stain-ing, makes possible analysis of which ions are present and to what extent.(See photo.) This process is called gel electrophoresis.

DNA fragments typically occur in an ionized state in certain kinds of solu-tions and can therefore be analyzed by gel electrophoresis. Performing gel elec-trophoresis on DNA, however, is complicated by the fact although the dragcoefficient increases with fragment size, so does the total charge q. This isbecause there is a charge on each of the base pairs making up the rungs ofthe DNA ladder, so the total charge depends on how many pairs are strungtogether to make up the total length of the fragment. To the extent that both

and q are proportional to length, the ratio doesn’t change with length, andqzz

z

zvt � qE, and vt �qEz .

vt, zvt � mg, and vt �mgz .

z,1Fd � zv,

the terminal speed remains the same for fragments of all lengths in afield E. In this case, a gel with smaller pores is used, so that larger fragmentsare likely to become trapped over a shorter distance. Gel electrophoresis is astandard procedure for DNA analysis. It is used for such purposes as estab-lishing who is the father of a child, or identifying blood samples in legal casessuch as rape or murder trials.

vt �qEz

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19-5 Electrical and Gravitational Systems: Similaritiesand Differences; Electrical Potential Energy

Gravitational and electrostatic interactions are due to different inherent propertiesof the particles that constitute matter: gravitational mass and electric charge. Butmathematically, they obey similar inverse square laws (Coulomb’s law and Newton’suniversal law of gravitation).

Just as the total electrostatic force on a charged body is the vector sum ofthe forces exerted on it by all other charged bodies, the total gravitational forceon a body is the sum of the gravitational forces exerted on it by all other masses.For example, the sun and Earth both exert significant gravitational forces on themoon, and the paths of the chunks of debris that make up the rings of Saturnare affected by several significant gravitational forces. These include the forcesexerted by Saturn, by the rest of the debris in the ring, and quite significantly bysmall moons that rotate with the rings and help keep the debris in ring-shapedconfigurations (Figure 19-11). Because they “herd” the debris in this way, theyare called shepherd moons.

Suppose we find the force on one chunk of debris, and moments latera chunk with different mass is at the same point. Rather than recalculate thetotal force from scratch, we can be guided by what we did for electrostaticforces. We can calculate the contribution to the total force at a point fromall the surrounding bodies—the force per unit of mass—and then multiply itby the mass of whatever we place at that point. Analogous to Equations19-4 and 19-7, we can again write equations we used to describe gravitationalforces:

g (5-1, like 19-4)strength of property propertygravitational of A of A’sinteraction environment

between A and itsenvironment

or

where has magnitude (8-8 or 8-9, like 19-7)

Continuing the analogy, we call the gravitational field. Near Earth’s surfacehas the roughly constant magnitude It is the gravita-

tional acceleration of a freely falling body. We could treat this value as con-stant over trajectories near Earth’s surface small enough so we can regard Earthas flat. How does this compare with electric fields? Viewed from far away,Earth looks like a point object, and its gravitational field lines (Figure 19-12a)look like the electric field lines due to a negative point charge. STOP&ThinkWhy not a positive point charge? ◆ Viewed close up, Earth looks flat, andthe field lines (Figure 19-12b)—everywhere perpendicular to the surface—look like those near a plane of charge (Figure 19-8b). The lines are parallel,not spreading out or converging, so the density of lines, indicating the mag-nitude of the field, doesn’t change. The field is essentially constant in a flatEarth approximation. (Continuing to treat gravitational field lines as we didelectrical, we could deduce Gauss’s law for gravitational fields: The totalgravitational flux across an enclosing surface is proportional to the massenclosed.)

By considering the work done to move bodies in opposition to gravitationalforces within a system, we were able to derive expressions for gravitationalpotential energy when g is constant (e.g., over short distances from a planet’ssurface)

(6-7) PEgrav � mgy

g � G mR2

E� 9.8 m/s2.gS

gS

g � G

m

r2gS

FS

on A � mAgS

mA�Fon A

Figure 19-11 Small, irregularlyshaped shepherd moons keepthe material in Saturn’s ringsmoving in formation.

Figure 19-12 Gravitational fieldsfar from and near Earth’s surface.

(b) Near

(a) Far

Earth

0540T_c19_566-597.qxd 9/24/04 15:54 Page 580 EQA

19-5 Electrical and Gravitational Systems: Similarities and Differences; Electrical Potential Energy ◆ 581

and when we must consider g’s variation with r

(8-10)

In either case, we can think of the gravitational PE as theenergy stored by moving one body in a system against thegravitational field due to the rest of the system.

Purely by analogy (using the correspondences and ), we can obtain expressions for the elect-

rical PE (the energy stored by moving one charged bodyin a system against the electric field due to the rest of thesystem):

(19-8)(if we take the direction of the electric field to be the negative y

direction, as we did for the gravitational field )

When we must consider E’s variation with r: (19-9)

Equations 6-7 and 19-8 have the same sign because we chose our coordinatesystem so that the field direction is the same in both cases. Thus, in all parts ofFigure 19-13, the hand increases the PE of the body by lifting it. Note that Figure19-13c is really the same as Figure 19-14a or b: In each case the hand is pullingthe positive charge away from the plane of negative charge that is attracting it; ineach case we pick our coordinates so that the field is in the direction.

In the r-dependent situations described by Equations 8-10 and 19-9, however,we are stuck with keeping the object that is the source of the field at the cen-ter The hand in Figure 19-15 does positive work either by pulling twomutually attractive positive masses (masses are always positive) apart or by push-ing two mutually repulsive positive charges together. In the gravitational case PEincreases as r increases, but in the electrical case PE increases as r decreases, sothe signs in Equations 8-10 and 19-9 must be opposite.

1r � 02.

�y

PEelec � �k

q1q2

r

gS

PEelec � qEyWhen ES is constant:

G 4 kg 4 E,m 4 q,

PEgrav � �G

m1m2

r

0

+y

– – – – – –

+ +++++

(b) (c)(a)

+m

M

+

0

By rotating the page, you can turnFigure19-13c into either of these pictures

+y 0+y

– – – – – –

+

–

–

–

–

–

–

+

+

+ E

M

m

(a) (b)

g

Figure 19-13 Positive work isdone by lifting either a massor a positive charge against afield directed in the negativey direction.

Figure 19-14 The y direction ischosen so that the field is in thenegative y direction. Comparewith Figure 19-13c.

Figure 19-15 The hand doespositive work (r-dependent PEincreases) in both cases.

Example 19-6 Energy of a Proton–Electron System


A proton and an electron are a distance r apart.a. Compare the electric and gravitational PEs of the proton–electron system.b. If the distance r is increased, does the electrical PE increase or decrease?

Does the gravitational PE increase or decrease?

Brief SolutionRestating the question. The most meaningful way to compare two quantities ofvery different sizes is by a ratio. In a. we ask,

PEelec

PEgrav� ?

0540T_c19_566-597.qxd 9/24/04 17:21 Page 581 EQA


Example 19-6 shows that gravitational interactions are insignificantly weakcompared to electric interactions and are only evident between very massivebodies. Thus, on an atomic scale electrostatic forces are the binding forcesthat hold matter together, but on an astronomical scale it is mutual gravita-tional attraction that causes great clouds of hot gas to condense into stars andplanets.

Like the gravitational force, the electrostatic force is conservative, so Equa-tions 6-10 (conservation of mechanical energy) and 6-16 (the work-energy theo-rem) remain valid under the same conditions as before when the potential energyis partly or totally electrical.

Choice of approach. We can treat the proton and electron as point objects, sothe two PEs are (Equation 19-9) and (Equa-tion 8-8). The masses and charges are known constants.


The mathematical solution. a. By Equations 19-9 and 8-8,

is vastly greater than is negligibly small in comparison.b. Both PEs are negative, because is negative and because ithas a sign. As r increases, the absolute value of each PE decreases. Thenboth PEs become less negative, that is, both increase.

