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Chapter 23

Electric Fields

CHAPTE R OUTL I N E

23.1 Properties of Electric Charges

23.2 Charging Objects ByInduction

23.3 Coulomb’s Law

23.4 The Electric Field

23.5 Electric Field of a ContinuousCharge Distribution

23.6 Electric Field Lines

23.7 Motion of Charged Particlesin a Uniform Electric Field

706

! Mother and daughter are both enjoying the effects of electrically charging their bodies.Each individual hair on their heads becomes charged and exerts a repulsive force on theother hairs, resulting in the “stand-up’’ hairdos that you see here. (Courtesy of ResonanceResearch Corporation)

707

The electromagnetic force between charged particles is one of the fundamental forcesof nature. We begin this chapter by describing some of the basic properties of onemanifestation of the electromagnetic force, the electric force. We then discussCoulomb’s law, which is the fundamental law governing the electric force between anytwo charged particles. Next, we introduce the concept of an electric field associatedwith a charge distribution and describe its effect on other charged particles. We thenshow how to use Coulomb’s law to calculate the electric field for a given charge distrib-ution. We conclude the chapter with a discussion of the motion of a charged particlein a uniform electric field.

23.1 Properties of Electric Charges

A number of simple experiments demonstrate the existence of electric forces andcharges. For example, after running a comb through your hair on a dry day, you willfind that the comb attracts bits of paper. The attractive force is often strong enough tosuspend the paper. The same effect occurs when certain materials are rubbed together,such as glass rubbed with silk or rubber with fur.

Another simple experiment is to rub an inflated balloon with wool. The balloonthen adheres to a wall, often for hours. When materials behave in this way, they aresaid to be electrified, or to have become electrically charged. You can easily electrifyyour body by vigorously rubbing your shoes on a wool rug. Evidence of the electriccharge on your body can be detected by lightly touching (and startling) a friend.Under the right conditions, you will see a spark when you touch, and both of you willfeel a slight tingle. (Experiments such as these work best on a dry day because anexcessive amount of moisture in the air can cause any charge you build up to “leak”from your body to the Earth.)

In a series of simple experiments, it was found that there are two kinds of electriccharges, which were given the names positive and negative by Benjamin Franklin(1706–1790). We identify negative charge as that type possessed by electrons and posi-tive charge as that possessed by protons. To verify that there are two types of charge,suppose a hard rubber rod that has been rubbed with fur is suspended by a sewingthread, as shown in Figure 23.1. When a glass rod that has been rubbed with silk isbrought near the rubber rod, the two attract each other (Fig. 23.1a). On the otherhand, if two charged rubber rods (or two charged glass rods) are brought near eachother, as shown in Figure 23.1b, the two repel each other. This observation shows thatthe rubber and glass have two different types of charge on them. On the basis of theseobservations, we conclude that charges of the same sign repel one another andcharges with opposite signs attract one another.

Using the convention suggested by Franklin, the electric charge on the glass rod iscalled positive and that on the rubber rod is called negative. Therefore, any chargedobject attracted to a charged rubber rod (or repelled by a charged glass rod) must

have a positive charge, and any charged object repelled by a charged rubber rod (or at-tracted to a charged glass rod) must have a negative charge.

Attractive electric forces are responsible for the behavior of a wide variety of com-mercial products. For example, the plastic in many contact lenses, etafilcon, is made upof molecules that electrically attract the protein molecules in human tears. Theseprotein molecules are absorbed and held by the plastic so that the lens ends up beingprimarily composed of the wearer’s tears. Because of this, the lens does not behave as aforeign object to the wearer’s eye, and it can be worn comfortably. Many cosmetics alsotake advantage of electric forces by incorporating materials that are electricallyattracted to skin or hair, causing the pigments or other chemicals to stay put once theyare applied.

Another important aspect of electricity that arises from experimental observa-tions is that electric charge is always conserved in an isolated system. That is,when one object is rubbed against another, charge is not created in the process. Theelectrified state is due to a transfer of charge from one object to the other. Oneobject gains some amount of negative charge while the other gains an equal amountof positive charge. For example, when a glass rod is rubbed with silk, as in Figure23.2, the silk obtains a negative charge that is equal in magnitude to the posi-tive charge on the glass rod. We now know from our understanding of atomic struc-ture that electrons are transferred from the glass to the silk in the rubbing process.Similarly, when rubber is rubbed with fur, electrons are transferred from the fur tothe rubber, giving the rubber a net negative charge and the fur a net positivecharge. This process is consistent with the fact that neutral, uncharged mattercontains as many positive charges (protons within atomic nuclei) as negativecharges (electrons).

In 1909, Robert Millikan (1868–1953) discovered that electric charge alwaysoccurs as some integral multiple of a fundamental amount of charge e (see Section25.7). In modern terms, the electric charge q is said to be quantized, where q is thestandard symbol used for charge as a variable. That is, electric charge exists asdiscrete “packets,” and we can write q ! Ne, where N is some integer. Other experi-ments in the same period showed that the electron has a charge " e and the protonhas a charge of equal magnitude but opposite sign # e. Some particles, such asthe neutron, have no charge.

From our discussion thus far, we conclude that electric charge has the following im-portant properties:

708 C H A P T E R 2 3 • Electric Fields

Electric charge is conserved

RubberRubber

(a)

F F

(b)

F

F

Rubber

– – – – –

– – – – –––

– – – –

+ + + ++ +

Glass

–

+

Figure 23.1 (a) A negatively charged rubber rod suspended by a thread is attracted to a positively charged glass rod. (b) A negatively charged rubber rod is repelled by another negatively charged rubber rod.

–

++++++–

––

––

Figure 23.2 When a glass rod isrubbed with silk, electrons aretransferred from the glass to thesilk. Because of conservation ofcharge, each electron adds nega-tive charge to the silk, and an equalpositive charge is left behind onthe rod. Also, because the chargesare transferred in discrete bundles,the charges on the two objects are$ e, or $ 2e, or $ 3e, and so on.

23.2 Charging Objects By Induction

It is convenient to classify materials in terms of the ability of electrons to move throughthe material:

S E C T I O N 2 3 . 2 • Charging Objects By Induction 709

Quick Quiz 23.1 If you rub an inflated balloon against your hair, the twomaterials attract each other, as shown in Figure 23.3. Is the amount of charge presentin the system of the balloon and your hair after rubbing (a) less than, (b) the same as,or (c) more than the amount of charge present before rubbing?

Quick Quiz 23.2 Three objects are brought close to each other, two at atime. When objects A and B are brought together, they repel. When objects B and Care brought together, they also repel. Which of the following are true? (a) Objects Aand C possess charges of the same sign. (b) Objects A and C possess charges of oppo-site sign. (c) All three of the objects possess charges of the same sign. (d) One of theobjects is neutral. (e) We would need to perform additional experiments to determinethe signs of the charges.

Properties of electric charge• There are two kinds of charges in nature; charges of opposite sign attract oneanother and charges of the same sign repel one another.

• Total charge in an isolated system is conserved.

• Charge is quantized.

Electrical conductors are materials in which some of the electrons are freeelectrons1 that are not bound to atoms and can move relatively freely through thematerial; electrical insulators are materials in which all electrons are bound toatoms and cannot move freely through the material.

Figure 23.3 (Quick Quiz 23.1)Rubbing a balloon against yourhair on a dry day causes theballoon and your hair to becomecharged.

Char

les D

. Win

ters

1 A metal atom contains one or more outer electrons, which are weakly bound to the nucleus. Whenmany atoms combine to form a metal, the so-called free electrons are these outer electrons, which are notbound to any one atom. These electrons move about the metal in a manner similar to that of gas mole-cules moving in a container.

Materials such as glass, rubber, and wood fall into the category of electrical insulators.When such materials are charged by rubbing, only the area rubbed becomes charged,and the charged particles are unable to move to other regions of the material.

In contrast, materials such as copper, aluminum, and silver are good electrical con-ductors. When such materials are charged in some small region, the charge readily dis-tributes itself over the entire surface of the material. If you hold a copper rod in yourhand and rub it with wool or fur, it will not attract a small piece of paper. This mightsuggest that a metal cannot be charged. However, if you attach a wooden handle to therod and then hold it by that handle as you rub the rod, the rod will remain chargedand attract the piece of paper. The explanation for this is as follows: without the insu-lating wood, the electric charges produced by rubbing readily move from the copperthrough your body, which is also a conductor, and into the Earth. The insulatingwooden handle prevents the flow of charge into your hand.

Semiconductors are a third class of materials, and their electrical propertiesare somewhere between those of insulators and those of conductors. Silicon and

germanium are well-known examples of semiconductors commonly used in the fabrica-tion of a variety of electronic chips used in computers, cellular telephones, and stereosystems. The electrical properties of semiconductors can be changed over many ordersof magnitude by the addition of controlled amounts of certain atoms to the materials.

To understand how to charge a conductor by a process known as induction, con-sider a neutral (uncharged) conducting sphere insulated from the ground, as shown inFigure 23.4a. There are an equal number of electrons and protons in the sphere if thecharge on the sphere is exactly zero. When a negatively charged rubber rod is broughtnear the sphere, electrons in the region nearest the rod experience a repulsive forceand migrate to the opposite side of the sphere. This leaves the side of the sphere nearthe rod with an effective positive charge because of the diminished number ofelectrons, as in Figure 23.4b. (The left side of the sphere in Figure 23.4b is positivelycharged as if positive charges moved into this region, but remember that it is onlyelectrons that are free to move.) This occurs even if the rod never actually touches thesphere. If the same experiment is performed with a conducting wire connected fromthe sphere to the Earth (Fig. 23.4c), some of the electrons in the conductor are sostrongly repelled by the presence of the negative charge in the rod that they move outof the sphere through the wire and into the Earth. The symbol at the end of the

wire in Figure 23.4c indicates that the wire is connected to ground, which means areservoir, such as the Earth, that can accept or provide electrons freely with negligibleeffect on its electrical characteristics. If the wire to ground is then removed (Fig.23.4d), the conducting sphere contains an excess of induced positive charge because ithas fewer electrons than it needs to cancel out the positive charge of the protons.When the rubber rod is removed from the vicinity of the sphere (Fig. 23.4e), this in-duced positive charge remains on the ungrounded sphere. Note that the rubber rodloses none of its negative charge during this process.

Charging an object by induction requires no contact with the object inducing thecharge. This is in contrast to charging an object by rubbing (that is, by conduction),which does require contact between the two objects.

A process similar to induction in conductors takes place in insulators. In most neutralmolecules, the center of positive charge coincides with the center of negative charge.However, in the presence of a charged object, these centers inside each molecule in aninsulator may shift slightly, resulting in more positive charge on one side of the moleculethan on the other. This realignment of charge within individual molecules produces alayer of charge on the surface of the insulator, as shown in Figure 23.5a. Knowing aboutinduction in insulators, you should be able to explain why a comb that has been rubbedthrough hair attracts bits of electrically neutral paper and why a balloon that has beenrubbed against your clothing is able to stick to an electrically neutral wall.


– – –

(b)

+ –

–

––– ––

–+

+ +

+

++

+

(a)

– +–+

+ ++

+ ++

– ––

–

––

– – –

(c)

+

+

+

+

+

+ +

+

–

–

–

–

– – –

(d)

+–+

++

+ +

–

–

–+

+

(e)

–

+– –

–

+

++

+

+

++

Figure 23.4 Charging a metallicobject by induction (that is, the twoobjects never touch each other).(a) A neutral metallic sphere, withequal numbers of positive and neg-ative charges. (b) The electrons onthe neutral sphere are redistrib-uted when a charged rubber rod isplaced near the sphere. (c) Whenthe sphere is grounded, some of itselectrons leave through the groundwire. (d) When the ground connec-tion is removed, the sphere hasexcess positive charge that isnonuniformly distributed. (e) When the rod is removed, theremaining electrons redistributeuniformly and there is a net uniform distribution of positivecharge on the sphere.

+

+

+

+

+

+

+–

+–

+–

+–

+–

+–

Insulator

Inducedcharges

Chargedobject

(a) (b)

Figure 23.5 (a) The charged object on the left induces a charge distribution on thesurface of an insulator due to realignment of charges in the molecules. (b) A chargedcomb attracts bits of paper because charges in molecules in the paper are realigned.

