23

CHAPTER OUTLINE23.1 Properties of Electric Charges23.2 Charging Objects by Induction23.3 Coulomb’s Law23.4 The Electric Field23.5 Electric Field of a Continuous Charge Distribution23.6 Electric Field Lines23.7 Motion of Charged Particles in a Uniform Electric Field

Electric Fields

ANSWERS TO QUESTIONS

Q23.1 A neutral atom is one that has no net charge. This means that ithas the same number of electrons orbiting the nucleus as it hasprotons in the nucleus. A negatively charged atom has one ormore excess electrons.

Q23.2 When the comb is nearby, molecules in the paper are polarized,similar to the molecules in the wall in Figure 23.5a, and thepaper is attracted. During contact, charge from the comb istransferred to the paper by conduction. Then the paper has thesame charge as the comb, and is repelled.

Q23.3 The clothes dryer rubs dissimilar materials together as ittumbles the clothes. Electrons are transferred from one kind ofmolecule to another. The charges on pieces of cloth, or onnearby objects charged by induction, can produce strongelectric fields that promote the ionization process in thesurrounding air that is necessary for a spark to occur. Then youhear or see the sparks.

Q23.4 To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and coulddischarge with a spark, possibly causing an explosion of any flammable material in the oxygen-enriched atmosphere.

Q23.5 Electrons are less massive and more mobile than protons. Also, they are more easily detached fromatoms than protons.

Q23.6 The electric field due to the charged rod induces charges on near and far sides of the sphere. Theattractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is largerthan the repulsive Coulomb force of the rod on the like charge on the far side of the sphere. Theresult is a net attraction of the sphere to the rod. When the sphere touches the rod, charge isconducted between the rod and the sphere, leaving both the rod and the sphere like-charged. Thisresults in a repulsive Coulomb force.

Q23.7 All of the constituents of air are nonpolar except for water. The polar water molecules in the air quitereadily “steal” charge from a charged object, as any physics teacher trying to perform electrostaticsdemonstrations in the summer well knows. As a result—it is difficult to accumulate large amounts ofexcess charge on an object in a humid climate. During a North American winter, the cold, dry airallows accumulation of significant excess charge, giving the potential (pun intended) for a shocking(pun also intended) introduction to static electricity sparks.

1

2 Electric Fields

Q23.8 Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) oftwo particles, and inversely proportional to the square of the separation distance. An electrical forceexhibits the same proportionalities, with charge as the intrinsic property.

Differences: The electrical force can either attract or repel, while the gravitational force asdescribed by Newton’s law can only attract. The electrical force between elementary particles isvastly stronger than the gravitational force.

Q23.9 No. The balloon induces polarization of the molecules in the wall, so that a layer of positive chargeexists near the balloon. This is just like the situation in Figure 23.5a, except that the signs of thecharges are reversed. The attraction between these charges and the negative charges on the balloonis stronger than the repulsion between the negative charges on the balloon and the negative chargesin the polarized molecules (because they are farther from the balloon), so that there is a net attractiveforce toward the wall. Ionization processes in the air surrounding the balloon provide ions to whichexcess electrons in the balloon can transfer, reducing the charge on the balloon and eventuallycausing the attractive force to be insufficient to support the weight of the balloon.

Q23.10 The electric field due to the charged rod induces a charge in the aluminum foil. If the rod is broughttowards the aluminum from above, the top of the aluminum will have a negative charge induced onit, while the parts draping over the pencil can have a positive charge induced on them. Thesepositive induced charges on the two parts give rise to a repulsive Coulomb force. If the pencil is agood insulator, the net charge on the aluminum can be zero.

Q23.11 So the electric field created by the test charge does not distort the electric field you are trying tomeasure, by moving the charges that create it.

Q23.12 With a very high budget, you could send first a proton and then an electron into an evacuatedregion in which the field exists. If the field is gravitational, both particles will experience a force inthe same direction, while they will experience forces in opposite directions if the field is electric.

On a more practical scale, stick identical pith balls on each end of a toothpick. Charge one pithball + and the other –, creating a large-scale dipole. Carefully suspend this dipole about its center ofmass so that it can rotate freely. When suspended in the field in question, the dipole will rotate toalign itself with an electric field, while it will not for a gravitational field. If the test device does notrotate, be sure to insert it into the field in more than one orientation in case it was aligned with theelectric field when you inserted it on the first trial.