◆ Related homework: Problems 19-31, 19-32, 19-33, 19-36, and 19-37.

�PEgravqePEelec

PEgrav; PEgravPEelec

� 2.30 � 1039 1unitless2

�

a9 � 109

N � m2

C2 b 11.6 � 10�19 C2 1�1.6 � 10�19 C2�a6.67 � 10�11N � m2

kg2 b 11.67 � 10�27 kg2 19.11 � 10�31 kg2

PEelec

PEgrav� PEgrav �

�k

q1q2

r

�G

m1m2

r

��kq1q2

�Gm1m2

qp � e � 1.6 � 10�19 C qe � �e � �1.6 � 10�19 C k � 9 � 109

N � m2

C2

mp � 1.67 � 10�27 kg me � 9.11 � 10�31 kg G � 6.67 � 10�11

N � m2

kg2

PEelec

PEgrav� ?r � ?

PEgrav � �G

m1m2

rPEelec � �k

q1q2

r

Knownconstants: e

EXAMPLE 19-6 continued

Example 19-7 Repelling Klingon Space Rats

A Klingon spacecraft hovers 18 m from the vast flat surface of a space station.Klingon attack rats, each with a mass of 2.00 kg, can propel themselves acrossthe gap at speeds of up to 12 m/s. But space station engineers have figuredout that in the process of propelling themselves, the rats acquire a positivecharge of They therefore devise a way of charging up the planesurface of the station so that it will have a uniform electric field. Find the mag-nitude and direction of the minimum electric field required to repel all rats.

5.4 � 10�4 C.

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19-6 Potential and Potential Differences ◆ 583

19-6 Potential and Potential DifferencesJust as the electrostatic force on any body may be treated as the product of twocontributions, one from the body itself (its charge Q) and the other from the elec-trical environment (the field), the electric PE may likewise be separated into twocontributions:

(19-10)

body’s environment’scontribution contribution

Suppose the body is small enough for us to consider Q a point charge. As theelectric field is a function of position giving the force on each unit of charge placedat that position, so the quantity which is called the potential, is a functionof position giving the potential energy of each unit of charge at that position. Theunits of potential are units of energy (joules) divided by units of charge (coulombs)and are therefore J/C.

It follows from Equations 19-8 and 19-9 that

(19-11)

(in a region of constant field in the direction, that is, near an effectively infinite plane ofcharge or close enough to a uniformly charged surface so that it appears flat)

�y

potential PEelec

Q�

QEy

Q� Ey

PEelec

Q ,

PEelec � Q aPEelec

Qb

SolutionChoice of approach. We apply conservation of mechanical energy(Equation 6-10) to the case when the PE is purely electrical and is dueto a uniform field. (Remember that for other arrangements of charge,like a point charge, the field is not uniform and we cannot use Equa-tion 19-8.) We desire a final state in which the rat’s velocity is reducedto zero. We first sketch the situation (Figure 19-16) to show the nec-essary direction of the field so that the force on the positively chargedrat opposes its motion. We choose our coordinates so that, consistentwith Equation 19-8, the positive y direction is opposite to the electricfield.


State A State B Constants

The mathematical solution.

or (6-10)

becomes

Solving for E,


E �mvA

2

2q 1yB � yA2 �12.00 kg2 112 m/s22

215.4 � 10�4 C2 118 m � 02 � 1.5 � 104 N/C

qEyA � 12 mv2

A � qEyB � 0

PEA � KEA � PEB � KEB

1Total Energy2A � 1Total Energy2B

E � ? E � ?

PEA � qEyA KEA � 12 mv2

A PEB � qEyB KEB � 0 q � 5.4 � 10�4 C

yA � 0 vA � 12 m/s yB � 18 m vB � 0 m � 2.00 kg

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

++

+++

CM

+++

+

++

+

++

+

++

O +y18 m

PlaneE

Figure 19-16 A sketch ofthe physical situation forExample 19-7.

➥ We will shortly introduce anothername for J/C, but not yet, becausewe want the units to serve as areminder that we are talking aboutPE per unit charge.

0540T_c19_566-597.qxd 9/24/04 15:55 Page 583 EQA


and (19-12)

(at a great enough distance r from a body with charge q so thatthe body may be regarded as a point object)

potential PEelec

Q�

kQq

r

Q�

kq

r

In Chapter 6, we pointed out that it is changes in potential energy rather thanPE values at a single point that have physical meaning. Likewise, the change inpotential energy per unit charge—the change in potential from one point toanother—is what has physical significance, not its value at either point. That’swhy we have not assigned potential its own symbol, but we now choose asour symbol for the potential difference between two points A and B:

(19-14)

Units of potential difference, like potential, are J/C. A more common name forthese units, especially in the context of electric circuits, is volts (V):

(19-15)1 volt � 1joule

coulomb 11 V � 1 J/C2

VAB � aPEelec

Qb

B� aPEelec

Qb

A�

¢PEelec

Q

VAB

Example 19-8 Change of Potential Experienced by an Electron

An electron is released at rest at a point where the potential is 3.0 J/C. Thepoint is in a region of uniform electric field of magnitude 75 N/C. Set in motionby the field, the electron reaches a point 0.020 m from its starting point. Whatis the potential at this point?

SolutionChoice of approach. We sketch the situation with in the negative y direction(Figure 19-17) so that we can apply Equation 19-11. Because the force on anegative charge is opposite to the field, the electron moves inthe positive y direction. Because

(E constant) (19-13)



Making sense of the results. To lose potential energy, negatively charged bod-ies must move from lower to higher potential. Symbolically, must bepositive for to be negative when Q is negative.


¢PEelec � Q¢ 1PEelec

Q 2¢ 1PEelec

Q 2

aPEelec

Qb

2� aPEelec

Qb

1� ¢aPEelec

Qb � 3.0 J/C � 1.5 J/C � 4.5 J/C

¢aPEelec

Qb � E ¢y � 175 N/C2 10.020 m2 � 1.5 J/C 11 J � 1 N � m2

aPEelec

Qb

2� ?aPEelec

Qb

1� 3.0 J/C

E � 75 N/C¢y � 0.020 m

¢aPEelec

Qb � E ¢y

PEelec

Q � Ey,¢y � 0.020 m

ES

E

y2

+y

e–

Felec

y1

∆y = 0.020 m

Figure 19-17 A sketch ofthe physical situation forExample 19-8.

0540T_c19_566-597.qxd 9/24/04 15:55 Page 584 EQA

19-6 Potential and Potential Differences ◆ 585

Thus, voltage is a common expression for potential difference and motivates ourchoice of symbol.

Another unit of energy, often used when applying these ideas on an atomicscale, is the electron volt (eV). Because an amount of potential difference mul-tiplied by an amount of charge gives us an amount of energy, the electron voltis defined as 1 volt multiplied by the magnitude of the charge of one electron.

(19-16)

We must emphasize that voltage is a difference; it does not tell us the PE perunit charge at one point but how much it changes between two different points.For example, the voltage of a battery is the increase (a positive difference) inenergy that each unit of charge gets in passing from one terminal to the other.Because the manufacturers don’t know what will be connected to a battery orfor how long, the useful thing to tell you is how much nonelectrostatic work thebattery does on each unit of charge (and how much energy it thus transfers toeach) passing between its endpoints.

1 eV � 11.6 � 10�19 C2 11 V2 � 1.6 � 10�19 J

In Figure 19-18, the escalator does the same amount ofnongravitational work on each unit of a population ofmice (we may takeour unit of mass inthis case to be onestandard mouse)and so increases thegravitational PE ofeach unit of mass bythe same amount,no matter howmany mice ride theescalator in all.