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23.3 Coulomb’s Law

Charles Coulomb (1736–1806) measured the magnitudes of the electric forces betweencharged objects using the torsion balance, which he invented (Fig. 23.6). Coulomb con-firmed that the electric force between two small charged spheres is proportional to theinverse square of their separation distance r —that is, Fe % 1/r 2. The operating principleof the torsion balance is the same as that of the apparatus used by Cavendish to measurethe gravitational constant (see Section 13.2), with the electrically neutral spheresreplaced by charged ones. The electric force between charged spheres A and B in Figure23.6 causes the spheres to either attract or repel each other, and the resulting motioncauses the suspended fiber to twist. Because the restoring torque of the twisted fiber isproportional to the angle through which the fiber rotates, a measurement of this angleprovides a quantitative measure of the electric force of attraction or repulsion. Once thespheres are charged by rubbing, the electric force between them is very large comparedwith the gravitational attraction, and so the gravitational force can be neglected.

From Coulomb’s experiments, we can generalize the following properties of theelectric force between two stationary charged particles. The electric force

• is inversely proportional to the square of the separation r between the particles anddirected along the line joining them;

• is proportional to the product of the charges q1 and q2 on the two particles;

• is attractive if the charges are of opposite sign and repulsive if the charges have thesame sign;

• is a conservative force.

We will use the term point charge to mean a particle of zero size that carriesan electric charge. The electrical behavior of electrons and protons is very welldescribed by modeling them as point charges. From experimental observations on theelectric force, we can express Coulomb’s law as an equation giving the magnitude ofthe electric force (sometimes called the Coulomb force) between two point charges:

(23.1)

where ke is a constant called the Coulomb constant. In his experiments, Coulomb wasable to show that the value of the exponent of r was 2 to within an uncertainty of a fewpercent. Modern experiments have shown that the exponent is 2 to within an uncer-tainty of a few parts in 1016.

The value of the Coulomb constant depends on the choice of units. The SI unit ofcharge is the coulomb (C). The Coulomb constant ke in SI units has the value

ke ! 8.987 5 & 109 N ' m2/C2 (23.2)

This constant is also written in the form

(23.3)ke !1

4()0

Fe ! ke ! q 1 !! q 2 !

r 2

S E C T I O N 2 3 . 3 • Coulomb’s Law 711

Quick Quiz 23.3 Three objects are brought close to each other, two at atime. When objects A and B are brought together, they attract. When objects B and Care brought together, they repel. From this, we conclude that (a) objects A and Cpossess charges of the same sign. (b) objects A and C possess charges of opposite sign.(c) all three of the objects possess charges of the same sign. (d) one of the objects isneutral. (e) we need to perform additional experiments to determine informationabout the charges on the objects.

Coulomb’s law

Coulomb constant

Suspensionhead

Fiber

B

A

Figure 23.6 Coulomb’s torsionbalance, used to establish theinverse-square law for the electricforce between two charges.

where the constant )0 (lowercase Greek epsilon) is known as the permittivity of freespace and has the value

)0 ! 8.854 2 & 10"12 C2/N ' m2 (23.4)

The smallest unit of charge e known in nature2 is the charge on an electron (" e)or a proton (# e) and has a magnitude

e ! 1.602 19 & 10"19 C (23.5)

Therefore, 1 C of charge is approximately equal to the charge of 6.24 & 1018 electronsor protons. This number is very small when compared with the number of freeelectrons in 1 cm3 of copper, which is on the order of 1023. Still, 1 C is a substantialamount of charge. In typical experiments in which a rubber or glass rod is charged byfriction, a net charge on the order of 10"6 C is obtained. In other words, only a verysmall fraction of the total available charge is transferred between the rod and therubbing material.

The charges and masses of the electron, proton, and neutron are given in Table23.1.


Example 23.1 The Hydrogen Atom

gravitational force is

!

The ratio Fe /Fg " 2 & 1039. Thus, the gravitational forcebetween charged atomic particles is negligible when com-pared with the electric force. Note the similarity of form ofNewton’s law of universal gravitation and Coulomb’s law ofelectric forces. Other than magnitude, what is a fundamen-tal difference between the two forces?

3.6 & 10"47 N

&(9.11 & 10"31 kg)(1.67 & 10"27 kg)

(5.3 & 10"11 m)2

!(6.67 & 10"11 N'm2/kg2)

Fg ! G memp

r2

The electron and proton of a hydrogen atom are separated(on the average) by a distance of approximately 5.3 &10"11 m. Find the magnitudes of the electric force and thegravitational force between the two particles.

Solution From Coulomb’s law, we find that the magnitudeof the electric force is

!

Using Newton’s law of universal gravitation and Table 23.1for the particle masses, we find that the magnitude of the

8.2 & 10"8 N

Fe ! ke !e ! !"e !

r 2 ! (8.99 & 109 N'm2/C2) (1.60 & 10"19 C)2

(5.3 & 10"11 m)2

Quick Quiz 23.4 Object A has a charge of # 2 *C, and object Bhas a charge of # 6 *C. Which statement is true about the electric forces on the objects? (a) FAB ! " 3FBA (b) FAB ! " FBA (c) 3FAB ! " FBA (d) FAB ! 3FBA(e) FAB ! FBA (f) 3FAB ! FBA

Particle Charge (C) Mass (kg)

Electron (e) " 1.602 191 7 & 10"19 9.109 5 & 10"31

Proton (p) # 1.602 191 7 & 10"19 1.672 61 & 10"27

Neutron (n) 0 1.674 92 & 10"27

Charge and Mass of the Electron, Proton, and Neutron

Table 23.1

2 No unit of charge smaller than e has been detected on a free particle; however, current theoriespropose the existence of particles called quarks having charges "e/3 and 2e/3. Although there isconsiderable experimental evidence for such particles inside nuclear matter, free quarks have neverbeen detected. We discuss other properties of quarks in Chapter 46 of the extended version of this text.

Charles CoulombFrench physicist (1736–1806)

Coulomb’s major contributionsto science were in the areas ofelectrostatics and magnetism.During his lifetime, he alsoinvestigated the strengths ofmaterials and determined theforces that affect objects onbeams, thereby contributing tothe field of structuralmechanics. In the field ofergonomics, his researchprovided a fundamentalunderstanding of the ways inwhich people and animals canbest do work. (Photo courtesy ofAIP Niels Bohr Library/E. ScottBarr Collection)

S E C T I O N 2 3 . 3 • Coulomb’s Law 713

Quick Quiz 23.5 Object A has a charge of #2 *C, and object B has acharge of # 6 *C. Which statement is true about the electric forces on the objects? (a) FAB ! " 3FBA (b) FAB ! " FBA (c) 3FAB ! " FBA (d) FAB ! 3FBA(e) FAB ! FBA (f) 3FAB ! FBA

Example 23.2 Find the Resultant Force

q2 ! " 2.0 *C, and a ! 0.10 m. Find the resultant forceexerted on q3.

Solution First, note the direction of the individual forcesexerted by q1 and q2 on q3. The force F23 exerted by q2 onq3 is attractive because q2 and q3 have opposite signs. Theforce F13 exerted by q1 on q3 is repulsive because bothcharges are positive.

The magnitude of F23 is

In the coordinate system shown in Figure 23.8, the attractiveforce F23 is to the left (in the negative x direction).

! 9.0 N

! (8.99 & 109 N'm2/C2) (2.0 & 10"6 C)(5.0 & 10"6 C)

(0.10 m)2

F23 ! ke ! q 2 !! q 3 !

a2

Consider three point charges located at the corners of a righttriangle as shown in Figure 23.8, where q1 ! q3 ! 5.0 *C,

When dealing with Coulomb’s law, you must remember that force is a vector quan-tity and must be treated accordingly. The law expressed in vector form for the electricforce exerted by a charge q1 on a second charge q2, written F12, is

(23.6)

where r is a unit vector directed from q1 toward q2, as shown in Figure 23.7a. Becausethe electric force obeys Newton’s third law, the electric force exerted by q2 on q1 isequal in magnitude to the force exerted by q1 on q2 and in the opposite direction;that is, F21 ! " F12. Finally, from Equation 23.6, we see that if q1 and q2 have thesame sign, as in Figure 23.7a, the product q1q2 is positive. If q1 and q2 are of oppositesign, as shown in Figure 23.7b, the product q1q2 is negative. These signs describe therelative direction of the force but not the absolute direction. A negative product indi-cates an attractive force, so that the charges each experience a force toward theother—thus, the force on one charge is in a direction relative to the other. A positiveproduct indicates a repulsive force such that each charge experiences a force awayfrom the other. The absolute direction of the force in space is not determined solely bythe sign of q1q2—whether the force on an individual charge is in the positive or nega-tive direction on a coordinate axis depends on the location of the other charge. Forexample, if an x axis lies along the two charges in Figure 23.7a, the product q1q2 ispositive, but F12 points in the # x direction and F21 points in the " x direction.

F12 ! ke q 1q 2

r 2 r

–

+r

(a)F21

F12

q1

q2

(b)

F21

F12

q1

q2

r

+

+

Active Figure 23.7 Two pointcharges separated by a distance rexert a force on each other that isgiven by Coulomb’s law. The forceF21 exerted by q2 on q1 is equal inmagnitude and opposite in directionto the force F12 exerted by q1 on q2.(a) When the charges are of thesame sign, the force is repulsive.(b) When the charges are ofopposite signs, the force is attractive.

At the Active Figures linkat http://www.pse6.com, youcan move the charges to anyposition in two-dimensionalspace and observe the electricforces on them.

F13

q3

q1

q2

a

a

y

x

–

+

+F23

2a√

Figure 23.8 (Example 23.2) The force exerted by q1 on q3 isF13. The force exerted by q2 on q3 is F23. The resultant force F3exerted on q3 is the vector sum F13 # F23.

Vector form of Coulomb’s law

When more than two charges are present, the force between any pair of them is givenby Equation 23.6. Therefore, the resultant force on any one of them equals the vectorsum of the forces exerted by the various individual charges. For example, if four chargesare present, then the resultant force exerted by particles 2, 3, and 4 on particle 1 is

F1 ! F21 # F31 # F41


Example 23.4 Find the Charge on the Spheres

Solution Figure 23.10a helps us conceptualize thisproblem—the two spheres exert repulsive forces on eachother. If they are held close to each other and released, theywill move outward from the center and settle into the con-figuration in Figure 23.10a after the damped oscillations

Two identical small charged spheres, each having a mass of3.0 & 10"2 kg, hang in equilibrium as shown in Figure23.10a. The length of each string is 0.15 m, and the angle +is 5.0°. Find the magnitude of the charge on each sphere.

The magnitude of the force F13 exerted by q1 on q3 is

The repulsive force F13 makes an angle of 45° with the xaxis. Therefore, the x and y components of F13 are equal,with magnitude given by F13 cos 45° ! 7.9 N.

Combining F13 with F23 by the rules of vector addition,we arrive at the x and y components of the resultant forceacting on q3:

! 11 N

! (8.99 & 109 N'm2/C2) (5.0 & 10"6 C)(5.0 & 10"6 C)

2(0.10 m)2

F13 ! ke! q 1 !! q 3 !(√2a)2

F3x ! F13x # F23x ! 7.9 N # (" 9.0 N) ! " 1.1 N

F3y ! F13y # F23y ! 7.9 N # 0 ! 7.9 N

We can also express the resultant force acting on q3 in unit-vector form as

What If? What if the signs of all three charges were changedto the opposite signs? How would this affect the result for F3?

Answer The charge q3 would still be attracted toward q2and repelled from q1 with forces of the same magnitude.Thus, the final result for F3 would be exactly the same.

("1.1i # 7.9 j) NF3 !

of the forces on q3 are equal, but both forces are in thesame direction at this location.

What If? Suppose charge q3 is constrained to move onlyalong the x axis. From its initial position at x ! 0.775 m, it ispulled a very small distance along the x axis. When released,will it return to equilibrium or be pulled further from equilib-rium? That is, is the equilibrium stable or unstable?

Answer If the charge is moved to the right, F13 becomeslarger and F23 becomes smaller. This results in a net force tothe right, in the same direction as the displacement. Thus,the equilibrium is unstable.