Q23.13 The student standing on the insulating platform is held at the same electrical potential as thegenerator sphere. Charge will only flow when there is a difference in potential. The student whounwisely touches the charged sphere is near zero electrical potential when compared to the chargedsphere. When the student comes in contact with the sphere, charge will flow from the sphere to himor her until they are at the same electrical potential.

Q23.14 An electric field once established by a positive or negative charge extends in all directions from thecharge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material atpoint A in Figure 23.23(a), so there is no charge, nor is there a force. There would be a force if acharge were present at point A, however. A field does exist at point A.

Q23.15 If a charge distribution is small compared to the distance of a field point from it, the chargedistribution can be modeled as a single particle with charge equal to the net charge of thedistribution. Further, if a charge distribution is spherically symmetric, it will create a field at exteriorpoints just as if all of its charge were a point charge at its center.

Chapter 23 3

Q23.16 The direction of the electric field is the direction in which a positive test charge would feel a forcewhen placed in the field. A charge will not experience two electrical forces at the same time, but thevector sum of the two. If electric field lines crossed, then a test charge placed at the point at whichthey cross would feel a force in two directions. Furthermore, the path that the test charge wouldfollow if released at the point where the field lines cross would be indeterminate.

Q23.17 Both figures are drawn correctly. E1 and E2 are the electric fields separately created by the pointcharges q1 and q2 in Figure 23.14 or q and –q in Figure 23.15, respectively. The net electric field is thevector sum of E1 and E2 , shown as E. Figure 23.21 shows only one electric field line at each pointaway from the charge. At the point location of an object modeled as a point charge, the direction ofthe field is undefined, and so is its magnitude.

Q23.18 The electric forces on the particles have the same magnitude, but are in opposite directions. Theelectron will have a much larger acceleration (by a factor of about 2 000) than the proton, due to itsmuch smaller mass.

Q23.19 The electric field around a point charge approaches infinity as r approaches zero.

Q23.20 Vertically downward.

Q23.21 Four times as many electric field lines start at the surface of the larger charge as end at the smallercharge. The extra lines extend away from the pair of charges. They may never end, or they mayterminate on more distant negative charges. Figure 23.24 shows the situation for charges +2q and –q.

Q23.22 At a point exactly midway between the two changes.

Q23.23 Linear charge density, λ, is charge per unit length. It is used when trying to determine the electricfield created by a charged rod.

Surface charge density, σ, is charge per unit area. It is used when determining the electric fieldabove a charged sheet or disk.

Volume charge density, ρ, is charge per unit volume. It is used when determining the electricfield due to a uniformly charged sphere made of insulating material.

Q23.24 Yes, the path would still be parabolic. The electrical force on the electron is in the downwarddirection. This is similar to throwing a ball from the roof of a building horizontally or at some anglewith the vertical. In both cases, the acceleration due to gravity is downward, giving a parabolictrajectory.

Q23.25 No. Life would be no different if electrons were + charged and protons were – charged. Oppositecharges would still attract, and like charges would repel. The naming of + and – charge is merely aconvention.

Q23.26 If the antenna were not grounded, electric charges in the atmosphere during a storm could place theantenna at a high positive or negative potential. The antenna would then place the television setinside the house at the high voltage, to make it a shock hazard. The wire to the ground keeps theantenna, the television set, and even the air around the antenna at close to zero potential.

Q23.27 People are all attracted to the Earth. If the force were electrostatic, people would all carry chargewith the same sign and would repel each other. This repulsion is not observed. When we changedthe charge on a person, as in the chapter-opener photograph, the person’s weight would changegreatly in magnitude or direction. We could levitate an airplane simply by draining away its electriccharge. The failure of such experiments gives evidence that the attraction to the Earth is not due toelectrical forces.

4 Electric Fields

Q23.28 In special orientations the force between two dipoles can be zero or a force of repulsion. In generaleach dipole will exert a torque on the other, tending to align its axis with the field created by the firstdipole. After this alignment, each dipole exerts a force of attraction on the other.

SOLUTIONS TO PROBLEMS

Section 23.1 Properties of Electric Charges

*P23.1 (a) The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces itsmass by a negligible amount, to

1 007 9 1 660 10 9 11 10 1 67 1027 31 27. . . .× − × = ×− − − kg kg kge j .

Its charge, due to loss of one electron, is

0 1 1 60 10 1 60 1019 19− − × = + ×− −. . C Ce j .