In the gravita-tional circuit, thereis a value of thecoefficient of fric-tion between mouseand slide for whichthe amount ofenergy dissipatedexactly equals the

gravitational PE lost, so that the mice move at constantspeed rather than gaining or losing KE. Somewhat anal-

ogously, becausethere is resistance inthe wire, the elec-trons travel at anunchanging driftvelocity. But like allanalogies, this onehas its limitations. Ifthe coefficient offriction is increased,the mice will beslowed down, ulti-mately causing aback-up of mice atthe top of the esca-lator because theescalator keeps car-rying the mice at thesame speed asbefore. But in the

Case 19-1 ◆ Electric and Gravitational Circuits: Exploring an Analogy

SlideEnoughfriction topreventspeeding up

Escalator

(a)a Gravitational circuit

Electrons

Conductingwire

Negativepole

Positivepole

Battery

(b) Electrical curcuit

Figure 19-18 Gravitational and electrical circuits compared.

(a) Gravitational circuit (b) Electrical circuitSimilarities

Differences

Each mouse gains PE when the electromechanical mechanismsof the escalator do work to move the mice in opposition togravitational forces.

The energy gained is dissipated because of friction along thepath.

We can characterize the escalator by how much PE in joules itprovides to each unit of mass (mouse or kg).

The current or flow of mice consists of a few mice doing com-plete laps of the circuit.

Each electron gains PE when the chemical mechanisms ofthe battery do work to move the electrons in opposition toelectrostatic forces.

The energy gained is dissipated because of resistance alongthe path.

We can characterize the battery by how much PE in joules itprovides to each unit of charge (electron or coulomb).

11 J�C � 1 volt2

The net flow of electrons is the result of vast numbers of elec-trons shifting very small distances at very slow drift velocities.

0540T_c19_566-597.qxd 9/24/04 18:26 Page 585 EQA


To further clarify the connection between voltage and energy concepts, workthrough Example 19-9.

electrical circuit, if electrons are slowed down by greaterresistance and begin to back up, they repel the elec-trons coming up behind them, slowing them down aswell. This feedback continues throughout the circuit sothe electrons travel the entire circuit at a slower speed(resulting in a smaller current).

There are other differences. Individual mice mayboard the escalator at the bottom and do repeated lapsof the gravitational circuit in a short time. In the elec-tric circuit, in contrast, all parts of the circuit contribute

electrons to the flow. These electrons actually driftextremely slowly through the circuit, typically travelingon the order of 10 cm in an hour (we’ll do the calcu-lation in the next chapter). But there are so many ofthem that, for example, under ordinary conditions of them will pass any point on the filament of a 75 Wbulb each second. Remember also that the individualelectrons do not travel at constant speed; they areaccelerated by the field and slowed down by collisions.The drift velocity is an average.

1019

Case 19-1 ◆ Electric and Gravitational Circuits: Exploring an Analogy (continued )

Example 19-9 A Photoelectron

For a guided interactive solution, go to Web Example 19-9 at www.wiley.com/college/touger

Light striking a metal surface may under certain conditions provide some sur-face electrons with sufficient energy to break away from the surface. This effectis called the photoelectric effect, and the resulting emission of electrons is calledphotoemission. Suppose that when light strikes one of two parallel plates, elec-trons are emitted with a maximum velocity of What voltagemust be maintained between the plates if none of the emitted electrons are toreach the opposite plate?

Brief SolutionChoice of approach. Because voltages are PE differences per unit charge, wefocus on energy. We can apply conservation of mechanical energy: The loss inKE as the electron is brought to a stop must equal the gain in PE. The force ona negative charge is opposite the field, so the field is toward the opposite plate.


State A State B Constants


(6-10)

becomes

Then

Making sense of the result. The decrease in potential, when multiplied by anegative charge, will yield an increase in PE.

◆ Related homework: Problems 19-46 and 19-48.

� 10.52 19.11 � 10�31 kg2 14.8 � 105m/s22

�1.6 � 10�19 C� �0.66 V

V AB �

PEB � PEA

q�

12 mvA

2

q

12 mvA2

� PEB � PEA

PEA � KEA � PEB � KEB

VAB �PEB � PEA

q� ?

q � �e � �1.6 � 10�19 CPEB � ? KEB � 0PEA � ? KEA � 12 mv2

A

me � 9.11 � 10�31 kgvB � 0 vA � 4.8 � 105 m/s

4.8 � 105 m/s.

0540T_c19_566-597.qxd 9/24/04 17:48 Page 586 EQA

19-7 Batteries ◆ 587

✦POTENTIAL DIFFERENCE AND SURFACE CHARGE DENSITYGRADIENTS In Chapter 18 we established that a current flows in a conduc-tor if the surface charge density changes along the conductor. In Example 19-4we saw that the surface charge density gradient along the conductor results inan electric field directed from higher to lower surface charge density along theconductor’s axis. The net field at any interior point depends on differencesbetween the surface charge upstream and downstream from the point. So wherethe density is changing uniformly (the gradient is constant), the field is likewiseconstant (except near the ends of the region of uniform change). Where the fieldis constant, the voltage or potential difference between any two points along theconductor is given by Equation 19-13, which we can rewrite as

(19-17)

If the surface charge densities at points A and B differ, there will be a field alongthe path between the points. If there is a field, there will be a potential differenceor voltage between the points. The voltage, then, is a measure of the surfacecharge density difference between A and B. We can think of it as a measure ofthe battery’s ability to “pump” charge to maintain this difference (somewhat asan ordinary air pump maintains a difference between the volume density of mol-ecules at opposite ends of the pump). Whatever we said of the surface chargedensity gradient in Chapter 18 also holds true for voltage. For example, if thereis a nonzero voltage (now replacing the words surface charge density gradient)between two points on a conductor, there will be a movement of charge (a cur-rent) between the points.

Again, a gravitational analogy is useful. In Figure 19-19we retain the convention that the electric and gravita-tional fields are both in the negative y direction. Then thegravitational and electrical PE of a body both decrease inthis direction. The change in PE per unit of mass is

It is a decrease when the object isfalling and is negative. We usually choose our coordi-nates so that when In the same way in theelectrical case, we choose our coordinates so that when Like ground level in the gravitational case, thisis electrical “ground.”

Analogies can also help in thinking about circuits. For example, as is true forgravitational PE, the change in electric PE is the same for any path between thesame two points (recall the discussion of Figures 6-17 and 6-18). So it must alsohold true for the voltage, the change in electric PE per unit charge. In electriccircuits, then, we know that any wires or devices connected between the sametwo points are subject to the same voltage. We will use this idea extensively inChapter 21.

19-7 BatteriesHow do batteries produce potential differences? Batteries increase electric PE byseparating opposite charges that are ordinarily drawn toward each other. Wheredoes this energy come from? STOP&Think Do batteries create the energy? ◆

Recall from Section 6-1 (Figures 6-6 and 6-7 or WebLink 6-2) that chemicalreactions can release energy that had been stored as chemical energy. What occursis an energy conversion, not a violation of energy conservation. Common reac-tions in batteries include the dissolving or ionization to differing degrees of twodifferent metals immersed in a solution, which is often acidic. In such a case thetwo pieces of metal, such as a piece of copper and a piece of zinc, form thepoles of the battery. As the positive ions go into solution, their abandoned outerelectrons remain behind on the metals. Because the two metals ionize to different

y � 0.

PEelec

q � 0y � 0.

PEgrav

m � 0¢y

¢PEgrav

m �mg¢y

m � g¢y;

1¢y � yB � yA2VAB � E¢y

yA

y

+yB Unit ofcharge

Unit ofmass

1 kg

are both negativeas body falls toward“ground” or Earth

0

∆y = yB – yA

∆Vab and

“Ground” Earth= 0)(PEq

= g∆y∆PEgrav

m

Figure 19-19 Change in PE when going from higher tolower potential in electrical and gravitational systems.

0540T_c19_566-597.qxd 9/24/04 15:55 Page 587 EQA


degrees, the net negative charges left behind on the pieces of metal are different.Typically only one metal (like the copper) dissolves significantly in an acid solu-tion. Random (thermal) motion carries some of the positive ions away from thedissolving pole. Then if a wire is connected between the two poles, mutual repul-sion will cause the electrons left behind to flow to the other pole (Figure 19-20).Ions from the first pole that make it to this one will bind to its surplus electrons,thus adhering to and “plating” the second pole. (Copper plating is done in thisway.)