Note that if the charge is constrained to stay at a fixed xcoordinate but allowed to move up and down in Figure 23.9,the equilibrium is stable. In this case, if the charge is pulledupward (or downward) and released, it will move backtoward the equilibrium position and undergo oscillation.

Three point charges lie along the x axis as shown in Figure23.9. The positive charge q1 ! 15.0 *C is at x ! 2.00 m,the positive charge q2 ! 6.00 *C is at the origin, and theresultant force acting on q3 is zero. What is the x coordi-nate of q3?

Solution Because q3 is negative and q1 and q2 are positive,the forces F13 and F23 are both attractive, as indicated in Fig-ure 23.9. From Coulomb’s law, F13 and F23 have magnitudes

For the resultant force on q3 to be zero, F23 must be equal inmagnitude and opposite in direction to F13. Setting themagnitudes of the two forces equal, we have

Noting that ke and are common to both sides and so canbe dropped, we solve for x and find that

This can be reduced to the following quadratic equation:

3.00x2 # 8.00x " 8.00 = 0

Solving this quadratic equation for x, we find that the positive

root is x ! There is also a second root, x !

" 3.44 m. This is another location at which the magnitudes

0.775 m.

(4.00 " 4.00x # x 2)(6.00 & 10"6 C) ! x 2(15.0 & 10"6 C)

(2.00 " x)2 ! q 2 ! ! x2 ! q 1 !

! q 3 !

ke ! q 2 !! q 3 !

x2 ! ke ! q 1 !! q 3 !

(2.00 " x)2

F13 ! ke ! q 1 !! q 3 !

(2.00 " x)2 F23 ! ke ! q 2 !! q 3 !

x 2

Example 23.3 Where Is the Resultant Force Zero?

2.00 m

x

q1

xq3

–q2 F13F23

2.00 – x

+ +

Figure 23.9 (Example 23.3) Three point charges are placedalong the x axis. If the resultant force acting on q3 is zero, thenthe force F13 exerted by q1 on q3 must be equal in magnitudeand opposite in direction to the force F23 exerted by q2 on q3.

At the Interactive Worked Example link at http://www.pse6.com, you can predict where on the x axis the electric force is zerofor random values of q1 and q2.

Interactive

23.4 The Electric Field

Two field forces have been introduced into our discussions so far—the gravitationalforce in Chapter 13 and the electric force here. As pointed out earlier, field forces canact through space, producing an effect even when no physical contact occurs betweeninteracting objects. The gravitational field g at a point in space was defined in Section13.5 to be equal to the gravitational force Fg acting on a test particle of mass m dividedby that mass: g # Fg/m. The concept of a field was developed by Michael Faraday(1791–1867) in the context of electric forces and is of such practical value that we shalldevote much attention to it in the next several chapters. In this approach, an electricfield is said to exist in the region of space around a charged object—the sourcecharge. When another charged object—the test charge—enters this electric field, an

S E C T I O N 2 3 . 4 • The Electric Field 715

The separation of the spheres is 2a ! 0.026 m.From Coulomb’s law (Eq. 23.1), the magnitude of the

electric force is

where r ! 2a ! 0.026 m and !q ! is the magnitude of thecharge on each sphere. (Note that the term !q !2 arises herebecause the charge is the same on both spheres.) This equa-tion can be solved for !q !2 to give

To finalize the problem, note that we found only themagnitude of the charge !q ! on the spheres. There is no waywe could find the sign of the charge from the informationgiven. In fact, the sign of the charge is not important. Thesituation will be exactly the same whether both spheres arepositively charged or negatively charged.

What If? Suppose your roommate proposes solving thisproblem without the assumption that the charges are ofequal magnitude. She claims that the symmetry of theproblem is destroyed if the charges are not equal, so that thestrings would make two different angles with the vertical, andthe problem would be much more complicated. How wouldyou respond?

Answer You should argue that the symmetry is not de-stroyed and the angles remain the same. Newton’s third lawrequires that the electric forces on the two charges be thesame, regardless of the equality or nonequality of thecharges. The solution to the example remains the samethrough the calculation of !q !2. In this situation, the value of1.96 & 10" 15 C2 corresponds to the product q1q2, where q1and q2 are the values of the charges on the two spheres. Thesymmetry of the problem would be destroyed if the masses ofthe spheres were not the same. In this case, the stringswould make different angles with the vertical and theproblem would be more complicated.

4.4 & 10"8 C! q ! !

!q !2 !Fe r 2

ke!

(2.6 & 10"2 N)(0.026 m)2

8.99 & 109 N'm2/C2 ! 1.96 & 10"15 C2

Fe ! ke ! q !2

r 2

due to air resistance have vanished. The key phrase “in equi-librium’’ helps us categorize this as an equilibrium problem,which we approach as we did equilibrium problems inChapter 5 with the added feature that one of the forces on asphere is an electric force. We analyze this problem bydrawing the free-body diagram for the left-hand sphere inFigure 23.10b. The sphere is in equilibrium under the appli-cation of the forces T from the string, the electric force Fefrom the other sphere, and the gravitational force mg.

Because the sphere is in equilibrium, the forces in the hor-izontal and vertical directions must separately add up to zero:

From Equation (2), we see that T ! mg/cos + ; thus, T canbe eliminated from Equation (1) if we make this substitu-tion. This gives a value for the magnitude of the electricforce Fe :

Fe ! mg tan + ! (3.0 & 10"2 kg)(9.80 m/s2) tan(5.0°)

! 2.6 & 10"2 NConsidering the geometry of the right triangle in Figure23.10a, we see that sin + ! a/L. Therefore,

a ! L sin + ! (0.15 m) sin(5.0°) ! 0.013 m

(2) $ Fy ! T cos + " mg ! 0

(1) $ Fx ! T sin + " Fe ! 0

(a) (b)

mg

LL

θ θ

L = 0.15 mθ = 5.0°

q a q

θTT cos θ

T sin θ

θ

Fe

θ

θ

θ

Figure 23.10 (Example 23.4) (a) Two identical spheres, eachcarrying the same charge q, suspended in equilibrium. (b) Thefree-body diagram for the sphere on the left.

electric force acts on it. As an example, consider Figure 23.11, which shows a small posi-tive test charge q0 placed near a second object carrying a much greater positive chargeQ. We define the electric field due to the source charge at the location of the testcharge to be the electric force on the test charge per unit charge, or to be more specific


This dramatic photograph captures alightning bolt striking a tree nearsome rural homes. Lightning is associ-ated with very strong electric fields inthe atmosphere.©

John

ny A

uter

y

+

+ ++ ++ ++ +

+ ++

++

+

Q

q0

E

Figure 23.11 A small positive testcharge q0 placed near an objectcarrying a much larger positivecharge Q experiences an electricfield E directed as shown.

the electric field vector E at a point in space is defined as the electric force Feacting on a positive test charge q0 placed at that point divided by the test charge:

(23.7)E # Fe

q 0

Note that E is the field produced by some charge or charge distribution separate fromthe test charge—it is not the field produced by the test charge itself. Also, note that theexistence of an electric field is a property of its source—the presence of the test chargeis not necessary for the field to exist. The test charge serves as a detector of the electricfield.

Equation 23.7 can be rearranged as

(23.8)

where we have used the general symbol q for a charge. This equation gives us the forceon a charged particle placed in an electric field. If q is positive, the force is in the same

Fe ! q E

! PITFALL PREVENTION 23.1 Particles OnlyEquation 23.8 is only valid for acharged particle —an object ofzero size. For a charged object offinite size in an electric field, thefield may vary in magnitude anddirection over the size of theobject, so the correspondingforce equation may be morecomplicated.

Definition of electric field

S E C T I O N 2 3 . 4 • The Electric Field 717

direction as the field. If q is negative, the force and the field are in opposite directions.Notice the similarity between Equation 23.8 and the corresponding equation for a par-ticle with mass placed in a gravitational field, Fg ! mg (Eq. 5.6).

The vector E has the SI units of newtons per coulomb (N/C). The direction of E,as shown in Figure 23.11, is the direction of the force a positive test charge experienceswhen placed in the field. We say that an electric field exists at a point if a testcharge at that point experiences an electric force. Once the magnitude and direc-tion of the electric field are known at some point, the electric force exerted on anycharged particle placed at that point can be calculated from Equation 23.8. Theelectric field magnitudes for various field sources are given in Table 23.2.

When using Equation 23.7, we must assume that the test charge q0 is small enoughthat it does not disturb the charge distribution responsible for the electric field. If avanishingly small test charge q0 is placed near a uniformly charged metallic sphere, asin Figure 23.12a, the charge on the metallic sphere, which produces the electric field,remains uniformly distributed. If the test charge is great enough (q0, -- q0), as inFigure 23.12b, the charge on the metallic sphere is redistributed and the ratio of theforce to the test charge is different: (Fe,/q0, . Fe /q0). That is, because of this redistribu-tion of charge on the metallic sphere, the electric field it sets up is different from thefield it sets up in the presence of the much smaller test charge q0.

To determine the direction of an electric field, consider a point charge q as asource charge. This charge creates an electric field at all points in space surround-ing it. A test charge q 0 is placed at point P, a distance r from the source charge, as inFigure 23.13a. We imagine using the test charge to determine the direction of theelectric force and therefore that of the electric field. However, the electric fielddoes not depend on the existence of the test charge—it is established solely by thesource charge. According to Coulomb’s law, the force exerted by q on the testcharge is

where r is a unit vector directed from q toward q0. This force in Figure 23.13a isdirected away from the source charge q. Because the electric field at P, the positionof the test charge, is defined by E ! Fe/q0, we find that at P, the electric field createdby q is

(23.9)

If the source charge q is positive, Figure 23.13b shows the situation with the test chargeremoved—the source charge sets up an electric field at point P, directed away from q.If q is negative, as in Figure 23.13c, the force on the test charge is toward the sourcecharge, so the electric field at P is directed toward the source charge, as in Figure23.13d.

E ! ke qr 2 r

Fe ! ke qq0

r 2 r

Source E (N/C)

Fluorescent lighting tube 10Atmosphere (fair weather) 100Balloon rubbed on hair 1 000Atmosphere (under thundercloud) 10 000Photocopier 100 000Spark in air -3 000 000Near electron in hydrogen atom 5 & 1011

Typical Electric Field Values

Table 23.2

(a) (b)

q0+ q′0>>q0+

–– –

–

–

–––

–

–

––

– – ––

–––

––

––

–

Figure 23.12 (a) For a smallenough test charge q0, the chargedistribution on the sphere is undis-turbed. (b) When the test chargeq0, is greater, the charge distribu-tion on the sphere is disturbed asthe result of the proximity of q0, .

Active Figure 23.13 A test chargeq0 at point P is a distance r from apoint charge q. (a) If q is positive,then the force on the test charge isdirected away from q. (b) For thepositive source charge, the electricfield at P points radially outwardfrom q. (c) If q is negative, then theforce on the test charge is directedtoward q. (d) For the negativesource charge, the electric field at Ppoints radially inward toward q.

At the Active Figures link athttp://www.pse6.com, you canmove point P to any position intwo-dimensional space and observe the electric field due to q.

(b)

E

q r

P

r

+

(a)

F

q

q0

rP

r

+

–(c)

Fq

q0

P

r

–(d)

E

qP

rr


To calculate the electric field at a point P due to a group of point charges, we firstcalculate the electric field vectors at P individually using Equation 23.9 and then addthem vectorially. In other words,

at any point P, the total electric field due to a group of source charges equals thevector sum of the electric fields of all the charges.

Quick Quiz 23.6 A test charge of # 3 *C is at a point P where an externalelectric field is directed to the right and has a magnitude of 4 & 106 N/C. If the testcharge is replaced with another test charge of " 3 *C, the external electric field at P(a) is unaffected (b) reverses direction (c) changes in a way that cannot be determined

Electric field due to a finitenumber of point charges

Example 23.5 Electric Field Due to Two Charges

Solution First, let us find the magnitude of the electric fieldat P due to each charge. The fields E1 due to the 7.0-*Ccharge and E2 due to the " 5.0-*C charge are shown inFigure 23.14. Their magnitudes are

The vector E1 has only a y component. The vector E2 has anx component given by E 2 cos + ! E2 and a negative y

component given by " E 2 sin + ! " E 2 . Hence, we can

express the vectors as

The resultant field E at P is the superposition of E1 and E2:

From this result, we find that E makes an angle / of 66° withthe positive x axis and has a magnitude of 2.7 & 105 N/C.