(b) By similar logic, charge = + × −1 60 10 19. C

mass = × − × = ×− − −22 99 1 66 10 9 11 10 3 82 1027 31 26. . . . kg kg kge j

(c) charge of Cl C− −= − ×1 60 10 19.

mass = × + × = ×− − −35 453 1 66 10 9 11 10 5 89 1027 31 26. . . . kg kg kge j

(d) charge of Ca C C++ − −= − − × = + ×2 1 60 10 3 20 1019 19. .e jmass = × − × = ×− − −40 078 1 66 10 2 9 11 10 6 65 1027 31 26. . . . kg kg kge j e j

(e) charge of N C C3 19 193 1 60 10 4 80 10− − −= − × = − ×. .e jmass = × + × = ×− − −14 007 1 66 10 3 9 11 10 2 33 1027 31 26. . . . kg kg kge j e j

(f) charge of N C C4 19 194 1 60 10 6 40 10+ − −= × = + ×. .e jmass = × − × = ×− − −14 007 1 66 10 4 9 11 10 2 32 1027 31 26. . . . kg kg kge j e j

(g) We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.

charge = × = ×− −7 1 60 10 1 12 1019 18. . C Ce jmass = × − × = ×− − −14 007 1 66 10 7 9 11 10 2 32 1027 31 26. . . . kg kg kge j e j

(h) charge = − × −1 60 10 19. C

mass = + × + × = ×− − −2 1 007 9 15 999 1 66 10 9 11 10 2 99 1027 31 26. . . . .b g kg kg kg

Chapter 23 5

P23.2 (a) N =FHG

IKJ ×FHG

IKJFHG

IKJ = ×

10 06 02 10 47 2 62 1023 24.. .

grams107.87 grams mol

atomsmol

electrons

atom

(b) #.

. electrons added C

1.60 10 C electron= =

××

= ×−

−Qe

1 00 106 25 10

3

1915

or 2 38. electrons for every 10 already present9 .

Section 23.2 Charging Objects by Induction

Section 23.3 Coulomb’s Law

P23.3 If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains

N ≅FHG

IKJ ×FHG

IKJFHG

IKJ ≅ ×

70 0006 02 10 10 2 3 1023 28 grams

18 grams mol

moleculesmol

protons

molecule protons. . .

With an excess of 1% electrons over protons, each person has a charge

q = × × = ×−0 01 1 6 10 2 3 10 3 7 1019 28 7. . . . C Ce je j .

So F kq qre= = ×

×= ×1 2

29

7 2

2259 10

3 7 10

0 64 10e j e j.

. N N ~10 N26 .

This force is almost enough to lift a weight equal to that of the Earth:

Mg = × = ×6 10 6 1024 25 kg 9.8 m s N ~10 N2 26e j .

*P23.4 The force on one proton is F =k q q

re 1 2

2 away from the other proton. Its magnitude is

8 99 101 6 102 10

57 5919

15

2

..

.× ⋅××

FHG

IKJ =

−

− N m C C

m N2e j .

P23.5 (a) Fk q q

ree= =

× ⋅ ×

×= ×

−

−

−1 22

9 19 2

10 29

8 99 10 1 60 10

3 80 101 59 10

. .

..

N m C C

m N repulsion

2 2e je je j

b g

(b) FGm m

rg = =× ⋅ ×

×= ×

− −

−

−1 22

11 27 2

10 245

6 67 10 1 67 10

3 80 101 29 10

. .

..

N m C kg

m N

2 2e je je j

The electric force is larger by 1.24 10 times36× .

(c) If kq qr

Gm m

re1 2

21 2

2= with q q q1 2= = and m m m1 2= = , then

qm

Gke

= =× ⋅

× ⋅= ×

−−6 67 10

8 99 108 61 10

11

911.

..

N m kg N m C

C kg2 2

2 2 .

6 Electric Fields

P23.6 We find the equal-magnitude charges on both spheres:

F kq qr

kqre e= =1 2

2

2

2 so q rFke

= =×

× ⋅= × −1 00

1 00 101 05 10

43.

.. m

N8.99 10 N m C

C9 2 2a f .

The number of electron transferred is then

Nexfer

C C

electrons=×

×= ×

−

− −1 05 10

1 60 106 59 10

3

1915.

.. .

The whole number of electrons in each sphere is

N e etot g

107.87 g mol atoms mol atom =

FHG

IKJ × = ×− −10 0

6 02 10 47 2 62 1023 24.. .e je j .