In some batteries, the regions near the two poles are separated by a mecha-nism like a porous barrier, which permits larger ions to pass less readily thansmaller ions. If the positive and negative ions are of substantially differentsizes, the differential flow of the two kinds of ions through the barrier willgive rise to a difference in charge density across the barrier. In the sameway, the fact that the membranes of living cells are not equally permeableto different ions can give rise to potential differences across cell membranes.

Figure 19-21 shows typical movements of charge carriers and theresulting conventional currents for a type of battery called a galvanic orvoltaic cell. Note that although the charge carriers in the external wire are

electrons, the carriers within the battery may be ions. Althoughthe current may be equal everywhere in the circuit, the carri-ers comprising the current can be different in different parts ofthe circuit.

The increase in PE from one pole to the other can beequated to an amount of work, commonly described as workdue to non-electrostatic forces. Actually, electrostatic forces are

indirectly involved because they govern chemical reactions, but no electrostaticforces due to external application of an electric field are involved.

(b)

Zinc

0

Copper

Milliammeter

(a)

Figure 19-20 Two examples ofbatteries. (a) “Pickle power”: Thepickle’s juice is an acid solution. Thedeflection of the meter indicates asmall current between the dissimilarmetal poles of this battery. (b) A12.6-V lead-acid storage battery foran automobile.

More e–’sleft behindby ionization

Metal A dissolvesmore (more + ionsinto solution)

Porousmembraneor material

Metal B dissolvesless (fewer + ionsinto solution

Fewer e–’sleft behindby ionization

Flow of e–’s through external wire

Flow of – ions

Flow of + ions

Conventional current

Conventional current

Figure 19-21 Principle features inthe functioning of a galvanic orvoltaic cell.

➥A note on language: Originally abattery of cells referred to a numberof cells used in combination. Todaywe commonly speak even of a sin-gle cell as a battery.

✦ S U M M A RY ✦

The electrostatic force between point charges and isdescribed by Coulomb’s law.

Coulomb’s law: The magnitude

(19-1)

The direction of the force on either body is toward the otherbody if the two have opposite charges, and away if they havelike charges.

The force per unit charge, or electric field , is definedby and is a function of position. If a chargedF

Son A � qA E

SES

1k � 9 � 109 N � m2�C22F � k

�q1� �q2�

r 122

q2q1 body A is placed at that position,

(19-4)strength of property properties ofelectrical of A environment

interaction of point wherebetween A and its A is placed

environment

Electric field at a position a distance r from a point charge q

Magnitude: (19-7)

Direction: away from q if q positive, toward q if q negative

E � k

0q 0r2

Fon A � 0qA 0 E

0540T_c19_566-597.qxd 9/24/04 17:21 Page 588 EQA

and (19-9)

for point charges a distance r apart

The electric potential energy per unit charge is called theelectric potential or simply the potential:

(19-11)

(in a region of constant field in the direction, that is, near an effectively infinite plane of charge or close enough to a

uniformly charged surface so that it appears flat)

and (19-12)

(at a great enough distance r from a body with charge q sothat the body may be regarded as a point object)

As is true of PE, only differences in potential have phys-ical meaning. The potential difference or voltage betweentwo points A and B is

(19-14)

Units: (19-15)

The electron volt (eV), a unit of energy, is then defined as

(19-16)

In a constant field (taken to be in the negative y direction)

(19-17)

From this it follows that the electric field is the negativegradient of the potential. The table below summarizes the con-ditions in a conductor.

1¢y � yB � yA2VAB � E¢y

1 eV � 11.6 � 10�19 C2 11 V2 � 1.6 � 10�19 J

11 V � 1 J�C21 volt � 1

joule

coulomb

VAB � aPEelec

Qb

B� aPEelec

Qb

A�

¢PEelec

Q

Potential PEelec

Q�

kq

r

�y

Potential PEelec

Q� Ey

PEelec � �k

q1q2

r

Summary ◆ 589

Because electric fields are vectors, finding the electric field ata point P due to an array of point charges (Procedure 19-1)involves vector addition.

For a conductor in electrostatic equilibrium,

• there can be no electric field at any point in its interior;

• at all points on its surface, the electric field must be perpen-dicular to the surface.

But if a conducting wire is not in electrostatic equilibrium, theinterior field is in the direction of decreasing surface chargedensity (the direction of the negative gradient).

Electric field lines communicate information becausethey follow a few basic rules:

• Rule 1: The direction of the field line at any point (or ofthe tangent at that point if the field line is curved) is sim-ply the direction of the total field vector at that point.

• Rule 2: The density of field lines (how closely packed theyare) in a small region about a point is drawn to be pro-portional to the magnitude of the total field vector at thatpoint.

• Rule 3: Field lines start and end at charge carriers (or elsego to infinity). They are directed away from positive chargeand towards negative charge.

Because Coulomb’s law and Newton’s universal law ofgravitation have similar forms, we can treat many aspects ofelectrical systems like gravitational systems (and vice versa—we can treat as a gravitational field, map gravitational fields,etc.). Using the correspondences and we obtain

(19-8)if constant in the negative y directionE

SPEelec � qEy

G 4 k,m 4 q, g 4 E,gS

Conditions

Rate of change (in Response of positive charge carriersTotal change direction of decrease) to conditions (negative do the opposite)

Underlying picture Surface charge density Negative density gradient Movement of positive charge in direction ofdifference negative gradient

Formal concept Potential difference Negative potential Conventional current in direction of fieldgradient field�

Batteries establish potential differences by a combination ofchemical means and mechanisms resulting in unequal flow ofpositive and negative ions.

The table below contains some important things toremember about the quantities we introduced in this chapter.

Vector: Scalar:adds by vector addition adds by scalar addition

Work done moving charge a distanceTotal amount Electrostatic force Electric potential energy in one dimension: ¢W � F¢s¢sPEelecF

S

Potential difference between endpoints

Amount per unit of charge Electric field Electric potential of Vab �¢W

q�

F¢s

q� E¢s¢s:

PEelec

qES

0540T_c19_566-597.qxd 9/24/04 17:21 Page 589 EQA


The questions and activities in this group are particularly suitable forin-class use.

19-1. Discussion Question.Exploring an analogy: In Fig-ure 19-22a, the paint from thepainter’s spray can spread outto board A in a cone-shapedregion. If board A is removed,the cone will continue tospread to board B, twice as farfrom the nozzle of the can.a. When board A is removed,

how will the total amount ofpaint reaching board B each sec-ond compare to the total amountof paint that previously reachedboard A each second?

b. How will the area that getspainted on board B comparewith the area that gets painted onboard A? Express as a ratio

c. If each board got sprayed for1 s, how will the thickness ofthe paint on the two boards (theamount of paint on each squarecentimeter of board) compare? Express as a ratio

d. Now compare the spray paint situation to the situation ofelectric field lines directed outward from a positive point

thickness Bthickness A.

area Barea A.

charge in Figure 19-22b. The field lines cross two imaginaryspheres, both centered at the point charge. To which quan-tities in the paint situation can each of the following becompared: the areas of the two spheres? the total numberof field lines crossing each sphere? the magnitude of theelectric field? Briefly explain each.

e. How does the magnitude of the electric field compare atthe surfaces of the two spheres? Express as a ratio and briefly explain your reasoning.

19-2. Discussion Question.a. Other than in superconductors, can there ever be a current

between two points A and B if the potential difference is zero?

b. Figure 18-8 shows the anode and cathode in a cathode raytube as two parallel plates, with a hole in the negative cath-ode that permits electrons to pass through. Assume theplates are effectively infinite in size compared to the widthof the gap between them. The electron is accelerated fromthe negative cathode to the positive anode. Why isn’t theelectron slowed down by attraction to the anode once itmoves to the right of it?

c. Is there a potential difference between the anode and thescreen?

d. Is there an electric field to the right of the anode? Whatdoes Equation 19-16 tell you about your answer to c ?

e. Is there a current between the anode and the screen?

f. Are your answers to c and e consistent with your answerto a? If not, make any necessary corrections.