(1.1 & 105 i # 2.5 & 105 j) N/CE ! E1 # E2 !

E2 ! (1.1 & 105 i " 1.4 & 105 j) N/C

E1 ! 3.9 & 105 j N/C

45

35

! 1.8 & 105 N/C

E2 ! ke ! q 2 !r 2

2 ! (8.99 & 109 N'm2/C2) (5.0 & 10"6 C)

(0.50 m)2

! 3.9 & 105 N/C

E1 ! ke ! q 1 !r 1

2 ! (8.99 & 109 N'm2/C2) (7.0 & 10"6 C)

(0.40 m)2

A charge q1 ! 7.0 *C is located at the origin, and a secondcharge q2 ! " 5.0 *C is located on the x axis, 0.30 m fromthe origin (Fig. 23.14). Find the electric field at the point P,which has coordinates (0, 0.40) m.

0.40 m

Pθ

E

E2

0.50 m

E1

y

θx

q2q10.30 m

–

φ

+

Figure 23.14 (Example 23.5) The total electric field E at Pequals the vector sum E1 # E2, where E1 is the field due to thepositive charge q1 and E2 is the field due to the negativecharge q2.

This superposition principle applied to fields follows directly from the superpositionproperty of electric forces, which, in turn, follows from the fact that we know thatforces add as vectors from Chapter 5. Thus, the electric field at point P due to a groupof source charges can be expressed as the vector sum

(23.10)

where ri is the distance from the i th source charge qi to the point P and ri is a unitvector directed from qi toward P.

E ! ke $ i

qi

r i2 ri

S E C T I O N 2 3 . 5 • Electric Field of a Continuous Charge Distribution 719

The y components of E1 and E2 cancel each other, andthe x components are both in the positive x direction andhave the same magnitude. Therefore, E is parallel to thex axis and has a magnitude equal to 2E1 cos +. FromFigure 23.15 we see that cos + ! a/r ! a/(y 2 # a2)1/2.Therefore,

Because y -- a, we can neglect a2 compared to y2 and write

Thus, we see that, at distances far from a dipole but alongthe perpendicular bisector of the line joining the twocharges, the magnitude of the electric field created by thedipole varies as 1/r 3, whereas the more slowly varying fieldof a point charge varies as 1/r 2 (see Eq. 23.9). This isbecause at distant points, the fields of the two charges ofequal magnitude and opposite sign almost cancel eachother. The 1/r 3 variation in E for the dipole also is obtainedfor a distant point along the x axis (see Problem 22) and forany general distant point.

The electric dipole is a good model of many mole-cules, such as hydrochloric acid (HCl). Neutral atoms andmolecules behave as dipoles when placed in an externalelectric field. Furthermore, many molecules, such as HCl,are permanent dipoles. The effect of such dipoles on thebehavior of materials subjected to electric fields is dis-cussed in Chapter 26.

ke 2qay 3E "

! ke 2qa

(y 2 # a2)3/2

E ! 2E1 cos + ! 2ke q

(y 2 # a2)

a

(y 2 # a2)1/2

An electric dipole is defined as a positive charge q and anegative charge " q separated by a distance 2a. For the dipoleshown in Figure 23.15, find the electric field E at P due to thedipole, where P is a distance y -- a from the origin.

Solution At P, the fields E1 and E2 due to the two chargesare equal in magnitude because P is equidistant from thecharges. The total field is E ! E1 # E2, where

E1 ! E2 ! ke qr 2 ! ke

qy2 # a2

Example 23.6 Electric Field of a Dipole

P Eθ

θ

y

E1

E2y

r

θ

aq

θ

a–q– x+

Figure 23.15 (Example 23.6) The total electric field E at P dueto two charges of equal magnitude and opposite sign (an elec-tric dipole) equals the vector sum E1 # E2. The field E1 is dueto the positive charge q, and E2 is the field due to the negativecharge " q.

23.5 Electric Field of a ContinuousCharge Distribution

Very often the distances between charges in a group of charges are much smaller than thedistance from the group to some point of interest (for example, a point where the electricfield is to be calculated). In such situations, the system of charges can be modeled as con-tinuous. That is, the system of closely spaced charges is equivalent to a total charge that iscontinuously distributed along some line, over some surface, or throughout some volume.

To evaluate the electric field created by a continuous charge distribution, we use thefollowing procedure: first, we divide the charge distribution into small elements, each ofwhich contains a small charge 0q, as shown in Figure 23.16. Next, we use Equation 23.9to calculate the electric field due to one of these elements at a point P. Finally, we evalu-ate the total electric field at P due to the charge distribution by summing the contribu-tions of all the charge elements (that is, by applying the superposition principle).

The electric field at P due to one charge element carrying charge 0q is

0E ! ke 0qr 2 r

r

∆qr

P

∆E

Figure 23.16 The electric field atP due to a continuous charge distri-bution is the vector sum of thefields 0E due to all the elements0q of the charge distribution.

where r is the distance from the charge element to point P and r is a unit vector di-rected from the element toward P. The total electric field at P due to all elements inthe charge distribution is approximately

where the index i refers to the i th element in the distribution. Because the charge dis-tribution is modeled as continuous, the total field at P in the limit 0qi : 0 is

(23.11)

where the integration is over the entire charge distribution. This is a vector operationand must be treated appropriately.

We illustrate this type of calculation with several examples, in which we assume thecharge is uniformly distributed on a line, on a surface, or throughout a volume. Whenperforming such calculations, it is convenient to use the concept of a charge densityalong with the following notations:

• If a charge Q is uniformly distributed throughout a volume V, the volume chargedensity 1 is defined by

where 1 has units of coulombs per cubic meter (C/m3).

• If a charge Q is uniformly distributed on a surface of area A, the surface chargedensity 2 (lowercase Greek sigma) is defined by

where 2 has units of coulombs per square meter (C/m2).

• If a charge Q is uniformly distributed along a line of length !, the linear chargedensity 3 is defined by

where 3 has units of coulombs per meter (C/m).

• If the charge is nonuniformly distributed over a volume, surface, or line, theamounts of charge dq in a small volume, surface, or length element are

dq ! 1 dV dq ! 2 dA dq ! 3 d!

3 # Q!

2 # QA

1 # QV

E ! ke lim0qi : 0

$

i

0qi

ri 2 ri ! ke %

dqr 2 r

E " ke $ i

0qi

ri 2 r i


P R O B L E M - S O LV I N G H I N T S

Finding the Electric Field• Units: in calculations using the Coulomb constant ke (! 1/4()0), charges must

be expressed in coulombs and distances in meters.

• Calculating the electric field of point charges: to find the total electric field at agiven point, first calculate the electric field at the point due to each individualcharge. The resultant field at the point is the vector sum of the fields due tothe individual charges.

• Continuous charge distributions: when you are confronted with problems thatinvolve a continuous distribution of charge, the vector sums for evaluating the

Volume charge density

Surface charge density

Linear charge density

Electric field due to acontinuous charge distribution

S E C T I O N 2 3 . 5 • Electric Field of a Continuous Charge Distribution 721

total electric field at some point must be replaced by vector integrals. Dividethe charge distribution into infinitesimal pieces, and calculate the vector sumby integrating over the entire charge distribution. Examples 23.7 through 23.9demonstrate this technique.

• Symmetry: with both distributions of point charges and continuous charge dis-tributions, take advantage of any symmetry in the system to simplify yourcalculations.

Example 23.7 The Electric Field Due to a Charged Rod

is particularly simple in this case. The total field at P due to allsegments of the rod, which are at different distances from P,is given by Equation 23.11, which in this case becomes3

where the limits on the integral extend from one end of therod (x ! a) to the other (x ! ! # a). The constants ke and 3can be removed from the integral to yield

where we have used the fact that the total charge Q ! 3!.

What If? Suppose we move to a point P very far away fromthe rod. What is the nature of the electric field at such a point?

Answer If P is far from the rod (a -- !), then ! in thedenominator of the final expression for E can be neglected,and E " keQ /a2. This is just the form you would expectfor a point charge. Therefore, at large values of a/!, thecharge distribution appears to be a point charge of magni-tude Q —we are so far away from the rod that we cannot dis-tinguish that it has a size. The use of the limiting technique( ) often is a good method for checking a mathe-matical expression.a/! : 4

ke Qa(! # a)

! ke 3 & 1a

"1

! # a ' !

E ! ke 3%!#a

a

dxx 2 ! ke 3 ("

1x )a

!#a

E ! %!#a

a ke 3

dxx 2

A rod of length ! has a uniform positive charge per unitlength 3 and a total charge Q. Calculate the electric field ata point P that is located along the long axis of the rod and adistance a from one end (Fig. 23.17).

Solution Let us assume that the rod is lying along thex axis, that dx is the length of one small segment, and thatdq is the charge on that segment. Because the rod has acharge per unit length 3, the charge dq on the smallsegment is dq ! 3 dx.

The field d E at P due to this segment is in the negativex direction (because the source of the field carries a positivecharge), and its magnitude is

Because every other element also produces a field in the neg-ative x direction, the problem of summing their contributions

dE ! ke dqx

2 ! ke 3 dxx 2

3 It is important that you understand how to carry out integrations such as this. First, express thecharge element dq in terms of the other variables in the integral. (In this example, there is one vari-able, x, and so we made the change dq ! 3 dx.) The integral must be over scalar quantities; therefore,you must express the electric field in terms of components, if necessary. (In this example the field hasonly an x component, so we do not bother with this detail.) Then, reduce your expression to an inte-gral over a single variable (or to multiple integrals, each over a single variable). In examples that havespherical or cylindrical symmetry, the single variable will be a radial coordinate.

x

y

!a

Px

dxdq = dx

E

3

Figure 23.17 (Example 23.7) The electric field at P due to auniformly charged rod lying along the x axis. The magnitude ofthe field at P due to the segment of charge dq is kedq/x2. Thetotal field at P is the vector sum over all segments of the rod.

This field has an x component dEx ! dE cos + along the xaxis and a component dE⊥ perpendicular to the x axis. Aswe see in Figure 23.18b, however, the resultant field at Pmust lie along the x axis because the perpendicular com-

d E ! ke dqr 2

A ring of radius a carries a uniformly distributed positivetotal charge Q. Calculate the electric field due to the ring ata point P lying a distance x from its center along the centralaxis perpendicular to the plane of the ring (Fig. 23.18a).

Solution The magnitude of the electric field at P due tothe segment of charge dq is

Example 23.8 The Electric Field of a Uniform Ring of Charge


This result shows that the field is zero at x = 0. Does thisfinding surprise you?

What If? Suppose a negative charge is placed at the centerof the ring in Figure 23.18 and displaced slightly by adistance x !! a along the x axis. When released, what type ofmotion does it exhibit?

Answer In the expression for the field due to a ring ofcharge, we let x 55 a, which results in

Thus, from Equation 23.8, the force on a charge " q placednear the center of the ring is

Because this force has the form of Hooke’s law (Eq. 15.1),the motion will be simple harmonic !

Fx ! "ke qQ

a3 x

Ex !ke Qa3 x

ponents of all the various charge segments sum to zero.That is, the perpendicular component of the field createdby any charge element is canceled by the perpendicularcomponent created by an element on the opposite side ofthe ring. Because r ! (x 2 # a2)1/2 and cos + ! x/r, we findthat

All segments of the ring make the same contributionto the field at P because they are all equidistant fromthis point. Thus, we can integrate to obtain the total fieldat P :

ke x(x2 # a

2)3/2 Q!

Ex ! % ke x

(x 2 # a

2)3/2 dq !ke x

(x 2 # a

2)3/2 % dq

d Ex ! d E cos + ! &ke dqr 2 '

xr

!ke x

(x2 # a2)3/2 dq

(a)

+ +

+

+

++

+

++ +

++++

++

θ P dEx

dEdE⊥

x

r

dq

a

(b)

+ +

+

+

++

+

++

+

++

++ +

+

θ

dE2

1

dE1

2

Figure 23.18 (Example 23.8) A uniformly charged ring of radius a. (a) The field at Pon the x axis due to an element of charge dq. (b) The total electric field at P is alongthe x axis. The perpendicular component of the field at P due to segment 1 is canceledby the perpendicular component due to segment 2.