The fraction transferred is then

fNN

= =××

FHG

IKJ = × =−xfer

tot

6 59 102 62 10

2 51 10 2 5115

249.

.. . charges in every billion.

P23.7 F kq qre11 2

2

9 6 6

2

8 99 10 7 00 10 2 00 10

0 5000 503= =

× ⋅ × ×=

− −. . .

..

N m C C C

m N

2 2e je je ja f

F kq qre21 2

2

9 6 6

2

8 99 10 7 00 10 4 00 10

0 5001 01= =

× ⋅ × ×=

− −. . .

..

N m C C C

m N

2 2e je je ja f

FF

x

y

= °+ °== °− °= −

= − = °

0 503 60 0 1 01 60 0 0 7550 503 60 0 1 01 60 0 0 436

0 755 0 436 0 872

. cos . . cos . .

. sin . . sin . .

. . .

N N

N N N at an angle of 330F i ja f a f

FIG. P23.7

P23.8 F kq qre= =

× ⋅ × ×

×=

−1 2

2

9 19 2 23 2

6 2

8 99 10 1 60 10 6 02 10

2 6 37 10514

. . .

.

N m C C

m kN

2 2e je j e je j

P23.9 (a) The force is one of attraction . The distance r in Coulomb’s law is the distance between

centers. The magnitude of the force is

Fk q q

re= = × ⋅

× ×= ×

− −−1 2

29

9 9

258 99 10

12 0 10 18 0 10

0 3002 16 10.

. .

.. N m C

C C

m N2 2e j e je j

a f .

(b) The net charge of − × −6 00 10 9. C will be equally split between the two spheres, or− × −3 00 10 9. C on each. The force is one of repulsion , and its magnitude is

Fk q q

re= = × ⋅

× ×= ×

− −−1 2

29

9 9

278 99 10

3 00 10 3 00 10

0 3008 99 10.

. .

.. N m C

C C

m N2 2e j e je j

a f .

Chapter 23 7

P23.10 Let the third bead have charge Q and be located distance x from the left end of the rod. This beadwill experience a net force given by

F i i= +−

−k q Q

x

k q Q

d xe e3

2 2

b g b ga f e j .

The net force will be zero if 3 12 2x d x=

−a f , or d xx

− =3

.

This gives an equilibrium position of the third bead of x d= 0 634. .

The equilibrium is stable if the third bead has positive charge .

P23.11 (a) Fk ere= = × ⋅

×

×= ×

−

−

−2

29

19 2

10 288 99 10

1 60 10

0 529 108 22 10.

.

.. N m C

C

m N2 2e j e j

e j

(b) We have Fmv

r=

2

from which

vFrm

= =× ×

×= ×

− −

−

8 22 10 10

9 11 102 19 10

8 10

316

.

..

N 0.529 m

kg m s

e j.

P23.12 The top charge exerts a force on the negative charge k qQ

xe

d2

2 2c h + which is directed upward and to the

left, at an angle of tan− FHGIKJ

1

2dx

to the x-axis. The two positive charges together exert force

22 2

42

42 1 2

k qQ

x

x

xme

d d+

F

HGG

I

KJJ

−

+

F

HGGG

I

KJJJ=

e ja fe j

ia or for x

d<<

2, a x≈

−283

k qQmd

e .

(a) The acceleration is equal to a negative constant times the excursion from equilibrium, as in

a x= −ω 2 , so we have Simple Harmonic Motion with ω 23

16=

k qQmd

e .

Tmdk qQe

= =2

2

3πω

π, where m is the mass of the object with charge −Q .

(b) v A ak qQmd

emax = =ω 4 3

8 Electric Fields

Section 23.4 The Electric Field

P23.13 For equilibrium, F Fe g= −

or q mgE j= − −e j .

Thus, E j=mgq

.

(a) E j j j= =×

− ×= − ×

−

−−mg

q

. .

..

9 11 10 9 80

1 60 105 58 10

31

1911

kg m s

C N C

2e je je j e j

(b) E j j j= =×

×= ×

−

−−mg

q

. .

..

1 67 10 9 80

1 60 101 02 10

27

197

kg m s

C N C

2e je je j e j

P23.14 Fy∑ = 0 : QE mgj j+ − =e j 0

∴ = =×

=−

mQEg

24 0 10 610

9 801 49

6.

..