VAB

E at BE at A,

✦ Q UA L I TAT I V E A N D Q UA N T I TAT I V E P RO B L E M S ✦

H a n d s - O n A c t i v i t i e s a n d D i s c u s s i o n Q u e s t i o n s

(a)a

Board A

Board B

r

2r

AB

rr

22r

A

Section 19-1 Making Electrostatic Force and ChargeConcepts Quantitative

19-3. Two point charges exert an attractive force of 5.0 N oneach other. How much force would they exert on each otherif the distance between them werea. ten times as great?

b. one-tenth as great?

c. How would your answers to a and b be altered, if at all,if the forces the charges exert on each other were repulsiverather than attractive?

19-4. How many electrons must be brought together to con-stitute a total charge (ignoring sign) of one coulomb?

19-5.a. Find the electrostatic force that two electrons exert on each

other when they are 1 mm apart.

b. Compare this with the gravitational force they exert oneach other at the same separation Express your comparison as a ratio

19-6. What must be the charge on each of two identical bod-ies if they are to exert a repulsive force of 1 N on each otherat a separation of 5 cm?

19-7. SSM WWW A point object with a charge of is placed at the origin.

1.00 � 10�8 C

Felec

Fgrav.

1me � 9.11 � 10�31 kg2.

a. Where must a proton be positioned to experience a forceof directed to the left?

b. Where must an electron be positioned to experience a forceof directed to the left?

19-8. Three charged point objects A, B, and C are allplaced along the x axis. A is at the origin, and B is at

A and C have charges of B’s chargeis Find the magnitude and direction of the totalelectrostatic force on C if C is placed 1 cm a. to the leftof B. b. to the right of B. c. to the left of A. d. to the rightof A.

19-9. Suppose that in Problem 9-8, object C has a mass ofWhat is its acceleration when it is positioned

1 cm to the left of A?

19-10. Charged point objects A and B are placed along thex axis at and respectively. Themagnitude of the charge on A is double that of B. Where onthe x axis will a positively charged point object experiencezero total force if a. both A and B are positive? b. A is positiveand B is negative?

19-11. Repeat Problem 19-8 for the case in which A and Bremain where they were but C is placed along the y axis aty � 4 cm.

x � �0.04 m,x � �0.04 m

5.0 � 10�21 kg.

1.6 � 10�6 C.1.0 � 10�6 C;x � 3 cm.

1.44 � 10�15 N

1.44 � 10�15 N

R e v i ew a n d P ra c t i c e

SSM Solution is in the Student Solutions Manual WWW Solution is at http://www.wiley.com/college/touger

Figure 19-22 Problem19-1

(b)

0540T_c19_566-597.qxd 9/24/04 15:55 Page 590 EQA

19-12. At a certain instant, a proton approaching another pro-ton has an acceleration of What is its acceler-ation at a later distance when it is only half as far away?

Section 19-2 Electric Fields

19-13. Two charged point objects A and B are placed alongthe x axis. A is at the origin and B is at A’s chargeis and B’s charge is Find the mag-nitude and direction of the total electric field due to thesecharges at a point 1 cm a. to the left of B. b. to the right of B.c. to the left of A. d. to the right of A.

19-14. Use the results of Problem 19-13 to find the force ona point object C placed 1 cm to the right of B if C has acharge of a. b.

19-15. SSM Consider charges A and B in Problem 19-13.a. Find the magnitude and direction of the total electric fielddue to these charges at the point b. Findthe force on an electron placed at this point.

19-16. Find the magnitude and direction of the electric fieldat a certain point if a force of to the right isexerted on a. a sodium ion placed at that point. b. a chlorideion placed at that point.

19-17. Suppose an oil droplet in the Millikan experiment hastwo excess electrons on it. If the total non-electrostatic forceon the droplet is what electric field (magnitudeand direction) must be applied if the droplet is to descend atconstant velocity?

19-18. A 0.0050-kg ball of metal foil is suspended from aninsulating thread. When a charged rod is held at a fixed dis-tance directly to the left of the ball, the thread hangs at anangle of to the right of vertical. Calculate the electrostaticforce that the rod exerts on the ball.

19-19. SSM WWW For each of the following questions, eitheranswer it or tell why it cannot be answered.a. A point object has a charge of Find the mag-

nitude of the electric field at a position 1 m from this objectif there is nothing at that position.

b. Repeat a for the case when there is an electron at the posi-tion in question.

c. Two point objects, each with a charge of 0.003 C, are posi-tioned 0.40 m apart. Find the magnitude of the electric field.

19-20. Figure 19-23 shows the positions of two point objectswith equal positive charges Four points in the vicinity ofthese objects are labeled A, B, C, and D.a. Rank these points in order according to the magnitude of

the total electric field at each point. Rank them from leastto greatest, making sure to indicate any equalities.

b. Repeat a for the case when the charge at is neg-ative instead of positive.

c. Suppose now that the charge at is negative andthe charge at is positive. If you listed the pointsx � 5 cm

x � 1 cm

x � 5 cm

�Q.

�2 � 10�18 C.

15°

4.0 � 10�15 N,

8.0 � 10�11 N

1x � 0, y � 4 cm2.

�2.0 � 10�6 C.1.0 � 10�6 C.

1.6 � 10�6 C.1.0 � 10�6 C,x � 3 cm.

4.0 � 105 m�s2.in rank order as in a and b, how would the order compareto the orderings you listed for those two situations?

19-21. A singly charged positive ion is located at the originand a singly charged negative ion is located on the x axis at

a. Find the electric field (magnitude and direction) at the pointon the y axis.

b. Find the instantaneous acceleration (magnitude and direc-tion) of an electron as it passes through this point.

Section 19-3 Fields Due to Continuous Charge Distributions

19-22. Consider a square loop of an imaginary material hav-ing the peculiar property that any charge placed on it remainswhere it is put. Suppose that a certain amount of positivecharge is uniformly distributed over two adjacent sides of thissquare and an equal magnitude of negative charge is uniformlydistributed over the other two sides. By symmetry arguments,determine the direction of the electric field at the center of thesquare.

19-23. Figure 19-24 shows part of a closed circuit with thewires connected to the terminals of the battery greatlyenlarged. The diagram focuses on the surface charge densityon four ring-shaped regions of the wire surfaces, althoughthere is charge distributed over the remainder of the wire sur-faces as well. Ring B has a greater density of positive chargethan ring A; ring C has a greater density of negative chargethan ring D. Point is midway between rings A and B; point

is midway between rings C and D.a. What is the direction of the electric field at point due to

ring A alone? (Call this )

b. What is the direction of the electric field at point due toring B alone? (Call this

c. At point what is the direction of the total electric field

d. The rest of the surface charge distribution on the wire to thebattery’s left can be divided into rings of charge similar to Aand B. What is the direction of the total field at point P1 dueto any one pair of rings? What is the direction of the totalfield at point due to all such pairs of rings combined?

e. What is the direction of the electric field at point due toring C alone? (Call this )

f. What is the direction of the electric field at point due toring D alone? (Call this )

g. At point what is the direction of the total electric field

h. The rest of the surface charge distribution on the wire to thebattery’s right can be divided into rings of charge similar toC and D. What is the direction of the total field at point P2

due to any one pair of rings? What is the direction of thetotal field at point due to all such pairs of rings combined?

i . Write a sentence to describe the direction of the electricfield along the entire closed loop of the circuit, external tothe battery.

P2

ES

C � ES

D?P2,

ES

D.P2

ES

C.P2

P1

ES

A � ES

B?P1,

ES

B.2P1

ES

A.P1

P2

P1

y � 1.44 � 10�2 m

x � 6.00 � 10�3 m.