Example 23.9 The Electric Field of a Uniformly Charged Disk

A disk of radius R has a uniform surface charge density 2.Calculate the electric field at a point P that lies along thecentral perpendicular axis of the disk and a distance x fromthe center of the disk (Fig. 23.19).

Solution If we consider the disk as a set of concentric rings,we can use our result from Example 23.8—which gives thefield created by a ring of radius a—and sum the contribu-tions of all rings making up the disk. By symmetry, the fieldat an axial point must be along the central axis.

The ring of radius r and width dr shown in Figure 23.19has a surface area equal to 2(r dr. The charge dq on thisring is equal to the area of the ring multiplied by the surfacecharge density: dq ! 2(2r dr. Using this result in the equa-tion given for Ex in Example 23.8 (with a replaced by r), wehave for the field due to the ring

d Ex !ke x

(x2 # r 2)3/2 (2(2r dr) To obtain the total field at P, we integrate this expressionover the limits r ! 0 to r ! R, noting that x is a constant.

Px

r

R

dq

dr

Figure 23.19 (Example 23.9) A uniformly charged disk of ra-dius R. The electric field at an axial point P is directed alongthe central axis, perpendicular to the plane of the disk.

S E C T I O N 2 3 . 6 • Electric Field Lines 723

This gives

2(ke 2 &1 "x

(x 2 # R2)1/2 '!

! ke x(2 ( (x2 # r 2)"1/2

"1/2 )0

R

! ke x(2 %R

0 (x

2 # r 2)"3/2 d(r

2)

Ex ! ke x(2 %R

0

2r dr(x 2 # r 2)3/2

This result is valid for all values of x - 0. We can calculatethe field close to the disk along the axis by assuming thatR -- x ; thus, the expression in parentheses reduces to unityto give us the near-field approximation:

where )0 is the permittivity of free space. In the next chapterwe shall obtain the same result for the field created by auniformly charged infinite sheet.

Ex ! 2(ke 2 !2

2)0

23.6 Electric Field Lines

We have defined the electric field mathematically through Equation 23.7. We now explorea means of representing the electric field pictorially. A convenient way of visualizingelectric field patterns is to draw curved lines that are parallel to the electric field vector atany point in space. These lines, called electric field lines and first introduced by Faraday, arerelated to the electric field in a region of space in the following manner:

• The electric field vector E is tangent to the electric field line at each point. Theline has a direction, indicated by an arrowhead, that is the same as that of theelectric field vector.

• The number of lines per unit area through a surface perpendicular to the lines isproportional to the magnitude of the electric field in that region. Thus, the fieldlines are close together where the electric field is strong and far apart where thefield is weak.

These properties are illustrated in Figure 23.20. The density of lines throughsurface A is greater than the density of lines through surface B. Therefore, the magni-tude of the electric field is larger on surface A than on surface B. Furthermore, the factthat the lines at different locations point in different directions indicates that the fieldis nonuniform.

Is this relationship between strength of the electric field and the density of fieldlines consistent with Equation 23.9, the expression we obtained for E using Coulomb’slaw? To answer this question, consider an imaginary spherical surface of radius r con-centric with a point charge. From symmetry, we see that the magnitude of the electricfield is the same everywhere on the surface of the sphere. The number of lines N thatemerge from the charge is equal to the number that penetrate the spherical surface.Hence, the number of lines per unit area on the sphere is N/4(r 2 (where the surfacearea of the sphere is 4(r 2). Because E is proportional to the number of lines per unitarea, we see that E varies as 1/r 2; this finding is consistent with Equation 23.9.

Representative electric field lines for the field due to a single positive point chargeare shown in Figure 23.21a. This two-dimensional drawing shows only the field linesthat lie in the plane containing the point charge. The lines are actually directed radiallyoutward from the charge in all directions; thus, instead of the flat “wheel’’ of linesshown, you should picture an entire spherical distribution of lines. Because a positivetest charge placed in this field would be repelled by the positive source charge, the linesare directed radially away from the source charge. The electric field lines representingthe field due to a single negative point charge are directed toward the charge (Fig.23.21b). In either case, the lines are along the radial direction and extend all the way toinfinity. Note that the lines become closer together as they approach the charge; this in-dicates that the strength of the field increases as we move toward the source charge.

BA

Figure 23.20 Electric field linespenetrating two surfaces. The mag-nitude of the field is greater on sur-face A than on surface B.

! PITFALL PREVENTION 23.2 Electric Field Lines

are not Paths ofParticles!

Electric field lines represent thefield at various locations. Exceptin very special cases, they do notrepresent the path of a chargedparticle moving in an electricfield.


The rules for drawing electric field lines are as follows:

• The lines must begin on a positive charge and terminate on a negative charge.In the case of an excess of one type of charge, some lines will begin or end infi-nitely far away.

• The number of lines drawn leaving a positive charge or approaching a negativecharge is proportional to the magnitude of the charge.

• No two field lines can cross.

We choose the number of field lines starting from any positively charged object tobe Cq and the number of lines ending on any negatively charged object to be C !q !,where C is an arbitrary proportionality constant. Once C is chosen, the number of linesis fixed. For example, if object 1 has charge Q 1 and object 2 has charge Q 2, then theratio of number of lines is N2/N1 ! Q 2/Q 1. The electric field lines for two pointcharges of equal magnitude but opposite signs (an electric dipole) are shown in Figure23.22. Because the charges are of equal magnitude, the number of lines that begin atthe positive charge must equal the number that terminate at the negative charge. Atpoints very near the charges, the lines are nearly radial. The high density of linesbetween the charges indicates a region of strong electric field.

(a)

+ –

! PITFALL PREVENTION 23.3 Electric Field Lines

are not RealElectric field lines are not mater-ial objects. They are used only asa pictorial representation toprovide a qualitative descriptionof the electric field. Only a finitenumber of lines from eachcharge can be drawn, whichmakes it appear as if the fieldwere quantized and exists only incertain parts of space. The field,in fact, is continuous—existing atevery point. You should avoid ob-taining the wrong impressionfrom a two-dimensional drawingof field lines used to describe athree-dimensional situation.

(b)

Figure 23.22 (a) The electric field lines for two point charges of equal magnitude andopposite sign (an electric dipole). The number of lines leaving the positive chargeequals the number terminating at the negative charge. (b) The dark lines are smallpieces of thread suspended in oil, which align with the electric field of a dipole.

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(c)(a)

q

(b)

–q+ –

Figure 23.21 The electric field lines for a point charge. (a) For a positive point charge,the lines are directed radially outward. (b) For a negative point charge, the lines are di-rected radially inward. Note that the figures show only those field lines that lie in theplane of the page. (c) The dark areas are small pieces of thread suspended in oil, whichalign with the electric field produced by a small charged conductor at the center.

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S E C T I O N 2 3 . 7 • Motion of Charged Particles in a Uniform Electric Field 725

Figure 23.23 shows the electric field lines in the vicinity of two equal positive pointcharges. Again, the lines are nearly radial at points close to either charge, and thesame number of lines emerge from each charge because the charges are equal in mag-nitude. At great distances from the charges, the field is approximately equal to that ofa single point charge of magnitude 2q.

Finally, in Figure 23.24 we sketch the electric field lines associated with a positivecharge # 2q and a negative charge " q. In this case, the number of lines leaving # 2q istwice the number terminating at " q. Hence, only half of the lines that leave the posi-tive charge reach the negative charge. The remaining half terminate on a negativecharge we assume to be at infinity. At distances that are much greater than the chargeseparation, the electric field lines are equivalent to those of a single charge # q.

(a)

+ +C

A

B

(b)

Figure 23.23 (a) The electric field lines for two positive point charges. (The locationsA, B, and C are discussed in Quick Quiz 23.7.) (b) Pieces of thread suspended in oil,which align with the electric field created by two equal-magnitude positive charges.

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+2q – –q+

Active Figure 23.24 The electricfield lines for a point charge # 2qand a second point charge " q.Note that two lines leave # 2q forevery one that terminates on " q.

At the Active Figures linkat http://www.pse6.com, youcan choose the values andsigns for the two charges andobserve the electric field linesfor the configuration that youhave chosen.

Quick Quiz 23.7 Rank the magnitudes of the electric field at points A, B,and C shown in Figure 23.23a (greatest magnitude first).

Quick Quiz 23.8 Which of the following statements about electric field linesassociated with electric charges is false? (a) Electric field lines can be either straight orcurved. (b) Electric field lines can form closed loops. (c) Electric field lines begin onpositive charges and end on negative charges. (d) Electric field lines can never inter-sect with one another.

23.7 Motion of Charged Particles in a UniformElectric Field

When a particle of charge q and mass m is placed in an electric field E, the electricforce exerted on the charge is q E according to Equation 23.8. If this is the only forceexerted on the particle, it must be the net force and causes the particle to accelerateaccording to Newton’s second law. Thus,

The acceleration of the particle is therefore

(23.12)

If E is uniform (that is, constant in magnitude and direction), then the acceleration isconstant. If the particle has a positive charge, its acceleration is in the direction of the

a !qEm

Fe ! q E ! ma


electric field. If the particle has a negative charge, its acceleration is in the directionopposite the electric field.

Example 23.10 An Accelerating Positive Charge

from which we can find the kinetic energy of the charge after it has moved a distance 0x ! xf " xi :

We can also obtain this result from the work–kinetic energytheorem because the work done by the electric force is Fe 0x ! qE0x and W ! 0K.

K ! 12mvf

2 ! 12m & 2q E

m ' 0x ! q E 0x

A positive point charge q of mass m is released from rest in auniform electric field E directed along the x axis, as shownin Figure 23.25. Describe its motion.

Solution The acceleration is constant and is given byqE/m. The motion is simple linear motion along the x axis.Therefore, we can apply the equations of kinematics in onedimension (see Chapter 2):

Choosing the initial position of the charge as xi ! 0 andassigning vi ! 0 because the particle starts from rest, theposition of the particle as a function of time is

The speed of the particle is given by

The third kinematic equation gives us

v 2f ! 2axf ! & 2q E

m ' xf

vf ! at !q Em

t

x f ! 12at

2 !q E2m

t 2

vf2 ! v 2

i # 2a(xf " xi)

vf ! vi # at

xf ! xi # vit # 12at2

–

–

–

–

–

–

+

+

+

+

+

+

E

vv = 0

q

x

+ +

Figure 23.25 (Example 23.10) A positive point charge q in auniform electric field E undergoes constant acceleration in thedirection of the field.

The electric field in the region between two oppositely charged flat metallic platesis approximately uniform (Fig. 23.26). Suppose an electron of charge " e is projectedhorizontally into this field from the origin with an initial velocity vi i at time t ! 0.Because the electric field E in Figure 23.26 is in the positive y direction, the accelera-tion of the electron is in the negative y direction. That is,

(23.13)

Because the acceleration is constant, we can apply the equations of kinematics in twodimensions (see Chapter 4) with vxi ! vi and vyi ! 0. After the electron has been in the

a ! "e Eme

j

(0, 0)

!

E

–

(x, y)

–v

x

y– – – – – – – – – – – –

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

vi i

Active Figure 23.26 An electron isprojected horizontally into a uniformelectric field produced by twocharged plates. The electron under-goes a downward acceleration (oppo-site E), and its motion is parabolicwhile it is between the plates.

At the Active Figures linkat http://www.pse6.com, youcan choose the strength of theelectric field and the mass andcharge of the projectedparticle.

S E C T I O N 2 3 . 7 • Motion of Charged Particles in a Uniform Electric Field 727

electric field for a time interval, the components of its velocity at time t are

(23.14)

(23.15)

Its position coordinates at time t are

(23.16)

(23.17)

Substituting the value t ! xf/vi from Equation 23.16 into Equation 23.17, we see that yfis proportional to xf

2. Hence, the trajectory is a parabola. This should not be a surprise—consider the analogous situation of throwing a ball horizontally in a uniformgravitational field (Chapter 4). After the electron leaves the field, the electric force vanishes and the electron continues to move in a straight line in the direction of v inFigure 23.26 with a speed v - vi .

Note that we have neglected the gravitational force acting on the electron. This is agood approximation when we are dealing with atomic particles. For an electric field of104 N/C, the ratio of the magnitude of the electric force eE to the magnitude of thegravitational force mg is on the order of 1014 for an electron and on the order of 1011

for a proton.

yf ! 12ayt2 ! "1

2 e Eme

t2

xf ! vi t

vy ! ayt ! "eEme

t

vx ! vi ! constant

! PITFALL PREVENTION 23.4 Just Another ForceElectric forces and fields mayseem abstract to you. However,once Fe is evaluated, it causes aparticle to move according to ourwell-established understanding offorces and motion from Chapters5 and 6. Keeping this link withthe past in mind will help yousolve problems in this chapter.

Solution The horizontal distance across the field is ! !0.100 m. Using Equation 23.16 with xf ! !, we find that thetime at which the electron exits the electric field is

(C) If the vertical position of the electron as it enters the fieldis yi ! 0, what is its vertical position when it leaves the field?

Solution Using Equation 23.17 and the results from parts(A) and (B), we find that

If the electron enters just below the negative plate in Figure23.26 and the separation between the plates is less than thevalue we have just calculated, the electron will strike the pos-itive plate.

"1.95 cm! "0.019 5 m !

yf ! 12ayt

2 ! "12(3.51 & 1013 m/s2)(3.33 & 10"8 s)2

3.33 & 10"8 st !!

vi!

0.100 m3.00 & 106 m/s

!

An electron enters the region of a uniform electric fieldas shown in Figure 23.26, with vi ! 3.00 & 106 m/s and E ! 200 N/C. The horizontal length of the plates is! ! 0.100 m.

(A) Find the acceleration of the electron while it is in theelectric field.

Solution The charge on the electron has an absolute valueof 1.60 & 10"19 C, and me ! 9.11 & 10"31 kg. Therefore,Equation 23.13 gives

(B) If the electron enters the field at time t ! 0, find thetime at which it leaves the field.

"3.51 & 1013 j m/s2!

a ! "e Eme

j ! "(1.60 & 10"19 C)(200 N/C)

9.11 & 10"31 kg j

Example 23.11 An Accelerated Electron

At the Interactive Worked Example link at http://www.pse6.com, you can predict the required initial velocity for the exitingelectron to just miss the right edge of the lower plate, for random values of the electric field.

The Cathode Ray Tube

The example we just worked describes a portion of a cathode ray tube (CRT). Thistube, illustrated in Figure 23.27, is commonly used to obtain a visual display of elec-tronic information in oscilloscopes, radar systems, television receivers, and computermonitors. The CRT is a vacuum tube in which a beam of electrons is acceleratedand deflected under the influence of electric or magnetic fields. The electron beam is

Interactive


produced by an assembly called an electron gun located in the neck of the tube. Theseelectrons, if left undisturbed, travel in a straight-line path until they strike the front ofthe CRT, the “screen,’’ which is coated with a material that emits visible light whenbombarded with electrons.

In an oscilloscope, the electrons are deflected in various directions by two sets ofplates placed at right angles to each other in the neck of the tube. (A television CRTsteers the beam with a magnetic field, as discussed in Chapter 29.) An external electriccircuit is used to control the amount of charge present on the plates. The placing ofpositive charge on one horizontal plate and negative charge on the other creates anelectric field between the plates and allows the beam to be steered from side to side.The vertical deflection plates act in the same way, except that changing the charge onthem deflects the beam vertically.

Electrongun

Verticaldeflection

plates

Horizontaldeflection

plates

Electronbeam

Fluorescentscreen

Horizontalinput

Verticalinput

C AFigure 23.27 Schematic diagram of acathode ray tube. Electrons leaving thecathode C are accelerated to the anode A.In addition to accelerating electrons, theelectron gun is also used to focus the beamof electrons, and the plates deflect thebeam.

Electric charges have the following important properties:

• Charges of opposite sign attract one another and charges of the same sign repelone another.

• Total charge in an isolated system is conserved.

• Charge is quantized.

Conductors are materials in which electrons move freely. Insulators are materialsin which electrons do not move freely.

Coulomb’s law states that the electric force exerted by a charge q1 on a secondcharge q2 is

(23.6)

where r is the distance between the two charges and r is a unit vector directed from q1toward q2. The constant ke , which is called the Coulomb constant, has the valueke ! 8.99 & 109 N ' m2/C2.

The smallest unit of free charge e known to exist in nature is the charge on anelectron (" e) or proton (# e), where e ! 1.602 19 & 10"19 C.

The electric field E at some point in space is defined as the electric force Fe thatacts on a small positive test charge placed at that point divided by the magnitude q0 ofthe test charge:

(23.7)

Thus, the electric force on a charge q placed in an electric field E is given by

(23.8)Fe ! qE

E # Fe

q 0

F12 ! ke q 1q 2

r 2 r

S U M M A RY

Take a practice test forthis chapter by clicking onthe Practice Test link athttp://www.pse6.com.

Questions 729

At a distance r from a point charge q, the electric field due to the charge is given by

(23.9)

where r is a unit vector directed from the charge toward the point in question. Theelectric field is directed radially outward from a positive charge and radially inwardtoward a negative charge.

The electric field due to a group of point charges can be obtained by using the su-perposition principle. That is, the total electric field at some point equals the vectorsum of the electric fields of all the charges:

(23.10)

The electric field at some point due to a continuous charge distribution is

(23.11)

where dq is the charge on one element of the charge distribution and r is the distancefrom the element to the point in question.

Electric field lines describe an electric field in any region of space. The numberof lines per unit area through a surface perpendicular to the lines is proportional tothe magnitude of E in that region.

A charged particle of mass m and charge q moving in an electric field E has anacceleration

(23.12)a !q Em

E ! ke % dqr

2 r

E ! ke $ i

qi

r i

2 ri

E ! ke q

r 2 r

1. Explain what is meant by the term “a neutral atom.’’Explain what “a negatively charged atom’’ means.

2. A charged comb often attracts small bits of dry paper thatthen fly away when they touch the comb. Explain.

3. Sparks are often seen or heard on a dry day when fabricsare removed from a clothes dryer in dim light. Explain.

4. Hospital personnel must wear special conducting shoeswhile working around oxygen in an operating room. Why?Contrast with what might happen if people wore rubber-soled shoes.

5. Explain from an atomic viewpoint why charge is usuallytransferred by electrons.

6. A light, uncharged metallic sphere suspended from athread is attracted to a charged rubber rod. After ittouches the rod, the sphere is repelled by the rod. Explain.

7. A foreign student who grew up in a tropical country but isstudying in the United States may have had no experiencewith static electricity sparks or shocks until he or she firstexperiences an American winter. Explain.

8. Explain the similarities and differences between Newton’slaw of universal gravitation and Coulomb’s law.

A balloon is negatively charged by rubbing and then clingsto a wall. Does this mean that the wall is positivelycharged? Why does the balloon eventually fall?

9.

10. A light strip of aluminum foil is draped over a horizontalwooden pencil. When a rod carrying a positive charge isbrought close to the foil, the two parts of the foil standapart. Why? What kind of charge is on the foil?

11. When defining the electric field, why is it necessary tospecify that the magnitude of the test charge be very small?

12. How could you experimentally distinguish an electric fieldfrom a gravitational field?

13. A large metallic sphere insulated from ground is chargedwith an electrostatic generator while a student standing onan insulating stool holds the sphere. Why is it safe to dothis? Why would it not be safe for another person to touchthe sphere after it had been charged?

14. Is it possible for an electric field to exist in empty space?Explain. Consider point A in Figure 23.23(a). Does chargeexist at this point? Does a force exist at this point? Does afield exist at this point?

15. When is it valid to approximate a charge distribution by apoint charge?

16. Explain why electric field lines never cross. Suggestion:Begin by explaining why the electric field at a particularpoint must have only one direction.

17. Figures 23.14 and 23.15 show three electric field vectorsat the same point. With a little extrapolation, Figure

Q U E S T I O N S


23.21 would show many electric field lines at thesame point. Is it really true that “no two field lines cancross’’? Are the diagrams drawn correctly? Explain youranswers.

18. A free electron and a free proton are released in identicalelectric fields. Compare the electric forces on the two par-ticles. Compare their accelerations.

19. Explain what happens to the magnitude of the electricfield created by a point charge as r approaches zero.

20. An object with negative charge is placed in a region ofspace where the electric field is directed vertically upward.What is the direction of the electric force exerted on thischarge?

21. A charge 4q is at a distance r from a charge " q. Comparethe number of electric field lines leaving the charge 4qwith the number entering the charge " q. Where do theextra lines beginning on 4q end?

22. Consider two equal point charges separated by somedistance d. At what point (other than 4) would a third testcharge experience no net force?

23. Explain the differences between linear, surface, andvolume charge densities, and give examples of when eachwould be used.

24. If the electron in Figure 23.26 is projected into the electricfield with an arbitrary velocity vi (at an arbitrary angle toE), will its trajectory still be parabolic? Explain.

Would life be different if the electron were positivelycharged and the proton were negatively charged? Does thechoice of signs have any bearing on physical and chemicalinteractions? Explain.

26. Why should a ground wire be connected to the metalsupport rod for a television antenna?

27. Suppose someone proposes the idea that people arebound to the Earth by electric forces rather than by grav-ity. How could you prove this idea is wrong?

28. Consider two electric dipoles in empty space. Each dipolehas zero net charge. Does an electric force exist betweenthe dipoles—that is, can two objects with zero net chargeexert electric forces on each other? If so, is the force oneof attraction or of repulsion?

25.

Section 23.1 Properties of Electric Charges1. (a) Find to three significant digits the charge and the mass

of an ionized hydrogen atom, represented as H#. Sugges-tion: Begin by looking up the mass of a neutral atom onthe periodic table of the elements. (b) Find the chargeand the mass of Na#, a singly ionized sodium atom.(c) Find the charge and the average mass of a chloride ionCl" that joins with the Na# to make one molecule of tablesalt. (d) Find the charge and the mass of Ca## ! Ca2#, adoubly ionized calcium atom. (e) You can model thecenter of an ammonia molecule as an N3" ion. Find itscharge and mass. (f) The plasma in a hot star containsquadruply ionized nitrogen atoms, N4#. Find their chargeand mass. (g) Find the charge and the mass of the nucleusof a nitrogen atom. (h) Find the charge and the mass ofthe molecular ion H2O".

2. (a) Calculate the number of electrons in a small, electri-cally neutral silver pin that has a mass of 10.0 g. Silver has47 electrons per atom, and its molar mass is 107.87 g/mol.(b) Electrons are added to the pin until the net negativecharge is 1.00 mC. How many electrons are added forevery 109 electrons already present?

Section 23.2 Charging Objects by InductionSection 23.3 Coulomb’s Law

The Nobel laureate Richard Feynman once said thatif two persons stood at arm’s length from each other andeach person had 1% more electrons than protons, the

3.

force of repulsion between them would be enough to lifta “weight’’ equal to that of the entire Earth. Carry out anorder-of-magnitude calculation to substantiate thisassertion.

4. Two protons in an atomic nucleus are typically separatedby a distance of 2 & 10"15 m. The electric repulsion forcebetween the protons is huge, but the attractive nuclearforce is even stronger and keeps the nucleus from burstingapart. What is the magnitude of the electric force betweentwo protons separated by 2.00 & 10"15 m?

5. (a) Two protons in a molecule are separated by 3.80 &10"10 m. Find the electric force exerted by one proton onthe other. (b) How does the magnitude of this force com-pare to the magnitude of the gravitational force betweenthe two protons? (c) What If? What must be the charge-to-mass ratio of a particle if the magnitude of the gravita-tional force between two of these particles equals themagnitude of electric force between them?

6. Two small silver spheres, each with a mass of 10.0 g, areseparated by 1.00 m. Calculate the fraction of the electronsin one sphere that must be transferred to the other inorder to produce an attractive force of 1.00 & 104 N(about 1 ton) between the spheres. (The number ofelectrons per atom of silver is 47, and the number of atomsper gram is Avogadro’s number divided by the molar massof silver, 107.87 g/mol.)