Figure 19-23 Problem 19-20 Figure 19-24 Problems 19-23 and 19-64

Qualitative and Quantitative Problems ◆ 591

1

y (in cm)

x (in cm)+Q +QDC

B

A + +0 1 3 42 5

A B++

+++ ++

++++

++++++++

+++

–+P1

C

––––––

––––––––

––

D––

––––

P2

Battery

0540T_c19_566-597.qxd 9/24/04 15:55 Page 591 EQA

Section 19-4 Picturing Electric Fields

19-24. Figure 19-25 showsthe electric field lines dueto two point objects 1 and2 having equal positivecharges. If points A and Bare equidistant from object1, sketch the electric fieldvectors at these two points.(Adapted from P. Shaffer, G. Fran-

cis, and L. C. McDermott, paper

abstracted in AAPT Announcer, 22,

no. 2, p. 51 [1992].)

19-25. Figure 19-25 shows the electric field lines due to twopoint objects 1 and 2 having equal positive charges. Points Aand B are equidistant from object 1.a. Is the magnitude of the field at A less than, equal to, or

greater than the field at B? Briefly explain.

b. If another positively charged point object is placed at pointA and released from rest, will the object pass to the leftof, to the right of, or directly through point C? Brieflyexplain.

19-26. Two tiny spheres with equal and opposite charges areseparated by a distance much greater than their diameters.Carefully sketch the electric field lines representing the totalfield due to these charges.

19-27.a. Carefully sketch the electric field lines due to a thin straight

rod of length L with a charge distributed uniformlyalong it.

b. Repeat a for a similar rod with a charge distributeduniformly along it. What are the significant differences?

19-28. Figure 19-26 shows a metal cylinder in both front andtop views. A negative charge is placed on this cylinder and isallowed to reach electrostatic equilibrium.a. Sketch what the electric field due to this cylinder looks like

in the front view.

b. Sketch what the field looks like in the top view.

�2Q

�Q

Figure 19-26 Problem 19-28

19-29. Figure 19-27 shows two conducting objects A and B.The two objects are charged so that the fields at their surfaceshave equal magnitude. Suppose a small test charge A is placedat the center of object A and then moved 1 cm to the right.An identical test charge B is placed at the center of object Band then likewise moved 1 cm to the right. C marks the center

in each case. On which of the two test charges does the forcechange by a larger percent? Briefly explain.

Section 19-5 Electrical and Gravitational Systems:Similarities and Differences; Electrical Potential Energy

19-30.a. In Figure 19-18b, in what part(s) of the electric circuit (if

any) does the conventional current of positive charges flowoppositely to the direction of the electric field?

b. In Figure 19-18a, in what part(s) of the gravitational circuit(if any) does mass flow oppositely to the direction of thegravitational field?

c. Is it gravitational forces or other forces that causes mass toflow opposite to the gravitational field in b ? Briefly explain.

d. Is it electrostatic (Coulomb) forces or other forces thatcauses conventional current to flow opposite to the electricfield in a ? Briefly explain.

19-31. Two electrons are a small distance apart. When werearrange this system to increase its electrical potential energy,does the gravitational potential energy of the system increase,decrease, or remain the same? Briefly explain why.

19-32. When the distance between a proton and an electronis doubled, the force on the electron changes in magnitudefrom to and the potential energy of the two-particle sys-tem changes from to The final value of each quan-tity is a certain fraction its initial value. When we compare thetwo fractions, do we find that is less than, equal to, orgreater than Briefly explain.

19-33. An atom of the most common isotope of hydrogen con-sists of a single proton and an electron. Suppose the electronstarts out at a distance of from the proton, andafter absorbing radiated energy it ends up at four times thatdistance. By how much (in J) does the potential energy of thisatom change?

19-34. The electric fielddue to the uniformplane of charge in Fig-ure 19-28 has a magni-tude of 400 N/C. At aparticular instant, acharged particle is in theposition shown and hasa speed of 800 m/stoward the plate.a. If the charged particle

is a singly chargedpositive ion and the

5.29 � 10�11 m

F2

F1?

PE2

PE1

PE2.PE1

F2,F1



Object 2

Object 1B

A

C

+

+

Figure 19-25 Problems 19-24and 19-25

x xx

y

z

y

y

Front view Top view

C C

1 m

1 m1 m1 cm

1 cm

v = 800 m/s

E

7.5 × 10–4 m

+

––––––––––––


0540T_c19_566-597.qxd 9/24/04 15:55 Page 592 EQA

plane’s charge is negative, find the change in potentialenergy as the particle travels from its present position to thepoint where it strikes the plane.

b. Find the change in kinetic energy.

c. If the ion has a mass of calculate the speedwith which it strikes the plane.

19-35. SSM WWWa. Calculate the speed with which the ion strikes the plane in

Problem 19-34 if everything is the same except that theplane’s charge is positive.

b. Calculate the speed with which the ion would strike thepositively charged plane if the ion were negative.

19-36. Write a problem involving gravitational rather than elec-trostatic forces and fields that is completely analogous to Prob-lem 19-34.

19-37.a. Find the change in the electrical potential energy of a sys-

tem of two electrons during an interval in which the dis-tance between them decreases from to

b. Find the change in the gravitational potential energy of thesystem over this same time interval.

19-38. The electric field between the twooppositely charged plates in Figure 19-29has a magnitude of 1500 N/C.a. If a proton is released from point P in

the figure, which plate would it strikeand at what speed?

b. If an electron is released from point P,which plate would it strike and at whatspeed?

19-39. Two protons are instantaneouslyat rest at a distance of from each other. If the protons are neveraffected by any objects except each other,what maximum speed will each of theprotons reach? Hint: How do the speedsof the two protons compare at anyinstant? Why?

Section 19-6 Potential and Potential Differences

19-40. In Figure 19-30, the charge Q on the object in the cen-ter is Calculate the change in potential in going1.00 � 10�6 C.

5.0 � 10�10 m

2.0 � 10�10 m.3.0 � 10�10 m

2.0 � 10�26 kg,

a. from A to B. b. from C to D. c. from A to D. Briefly tellwhat relationship there is between this last answer and youranswers to a and b.

19-41. SSM In Figure 19-30, the charge Q on the object in thecenter is In each of the following cases, cal-culate the change of potential energy and put a sign on eachanswer to indicate whether the change is an increase ora decrease (Hint: Think about whether you can use anyof your results from Problem 19-40.)a. A proton moves outward from A to B.

b. A proton moves outward from C to D.

c. An electron moves outward from C to D.

d. An electron moves inward from D to A.

19-42. The electric field due to a charged point object has amagnitude of 22.5 N/C at a distance of 0.020 m from the object.a. Find the potential due to this object at the same distance.

b. At what distance would the electric field be doubled? Atwhat distance would the potential be doubled?

19-43. Repeat Example 19-8 for the case in which a proton isreleased at the given point and reaches a point 0.020 m fromits starting point.

19-44. When we calculate the electric field by in SIunits, we obtain E in When we calculate the field from

in SI units, we obtain E in Show that and are really the same units.

19-45. An electron in a hydrogen atom gains 10.2 eV of energyin going from its lowest energy level to its first excited state.a. How much is this in joules?

b. How many electrons in a sample of hydrogen gas could beraised to their first excited states by an input of 1 J of energy?

19-46. How much potential difference must be establishedbetween two parallel plates if an electron leaving one of thetwo plates with negligible speed is to acquire a speed of

on reaching the opposite plate?

19-47. The potential difference between two parallel plates is12 V. Find the magnitude of the electric field between theplates if the plates are 0.0025 m apart.

19-48. If the electric field in Example 19-8 is produced by twoequally but oppositely charged plates separated by a distanceof 0.10 m, what potential difference must be applied betweenthe plates?