Three point charges are located at the corners of an equi-lateral triangle as shown in Figure P23.7. Calculate the re-sultant electric force on the 7.00-*C charge.

7.

1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide

= coached solution with hints available at http://www.pse6.com = computer useful in solving problem

= paired numerical and symbolic problems

P R O B L E M S

Problems 731

8. Suppose that 1.00 g of hydrogen is separated into elec-trons and protons. Suppose also that the protons areplaced at the Earth’s north pole and the electrons areplaced at the south pole. What is the resulting compres-sional force on the Earth?

9. Two identical conducting small spheres are placed with theircenters 0.300 m apart. One is given a charge of 12.0 nCand the other a charge of " 18.0 nC. (a) Find the electricforce exerted by one sphere on the other. (b) What If?The spheres are connected by a conducting wire. Findthe electric force between the two after they have come toequilibrium.

10. Two small beads having positive charges 3q and q are fixedat the opposite ends of a horizontal, insulating rod, ex-tending from the origin to the point x ! d. As shown inFigure P23.10, a third small charged bead is free to slideon the rod. At what position is the third bead in equilib-rium? Can it be in stable equilibrium?

Section 23.4 The Electric FieldWhat are the magnitude and direction of the electric fieldthat will balance the weight of (a) an electron and (b) aproton? (Use the data in Table 23.1.)

14. An object having a net charge of 24.0 *C is placed in a uni-form electric field of 610 N/C directed vertically. What isthe mass of this object if it “floats’’ in the field?

15. In Figure P23.15, determine the point (other than infin-ity) at which the electric field is zero.

13.

Figure P23.7 Problems 7 and 18.

0.50 m

7.00 C

–4.00 C

60°x

y

+

+ –

µ

µ2.00 Cµ

d

+3q +q

Figure P23.10

11. Review problem. In the Bohr theory of the hydrogenatom, an electron moves in a circular orbit about a proton,where the radius of the orbit is 0.529 & 10"10 m. (a) Findthe electric force between the two. (b) If this force causesthe centripetal acceleration of the electron, what is thespeed of the electron?

12. Review problem. Two identical particles, each havingcharge # q, are fixed in space and separated by a distanced. A third point charge " Q is free to move and lies initiallyat rest on the perpendicular bisector of the two fixedcharges a distance x from the midpoint between the twofixed charges (Fig. P23.12). (a) Show that if x is small com-pared with d, the motion of " Q will be simple harmonicalong the perpendicular bisector. Determine the period ofthat motion. (b) How fast will the charge " Q be movingwhen it is at the midpoint between the two fixed charges,if initially it is released at a distance a 55 d from themidpoint?

Figure P23.12

+q

+q

–Qx

y

d/2

d/2x

1.00 m

–2.50 µC 6.00 µCµ µFigure P23.15

16. An airplane is flying through a thundercloud at a height of2 000 m. (This is a very dangerous thing to do because ofupdrafts, turbulence, and the possibility of electric dis-charge.) If a charge concentration of # 40.0 C is above theplane at a height of 3 000 m within the cloud and a chargeconcentration of " 40.0 C is at height 1 000 m, what is theelectric field at the aircraft?

17. Two point charges are located on the x axis. The first isa charge # Q at x ! " a. The second is an unknowncharge located at x ! # 3a. The net electric field thesecharges produce at the origin has a magnitude of2keQ /a2. What are the two possible values of theunknown charge?

18. Three charges are at the corners of an equilateral triangleas shown in Figure P23.7. (a) Calculate the electric field atthe position of the 2.00-*C charge due to the 7.00-*C and" 4.00-*C charges. (b) Use your answer to part (a) to de-termine the force on the 2.00-*C charge.

19. Three point charges are arranged as shown in FigureP23.19. (a) Find the vector electric field that the 6.00-nCand " 3.00-nC charges together create at the origin.(b) Find the vector force on the 5.00-nC charge.


20. Two 2.00-*C point charges are located on the x axis. Oneis at x ! 1.00 m, and the other is at x ! " 1.00 m. (a) De-termine the electric field on the y axis at y ! 0.500 m.(b) Calculate the electric force on a " 3.00-*C chargeplaced on the y axis at y ! 0.500 m.

21. Four point charges are at the corners of a square of side aas shown in Figure P23.21. (a) Determine the magnitudeand direction of the electric field at the location of chargeq. (b) What is the resultant force on q?

24. Consider an infinite number of identical charges (each ofcharge q) placed along the x axis at distances a, 2a, 3a,4a, . . . , from the origin. What is the electric field at theorigin due to this distribution? Suggestion: Use the fact that

Section 23.5 Electric Field of a Continuous Charge Distribution

25. A rod 14.0 cm long is uniformly charged and has a totalcharge of " 22.0 *C. Determine the magnitude and direc-tion of the electric field along the axis of the rod at a point36.0 cm from its center.

26. A continuous line of charge lies along the x axis, extend-ing from x ! # x 0 to positive infinity. The line carriescharge with a uniform linear charge density 30. What arethe magnitude and direction of the electric field at theorigin?

A uniformly charged ring of radius 10.0 cm has a totalcharge of 75.0 *C. Find the electric field on the axis ofthe ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and(d) 100 cm from the center of the ring.

28. A line of charge starts at x ! # x0 and extends to positiveinfinity. The linear charge density is 3 ! 30x0/x. Deter-mine the electric field at the origin.

29. Show that the maximum magnitude Emax of the electricfield along the axis of a uniformly charged ring occurs at

(see Fig. 23.18) and has the value

30. A uniformly charged disk of radius 35.0 cm carries chargewith a density of 7.90 & 10" 3 C/m2. Calculate the electricfield on the axis of the disk at (a) 5.00 cm, (b) 10.0 cm,(c) 50.0 cm, and (d) 200 cm from the center of the disk.

31. Example 23.9 derives the exact expression for the electricfield at a point on the axis of a uniformly charged disk.Consider a disk, of radius R ! 3.00 cm, having a uniformlydistributed charge of # 5.20 *C. (a) Using the result of Ex-ample 23.9, compute the electric field at a point on theaxis and 3.00 mm from the center. What If? Compare thisanswer with the field computed from the near-field ap-proximation E ! 2/2)0. (b) Using the result of Example23.9, compute the electric field at a point on the axis and30.0 cm from the center of the disk. What If? Comparethis with the electric field obtained by treating the disk as a# 5.20-*C point charge at a distance of 30.0 cm.

32. The electric field along the axis of a uniformly chargeddisk of radius R and total charge Q was calculated in Exam-ple 23.9. Show that the electric field at distances x that arelarge compared with R approaches that of a point chargeQ ! 2(R2. (Suggestion: First show that x/(x2 # R2)1/2 !(1 # R2/x2)"1/2 and use the binomial expansion (1 # 6)n " 1 # n6 when 6 55 1.)

A uniformly charged insulating rod of length 14.0 cmis bent into the shape of a semicircle as shown in FigureP23.33. The rod has a total charge of " 7.50 *C. Find themagnitude and direction of the electric field at O, thecenter of the semicircle.

33.

Q /(6√3()0a 2).x ! a/√2

27.

1 #122 #

132 #

142 # ' ' ' !

(2

6

Figure P23.21

Figure P23.22

a a

a

a

q

3q 4q

2q

2a

x–q q

y

Figure P23.19

0.100 m

x

–3.00 nC

5.00 nC 0.300 m 6.00 nC

y

22. Consider the electric dipole shown in Figure P23.22. Showthat the electric field at a distant point on the # x axis isEx " 4ke qa/x 3.

23. Consider n equal positive point charges each of magnitudeQ /n placed symmetrically around a circle of radius R .(a) Calculate the magnitude of the electric field at a pointa distance x on the line passing through the center of thecircle and perpendicular to the plane of the circle. (b) Explain why this result is identical to that of the calcu-lation done in Example 23.8.

Problems 733

34. (a) Consider a uniformly charged thin-walled right circu-lar cylindrical shell having total charge Q , radius R , andheight h. Determine the electric field at a point a distanced from the right side of the cylinder as shown in FigureP23.34. (Suggestion: Use the result of Example 23.8 andtreat the cylinder as a collection of ring charges.) (b) WhatIf? Consider now a solid cylinder with the same dimen-sions and carrying the same charge, uniformly distributedthrough its volume. Use the result of Example 23.9 to findthe field it creates at the same point.

36. Three solid plastic cylinders all have radius 2.50 cm andlength 6.00 cm. One (a) carries charge with uniform den-sity 15.0 nC/m2 everywhere on its surface. Another(b) carries charge with the same uniform density on itscurved lateral surface only. The third (c) carries chargewith uniform density 500 nC/m3 throughout the plastic.Find the charge of each cylinder.

37. Eight solid plastic cubes, each 3.00 cm on each edge, areglued together to form each one of the objects (i, ii, iii, andiv) shown in Figure P23.37. (a) Assuming each object carriescharge with uniform density 400 nC/m3 throughout itsvolume, find the charge of each object. (b) Assuming eachobject carries charge with uniform density 15.0 nC/m2

everywhere on its exposed surface, find the charge on eachobject. (c) Assuming charge is placed only on the edgeswhere perpendicular surfaces meet, with uniform density80.0 pC/m, find the charge of each object.

Figure P23.33

Figure P23.34

Figure P23.35

O

Rd

dx

h

y

y

dxx

P

O!

θθ0θ

A thin rod of length ! and uniform charge per unit length3 lies along the x axis, as shown in Figure P23.35. (a) Showthat the electric field at P, a distance y from the rod alongits perpendicular bisector, has no x component and isgiven by E ! 2ke 3 sin +0/y. (b) What If? Using your resultto part (a), show that the field of a rod of infinite length isE ! 2ke 3/y. (Suggestion: First calculate the field at P due toan element of length dx, which has a charge 3 dx. Thenchange variables from x to +, using the relationships x !y tan + and dx ! y sec2 + d+, and integrate over +.)

35.

Figure P23.37

(i) (ii) (iii) (iv)

Section 23.6 Electric Field Lines38. A positively charged disk has a uniform charge per unit

area as described in Example 23.9. Sketch the electric fieldlines in a plane perpendicular to the plane of the diskpassing through its center.

A negatively charged rod of finite length carries chargewith a uniform charge per unit length. Sketch the electricfield lines in a plane containing the rod.

40. Figure P23.40 shows the electric field lines for two pointcharges separated by a small distance. (a) Determine theratio q1/q2. (b) What are the signs of q1 and q2?

39.

q2

q1

Figure P23.40


Three equal positive charges q are at the corners of anequilateral triangle of side a as shown in Figure P23.41.(a) Assume that the three charges together create an electricfield. Sketch the field lines in the plane of the charges. Findthe location of a point (other than 4) where the electricfield is zero. (b) What are the magnitude and direction ofthe electric field at P due to the two charges at the base?

41. and with a positive charge of 1.00 & 10"6 C leaves thecenter of the bottom negative plate with an initial speed of1.00 & 105 m/s at an angle of 37.0° above the horizontal.Describe the trajectory of the particle. Which plate does itstrike? Where does it strike, relative to its starting point?

49. Protons are projected with an initial speed vi ! 9.55 &103 m/s into a region where a uniform electric fieldE ! " 720 j N/C is present, as shown in Figure P23.49. Theprotons are to hit a target that lies at a horizontal distanceof 1.27 mm from the point where the protons cross theplane and enter the electric field in Figure P23.49. Find(a) the two projection angles + that will result in a hit and(b) the total time of flight (the time interval during whichthe proton is above the plane in Figure P23.49) for eachtrajectory.

Figure P23.52

qqa

q

a a

P +

+ +

Figure P23.41

Figure P23.49

ˆ

θvi

1.27 mm

Target

E = (–720 j) N/C

×

Protonbeam

0.800 m

y

3.00 nC5.00 nC

0.500 m

– 4.00 nCx

Section 23.7 Motion of Charged Particles in a Uniform Electric Field

42. An electron and a proton are each placed at rest in anelectric field of 520 N/C. Calculate the speed of each par-ticle 48.0 ns after being released.

A proton accelerates from rest in a uniform electric fieldof 640 N/C. At some later time, its speed is 1.20 & 106 m/s(nonrelativistic, because v is much less than the speedof light). (a) Find the acceleration of the proton. (b) Howlong does it take the proton to reach this speed? (c) Howfar has it moved in this time? (d) What is its kinetic energyat this time?