Section 19-7 Batteries

19-49. A student watches the instructor make a battery usingtwo metal electrodes and an acidic solution. The student latertries to explain it to a friend who missed class. “If I take thesetwo pieces of copper, I can use them as my electrodes. Then,when I insert them into opposite ends of this acid bath, I willbe able to measure a potential difference between the ends ofthe electrodes that stick out of the bath.” Will there be a poten-tial difference for the set-up this student is describing? Brieflyexplain.

19-50. How does the random motion of molecules (thermalmotion) contribute to the working of a battery?

19-51. The cell membrane of a living organism is not equallypermeable to different kinds of ions. How might this fact giverise to a potential difference across the cell membrane?

8.5 � 104 m�s

V/mN/CV/m.Vab � E¢y

N/C.E � F

q

1�2.1� 2

1.00 � 10�6 C.


Figure 19-30 Problems 19-40 and 19-41

–

6.0 × 10–4 m

P

2.0 × 10–4 m

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

––––––––––––

Figure 19-29Problem 19-38

r (in m)

A

0.10 0.20 0.30D C

B

Q

0540T_c19_566-597.qxd 9/24/04 15:55 Page 593 EQA


The questions and problems in this group are not organized by sec-tion heading, so you must determine for yourself which ideas apply.Some of them will be more challenging than the Review and Practicequestions and problems (especially those marked with a • or ••).

19-52. A positive ion is positioned at the origin. An electronapproaches it from the right. As the electron draws closer,a. does its speed increase, decrease, or remain the same?b. does its acceleration increase, decrease, or remain the same?

19-53. SSM Two spheres of equal size have charges and Which of the diagrams in Figure 19-31 most accu-rately shows the electrical forces that the two spheres exert oneach other?

�6 mC.�2 mC


19-54. Figure 19-32shows five points ina uniform electricfield directed towardthe right. Rank thesepoints according toeach of the follow-ing, listing them inorder from least togreatest and indicat-ing any equalities:a. the magnitude of the electric field at each point.

b. the electric potential at each point.

19-55. SSM Figure 19-33shows five points in thevicinity of an infiniteplane of uniformly dis-tributed positive charge.Rank these points accord-ing to each of the fol-lowing, listing them inorder from least to great-est and indicating anyequalities:a. the magnitude of the electric field at

each point.

b. the electric potential at each point.

19-56.a. In Figure 19-34, all three charges have

equal magnitude. If object A is instan-taneously at rest when it is at the ori-gin, in what direction will it moveaway from that point?

••b. If instead, object A passes through the origin with a veloc-ity of 5.0 m/s to the left, estimate the direction in which itwill be moving s later.

19-57. The “leaves” of a certain electro-scope are two very small metallic spheresof mass suspended from 0.20m lengths of fine conducting wire. Whenthe spheres hang as shown in Figure 19-35,what is the magnitude of the charge oneach of the spheres (assuming the twospheres are equally charged)?

19-58. A charged sphere (mass ) suspended by a thread hangs at

an angle of to the left of the verticalwhen in the presence of a 650 N/C electricfield directed to the right. Calculate thecharge on the sphere.

19-59. A parrot is on a dry wooden perch in a metal birdcage.The cage is struck by lightning. The parrot is unharmed. Why?

19-60. Redraw Figure 19-5c for the case where both rings arenegative but the lower ring is more negative. Sketch the fieldcontribution due to each ring and the resultant field due tothe combined effect of the two rings.

••19-61.a. A uniformly charged cir-

cular ring with radiusR has a total chargeof Q. The ring is posi-tioned perpendicular tothe x-axis with its cen-ter at the origin (Figure19-36). Show that at anyposition x along the pos-itive x-axis, the magnitude of the electric field is

State any assumptions you make in your reasoning.

b. Find the approximate value of E when x is so large that Ris negligibly small in comparison. Can you suggest a rea-son why you might have expected this value?

•19-62. A point object with a small positive charge is placedbetween equally and oppositely charged plates. Assume thepoint object’s charge has a negligible effect on the configura-tion of charge on the plates. Which of the sketches in Fig-ure 19-37 most accurately shows possible electric field linesdue to this arrangement of charge? Assume the plate dimen-sions are large compared to the distance between them, sothat the plates extend beyond the top and bottom of the fig-ure. Note: Only a few field lines are shown in each sketch.

E �kQx

1x2� R

223/2

3.6°10�4 kg

6.0 �

3.0 � 10�3 kg

1.0 � 10�3


G o i n g F u r t h e r

+6+2

+6+2

+6+2

+6+2

(a)

(b)

(c)

(d )

B E

AC D

E

x

y


y

xAC D

B E

++++++++++


y

xA C

B

+ –

+


5.8° 5.8°


R

xx


+++++++

–––––––

+

+++++++

–––––––

+

+++++++

–––––––

+

+++++++

–––––––

+

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

0540T_c19_566-597.qxd 9/24/04 17:21 Page 594 EQA

You might try including some of the others to see whether thesketch makes sense.

19-63. Student A says, “When I connect a bulb by two wiresto the terminals of a battery, the electric field that causescharge to flow through the filament of the bulb is due entirelyto the concentrations of positive and negative charge at theterminals of the battery. There are no concentrations of chargeon the wire itself; they are not needed to cause the current. Ipicture it like this (Figure 19-38a).”

Student B says, “I don’t agree. I think the battery willcause a change in surface charge density along the length ofthe wire, like this (Figure 19-38b). Most of the change in sur-face charge density takes place over the length of the filament,because that’s where most of the resistance is.”

The two circuits in Figure 19-38c provide us with a wayof seeing which student’s model is better. The circuits are iden-tical, except that in circuit I, the bulb is very near one termi-nal of the battery, whereas in circuit II the bulb is fairly dis-tant from either terminal.a. How should the electric field in the bulb filament compare

in the two circuits if we assume that student A’s model iscorrect? How should the brightness of the bulb then com-pare in the two circuits? Briefly explain.

b. How should the electric field in the bulb filament comparein the two circuits if we assume that student B’s model iscorrect? How should the brightness of the bulb then com-pare in the two circuits? Briefly explain.

c. In real circuits, does the bulb brightness depend on howfar it is from the battery? (Check this experimentally if you’renot sure.) Which conclusion agrees with what we actuallyobserve, the one based on student A’s model or the onebased on student B’s model? (Problem based on comments received

in private communication from Bruce Sherwood.)

19-65. A certain wire has a right-angle bend in it. Along eachstraight segment, away from the bend, the surface charge den-sity gradient is constant, so the field is constant along the axisof the wire. Roughly sketch the configuration of charge that isrequired in the region of the bend for the field lines to turnthe corner in this region.

19-66. An electron is given an initial veloc-ity parallel to a flat plate with a positivecharge distributed uniformly over its surface(Figure 19-39). The distance d in the figureis very small compared to the dimensions ofthe plate.a. Sketch the trajectory of the electron after it

is given its initial velocity. What is theshape of this trajectory? Why?

b. Describe a situation involving only gravita-tional forces that is analogous to this situ-ation. Tell what feature of the gravitationalsituation corresponds to each feature of this situation. Doesthe moving object in the situation you describe follow thesame shape trajectory as the electron?

19-67.a. Make a list of the correspondences between features of the

gravitational and electric circuits in Figure 19-18.

b. What aspects of the electric circuit do not have correspon-ding features in the gravitational circuit?

19-68. How can an electron be made to follow a parabolictrajectory in the presence of a negatively charged plate? Drawa sketch to clarify what you describe. Specify any conditionsthat you think are required.

19-69. Just as we can speak of non-electrostatic forces exertedon oppositely charged charge carriers in a battery, we canspeak of the non-electrostatic force per unit of charge as anon-electrostatic field. Estimate the magnitude of the non-electrostatic field inside a standard 1.5 V D-cell battery andgive its direction (toward which pole).

•19-70. A simple model of a hydrogen atom consists of anelectron of charge and mass held in a circular orbit ofradius r by a proton of charge and mass Show that

the total energy E of this arrangement is

19-71. The DNA molecule, often described as a double helix,can be pictured as a twisted ladder. The two long strands thatwind around each other are connected by bonds between pairsof bases, one base on each strand. These are the “rungs” ofthe ladder. People studying DNA often look at DNA fragments,which are themselves basically chains of these base pairs. DNAfragments in a neutral pH solution are negatively charged;there will be two excess electrons per base pair. If a 0.1 micron

DNA fragment has a charge of how many base pairs are there along this fragment?

19-72. Hemoglobin is the protein molecule found in redblood cells. Under suitable conditions, the ratio of charge q todrag coefficient for hemoglobin molecules in solution is

An electric field of 1000 N/C is appliedto the solution in an electrophoresis apparatus.a. What terminal speed is attained by the hemoglobin molecules?

b. About how far will the hemoglobin molecules travel in a day?

qz � �2.0 � 10�10

C � mN � s .z

�9.0 � 10�17 C,11.0 � 10�7 m2

E � �k e2

2r.

mp.�eme�e



19-64.a. In Figure 19-24, what is the direction of the electric field

associated with electrostatic force inside the battery?

b. In what direction does conventional current flow throughthe battery?

c. Under what circumstances, if any, can the electric field asso-ciated with electrostatic forces and the conventional currentbe in opposite directions?

Bulb

Battery

Circuit I

+++++++

+

+

– – –++

+

+

+

++

– ––––

––

––

– –

–

–

– –Bulb

Battery

Almost all ofresistance isin filament

Circuit II

(a) Student A’s model (b) Student B’s model

(c) Testing the models:

+++++

++++

–––––––––––

+

+

+

+

+

+

+

e–

d

vo


0540T_c19_566-597.qxd 9/24/04 15:55 Page 595 EQA

19-73. An electron passes a point where the electric field isdirected toward the right and has a magnitude of 300 N/C. Ifthe electron interacts only with the sources of this field, findthe magnitude and direction of the electron’s acceleration at theinstant it passes through this point.

19-74. An electric field of 150 N/C is maintained between twoequally and oppositely charged plates with a distance of

between them. The region between the plates isevacuated, and the negative plate is sufficiently heated so thatit occasionally emits electrons.a. If an electron is emitted from the negative plate at negli-

gible speed, with what speed does it reach the positiveplate?

b. How long does it take the electron to reach the positive plate?

19-75. SSM A potential difference of 120 V is maintainedbetween two equally and oppositely charged plates. Theregion between the plates is evacuated, and the negativeplate is sufficiently heated so that it occasionally emits elec-trons. Answer whichever of the following questions it is pos-sible to answer. If a question cannot be answered, explainwhy not.a. If an electron is emitted from the negative plate at negli-

gible speed, with what speed does it reach the positiveplate?

b. How long does it take the electron to reach the positive plate?

19-76. Two lengths of identical resistance wire are each con-nected between the terminals of a 1.5 V D-cell battery. (Thewires are said to be connected in parallel.) Length A is 25 cmlong, and length B is 50 cm long.a. How does the voltage between the ends of length A com-

pare to the voltage between the ends of length B?

b. How does the electric field in length A compare to the elec-tric field in length B?

c. How does the resistance of length A compare to the resist-ance of length B?

d. How does the resistivity of length A compare to the resis-tivity of length B?

e. How does the current through length A compare to the cur-rent through length B?

19-77. Two lengths of identical resistance wire are each con-nected between the terminals of a 9.0 V battery. Length A is0.30 m long, and length B is 0.90 m long. Calculate the elec-tric field in each wire.

19-78. We have spoken of setting up a uniform charge den-sity gradient along the surface of a wire. Each of the choicesbelow shows the surface charge density (in ) along thesurface of a wire. Which one shows a uniform charge densitygradient?

a. 2 2 2 2 0 �2 �2 �2 �2

b. 8 4 2 1

c. 6 5 4 3 2 1 0 �1 �2

d. 2 2 2 2 2 2 2 2 2

19-79. We have spoken of setting up a uniform charge den-sity gradient along the surface of a wire. In each of the graphs

116

18

14

12

C/m2

1.2 � 10�2 m

in Figure 19-40, surface charge density is plotted against dis-tance along the wire. Which one of these graphs shows a uni-form, nonzero charge density gradient?

19-80.a. From point B in Figure 19-9, in what direction would you

travel to experience the potential increasing most rapidlywith position?

b. The direction you found in a is defined to be the directionof the potential gradient at point B. How does the direc-tion of the electric field at B compare with the direction ofthe potential gradient?

19-81. A proton and an electron are a distance r apart. If thedistance r is increased, does the ratio increase, decrease,or remain the same? Briefly explain.

19-82. Figure 19-41 shows an infinite plane of uniformly dis-tributed negative charge. The arrows in the figure show fourpossible paths that can be traveled in the vicinity of the plane.Each path starts at the tail of the arrow and ends at the headof the arrow. Rank these paths according to each of the fol-lowing, listing them in order from least to greatest and indi-cating any equalities:a. the change in potential from the beginning to the end of

the path.b. the change in electrical potential energy that would occur if

an electron traveled from the beginning to the end of thepath.

PEelec

PEgrav

19-83. Gravitational and electrical systems compared. The elec-trical system in Figure 19-42a consists of two wires connectedbetween the terminals of a battery. Wire 2 is twice as long aswire 1. In the gravitational system in Figure 19-42b, an objectat point P can reach the ground by falling vertically downward




A

Surface charge density plotted vertically,distance along wire plotted horizontally

0B C D

0 0 0


A B C D

– – – – – – – – – – – – – – – – – – – – – – – – – – –

y

+ –

Wire 1 (L)

P Path 2 (2L)

Path

1 (

L)

Wire 2 (2L)

(a) Electrical system (b) Gravitational system

Ground

0540T_c19_566-597.qxd 9/24/04 15:55 Page 596 EQA

(path 1) or sliding down a frictionless ramp (path 2). Path 2 istwice as long as path 1.a. In the electrical system, suppose an electron travels from

the negative to the positive pole of the battery. Comparethe loss of potential energy when it travels by wire 1 andwhen it travels by wire 2.

b. In the electrical system, compare the change in potentialbetween the two terminals of the battery along wire 1 andalong wire 2.

c. In the electrical system, compare the electrical field alongwire 1 to the electrical field along wire 2.

d. In the gravitational system, suppose an object descendsfrom point P to the ground. Compare the loss of potentialenergy when it descends by path 1 and when it descendsby path 2.

e. Compare the accelerations that the object would experiencealong the two paths.

f. In the gravitational system, compare the gravitational fieldalong path 1 to the gravitational field along path 2.

g. Do the two systems behave analogously?


19-84. In WebLink 19-1, if we replace charge carrier A with acharge carrier with twice as much charge,a. will the magnitude of the electric field at the origin increase,

decrease, or remain the same?

b. will the magnitude of the force on the charge carrier at theorigin increase, decrease, or remain the same?

19-85. In WebLink 19-1, if we replace charge carrier A with acharge carrier with a charge of opposite sign, will the electricfield at the origin change direction? Briefly explain.

19-86. Suppose that when a sphere of radius 1 m is centeredat the positive point charge in WebLink 19-2, 720 field linesin all cross the surface of the sphere. If this sphere is nowreplaced by a sphere of radius 2 m centered at the same point

charge, how many field lines in all cross the surface of thenew sphere?

19-87. Suppose that when a sphere of radius 1 m is centeredat the positive point charge in WebLink 19-2, 60 field linescross each square unit of the sphere’s surface. If this sphere isnow replaced by a sphere of radius 2 m centered at the samepoint charge, how many field lines cross each square unit ofthe new sphere’s surface?

19-88. In WebLink 19-3, does the electric field increase,decrease, or remain the same as you go from a. point A topoint B? b. point B to a point very far to the right of point B?c. point A to a point in the interior of the charged plate?

P ro b l e m s o n We b L i n k s

0540T_c19_566-597.qxd 9/24/04 15:55 Page 597 EQA

electrical field and electrical potential chapter 19 electrical field and electrical potential we...

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