44. A proton is projected in the positive x direction into aregion of a uniform electric field E ! " 6.00 & 105i N/Cat t ! 0. The proton travels 7.00 cm before coming to rest.Determine (a) the acceleration of the proton, (b) its initialspeed, and (c) the time at which the proton comes to rest.

The electrons in a particle beam each have a kineticenergy K. What are the magnitude and direction of theelectric field that will stop these electrons in a distance d ?

46. A positively charged bead having a mass of 1.00 g fallsfrom rest in a vacuum from a height of 5.00 m in a uni-form vertical electric field with a magnitude of 1.00 &104 N/C. The bead hits the ground at a speed of 21.0 m/s.Determine (a) the direction of the electric field (up ordown), and (b) the charge on the bead.

A proton moves at 4.50 & 105 m/s in the horizontal direc-tion. It enters a uniform vertical electric field with a magni-tude of 9.60 & 103 N/C. Ignoring any gravitational effects,find (a) the time interval required for the proton to travel5.00 cm horizontally, (b) its vertical displacement duringthe time interval in which it travels 5.00 cm horizontally,and (c) the horizontal and vertical components of its ve-locity after it has traveled 5.00 cm horizontally.

48. Two horizontal metal plates, each 100 mm square, arealigned 10.0 mm apart, with one above the other. They aregiven equal-magnitude charges of opposite sign so that auniform downward electric field of 2 000 N/C exists in theregion between them. A particle of mass 2.00 & 10"16 kg

47.

45.

43.

Additional Problems

50. Two known charges, " 12.0 *C and 45.0 *C, and anunknown charge are located on the x axis. The charge" 12.0 *C is at the origin, and the charge 45.0 *C is at x !15.0 cm. The unknown charge is to be placed so that eachcharge is in equilibrium under the action of the electricforces exerted by the other two charges. Is this situation pos-sible? Is it possible in more than one way? Find the requiredlocation, magnitude, and sign of the unknown charge.

51. A uniform electric field of magnitude 640 N/C existsbetween two parallel plates that are 4.00 cm apart. A protonis released from the positive plate at the same instant thatan electron is released from the negative plate. (a) Deter-mine the distance from the positive plate at which the twopass each other. (Ignore the electrical attraction betweenthe proton and electron.) (b) What If? Repeat part (a) for asodium ion (Na#) and a chloride ion (Cl").

52. Three point charges are aligned along the x axis as shownin Figure P23.52. Find the electric field at (a) the position(2.00, 0) and (b) the position (0, 2.00).

Problems 735

53. A researcher studying the properties of ions in the upperatmosphere wishes to construct an apparatus with the fol-lowing characteristics: Using an electric field, a beamof ions, each having charge q, mass m, and initial velocityv i, is turned through an angle of 90° as each ion under-goes displacement R i # R j. The ions enter a chamber asshown in Figure P23.53, and leave through the exit portwith the same speed they had when they entered thechamber. The electric field acting on the ions is to haveconstant magnitude. (a) Suppose the electric field is pro-duced by two concentric cylindrical electrodes not shownin the diagram, and hence is radial. What magnitudeshould the field have? What If? (b) If the field is producedby two flat plates and is uniform in direction, what valueshould the field have in this case?

the angle +. Find (a) the charge on the ball and (b) thetension in the string.

Figure P23.53

Figure P23.54

+

R

R

y

x

v

v

y

x

15.0°

20.0 cm

m = 2.00 g

E = 1.00 × 103i N/Cˆ

x

y

E

q

θ


Figure P23.58

54. A small, 2.00-g plastic ball is suspended by a 20.0-cm-longstring in a uniform electric field as shown in FigureP23.54. If the ball is in equilibrium when the string makesa 15.0° angle with the vertical, what is the net charge onthe ball?

q q

qq

y

xL

W

Figure P23.57

58. Inez is putting up decorations for her sister’s quinceañera(fifteenth birthday party). She ties three light silk ribbonstogether to the top of a gateway and hangs a rubber bal-loon from each ribbon (Fig. P23.58). To include the

57. Four identical point charges (q ! # 10.0 *C) are locatedon the corners of a rectangle as shown in Figure P23.57.The dimensions of the rectangle are L ! 60.0 cm andW ! 15.0 cm. Calculate the magnitude and direction ofthe resultant electric force exerted on the charge at thelower left corner by the other three charges.

A charged cork ball of mass 1.00 g is suspended on alight string in the presence of a uniform electric fieldas shown in Figure P23.55. When E ! (3.00i # 5.00j) &105 N/C, the ball is in equilibrium at + ! 37.0°. Find(a) the charge on the ball and (b) the tension in the string.A charged cork ball of mass m is suspended on a lightstring in the presence of a uniform electric field as shownin Figure P23.55. When E ! (Ai # B j) N/C, where Aand B are positive numbers, the ball is in equilibrium at

56.

55.


effects of the gravitational and buoyant forces on it, eachballoon can be modeled as a particle of mass 2.00 g, withits center 50.0 cm from the point of support. To show offthe colors of the balloons, Inez rubs the whole surface ofeach balloon with her woolen scarf, to make them hangseparately with gaps between them. The centers of thehanging balloons form a horizontal equilateral trianglewith sides 30.0 cm long. What is the common charge eachballoon carries?

59. Review problem. Two identical metallic blocks resting on africtionless horizontal surface are connected by a lightmetallic spring having a spring constant k as shown in Figure P23.59a and an unstretched length Li . A totalcharge Q is slowly placed on the system, causing the springto stretch to an equilibrium length L , as shown in FigureP23.59b. Determine the value of Q , assuming that all thecharge resides on the blocks and modeling the blocks aspoint charges.

62. Two small spheres, each of mass 2.00 g, are suspended bylight strings 10.0 cm in length (Fig. P23.62). A uniformelectric field is applied in the x direction. The sphereshave charges equal to " 5.00 & 10"8 C and # 5.00 &10"8 C. Determine the electric field that enables thespheres to be in equilibrium at an angle + ! 10.0°.

Figure P23.59

(a)

(b)

m mk

m mk

60. Consider a regular polygon with 29 sides. The distancefrom the center to each vertex is a. Identical charges q areplaced at 28 vertices of the polygon. A single charge Q isplaced at the center of the polygon. What is the magnitudeand direction of the force experienced by the chargeQ ? (Suggestion: You may use the result of Problem 63 inChapter 3.)

61. Identical thin rods of length 2a carry equal charges # Quniformly distributed along their lengths. The rods liealong the x axis with their centers separated by a distanceb - 2a (Fig. P23.61). Show that the magnitude of the forceexerted by the left rod on the right one is given by

F ! & keQ 2

4a 2 ' ln & b2

b2 " 4a 2 '

b

y

a–a b – a b + ax

Figure P23.61

Figure P23.62

θ

E

θ

– +

63. A line of positive charge is formed into a semicircle of ra-dius R ! 60.0 cm as shown in Figure P23.63. The chargeper unit length along the semicircle is described by the ex-pression 3 ! 30 cos +. The total charge on the semicircle is12.0 *C. Calculate the total force on a charge of 3.00 *Cplaced at the center of curvature.

Figure P23.63

y

R

x

θ

64. Three charges of equal magnitude q are fixed in posi-tion at the vertices of an equilateral triangle (Fig. P23.64).A fourth charge Q is free to move along the positive x axis

x+Q

+q

+q

–q

a2–

a2–

s

ra

θ

y

a 32√

Figure P23.64

Problems 737

under the influence of the forces exerted by the threefixed charges. Find a value for s for which Q is in equilib-rium. You will need to solve a transcendental equation.

Two small spheres of mass m are suspended from strings oflength ! that are connected at a common point. Onesphere has charge Q ; the other has charge 2Q. The stringsmake angles +1 and +2 with the vertical. (a) How are +1 and+2 related? (b) Assume +1 and +2 are small. Show that thedistance r between the spheres is given by

66. Review problem. Four identical particles, each havingcharge # q, are fixed at the corners of a square of side L. Afifth point charge " Q lies a distance z along the line per-pendicular to the plane of the square and passing throughthe center of the square (Fig. P23.66). (a) Show that theforce exerted by the other four charges on " Q is

Note that this force is directed toward the center of thesquare whether z is positive (" Q above the square) or neg-ative (" Q below the square). (b) If z is small comparedwith L, the above expression reduces to F ≈ "(constant)z k.Why does this imply that the motion of the charge " Q issimple harmonic, and what is the period of this motion ifthe mass of " Q is m?

F ! "4k e q Qz

[z2 # (L2/2)]3/2 k

r " & 4ke Q 2!

mg '1/3

65.

69. Eight point charges, each of magnitude q, are located onthe corners of a cube of edge s, as shown in Figure P23.69.(a) Determine the x, y, and z components of the resultantforce exerted by the other charges on the charge locatedat point A. (b) What are the magnitude and direction ofthis resultant force?

R

R

m

R

m

Figure P23.68


PointA

x

y

z

q

q q

q

qq

q

q

s

s

s

L

L

+q +q

z

–Q

z

+q +q

Figure P23.66

67. Review problem. A 1.00-g cork ball with charge 2.00 *C issuspended vertically on a 0.500-m-long light string in thepresence of a uniform, downward-directed electric field ofmagnitude E ! 1.00 & 105 N/C. If the ball is displacedslightly from the vertical, it oscillates like a simple pendu-lum. (a) Determine the period of this oscillation.(b) Should gravity be included in the calculation for part(a)? Explain.

68. Two identical beads each have a mass m and charge q.When placed in a hemispherical bowl of radius R with fric-tionless, nonconducting walls, the beads move, and atequilibrium they are a distance R apart (Fig. P23.68). De-termine the charge on each bead.

70. Consider the charge distribution shown in Figure P23.69.(a) Show that the magnitude of the electric field at thecenter of any face of the cube has a value of 2.18keq/s2.(b) What is the direction of the electric field at the centerof the top face of the cube?

Review problem. A negatively charged particle " q isplaced at the center of a uniformly charged ring, wherethe ring has a total positive charge Q as shown in Example23.8. The particle, confined to move along the x axis, isdisplaced a small distance x along the axis (where x 55 a)and released. Show that the particle oscillates in simpleharmonic motion with a frequency given by

72. A line of charge with uniform density 35.0 nC/m liesalong the line y ! " 15.0 cm, between the points with co-ordinates x ! 0 and x ! 40.0 cm. Find the electric field itcreates at the origin.

73. Review problem. An electric dipole in a uniform electricfield is displaced slightly from its equilibrium position, asshown in Figure P23.73, where + is small. The separationof the charges is 2a, and the moment of inertia of thedipole is I. Assuming the dipole is released from this

f !1

2( & ke q Q

ma3 '1/2

71.


position, show that its angular orientation exhibits simpleharmonic motion with a frequency

f !1

2( √ 2qa E

I

cannot determine from this information, however,whether the charges are positive or negative.

23.3 (e). In the first experiment, objects A and B may havecharges with opposite signs, or one of the objects may beneutral. The second experiment shows that B and C havecharges with the same signs, so that B must be charged.But we still do not know if A is charged or neutral.

23.4 (e). From Newton’s third law, the electric force exerted byobject B on object A is equal in magnitude to the force ex-erted by object A on object B.

23.5 (b). From Newton’s third law, the electric force exerted byobject B on object A is equal in magnitude to the force ex-erted by object A on object B and in the opposite direction.

23.6 (a). There is no effect on the electric field if we assumethat the source charge producing the field is not dis-turbed by our actions. Remember that the electric field iscreated by source charge(s) (unseen in this case), not thetest charge(s).

23.7 A, B, C. The field is greatest at point A because this iswhere the field lines are closest together. The absence oflines near point C indicates that the electric field there iszero.

23.8 (b). Electric field lines begin and end on charges andcannot close on themselves to form loops.

Figure P23.73

Eθq+

––q

2a

Answers to Quick Quizzes23.1 (b). The amount of charge present in the isolated system

after rubbing is the same as that before because charge isconserved; it is just distributed differently.

23.2 (a), (c), and (e). The experiment shows that A and Bhave charges of the same sign, as do objects B and C.Thus, all three objects have charges of the same sign. We

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