C N C

m s grams2

e jb g

P23.15 The point is designated in the sketch. The magnitudes of the electric fields,E1 , (due to the − × −2 50 10 6. C charge) and E2 (due to the 6 00 10 6. × − Ccharge) are

Ek qr d

e1 2

9 6

2

8 99 10 2 50 10= =

× ⋅ × −. . N m C C2 2e je j(1)

Ek qr d

e2 2

9 6

2

8 99 10 6 00 10

1 00= =

× ⋅ ×

+

−. .

.

N m C C

m

2 2e je ja f (2)

FIG. P23.15

Equate the right sides of (1) and (2)

to get d d+ =1 00 2 402 2. . ma f

or d d+ = ±1 00 1 55. . m

which yields d = 1 82. m

or d = −0 392. m .

The negative value for d is unsatisfactory because that locates a point between the charges whereboth fields are in the same direction.

Thus, d = −1 82 2 50. . m to the left of the C chargeµ .

Chapter 23 9

P23.16 If we treat the concentrations as point charges,

E j j

E j j

E E E

+

−

+ −

= = × ⋅ − = × −

= = × ⋅ − = × −

= + = ×

kq

r

kq

r

e

e

29

25

29

25

5

8 99 1040 0

1 0003 60 10

8 99 1040 0

1 0003 60 10

7 20 10

..

.

..

.

.

N m C C

m N C downward

N m C C

m N C downward

N C downward

2 2

2 2

e j a fb g e j e ja f

e j a fb g e j e ja f

*P23.17 The first charge creates at the origin field k Qae2 to the right.

Suppose the total field at the origin is to the right. Then q mustbe negative:

k Qa

k q

a

k Qa

e e e2 2 23

2i i i+ − =a f e j q Q= −9 .

In the alternative, the total field at the origin is to the left:

k Qa

k qa

k Qa

e e e2 2 29

2i i i+ − = −e j e j q Q= +27 .

xq+Q x = 0

FIG. P23.17

P23.18 (a) Ek qr

e1 2

9 6

25

8 99 10 7 00 10

0 5002 52 10= =

× ×= ×

−. .

..

e je ja f N C

Ek qr

E E E

E E

e

x

y

2 2

9 6

25

2 15 5 3

15 3

3

8 99 10 4 00 10

0 5001 44 10

60 1 44 10 2 52 10 60 0 18 0 10

60 0 2 52 10 60 0 218 10

18 0 218 10 18 0 218

= =× ×

= ×

= − °= × − × °= ×

= − °= − × °= − ×

= − × = −

−. .

..

cos . . cos . .

sin . . sin .

. .

e je ja f N C

N C

N C

N C kN CE i j i j

FIG. P23.18

(b) F E i j i j i j= = × − × = − × = −− −q 2 00 10 18 0 218 10 36 0 436 10 36 0 4366 3 3. . . . C N C N mNe je j e j e j

P23.19 (a) E j j j11

12

9 9

23

8 99 10 3 00 10

0 1002 70 10= − =

× ×− = − ×

−k q

re . .

..e j e je j

a f e j e j N C

E i i i

E E E i j

22

22

9 9

22

2 12 3

8 99 10 6 00 10

0 3005 99 10

5 99 10 2 70 10

= − =× ×

− = − ×

= + = − × − ×

−k q

re . .

..

. .

e j e je ja f e j e j

e j e j

N C

N C N C FIG. P23.19

(b) F E i j= = × − −−q 5 00 10 599 2 7009. C N Ce je jF i j i j= − × − × = − −− −3 00 10 13 5 10 3 00 13 56 6. . . .e j e j N Nµ

10 Electric Fields

P23.20 (a) Ek qr

e= =× ×

=−

2

9 6

2

8 99 10 2 00 10

1 1214 400

. .

.

e je ja f N C

Ex = 0 and Ey = °= ×2 14 400 26 6 1 29 104b gsin . . N C

so E j= ×1 29 104. N C . FIG. P23.20

(b) F E j j= = − × × = − ×− −q 3 00 10 1 29 10 3 86 106 4 2. . .e je j N

P23.21 (a) E r r r i i j j= + + = + °+ ° +k qr

k qr

k qr

k q

a

k q

a

k q

ae e e e e e1

12 1

2

22 2

3

32 3 2 2 2

2 3

245 0 45 0

4cos . sin .

b g b g e j b g

E i j= + = °3 06 5 06 5 912 2 2. . .k qa

k qa

k qa

e e e at 58.8

(b) F E= = °qk qae5 91

2

2. at 58.8

P23.22 The electric field at any point x is

Ek q

x a

k q

x a

k q ax

x a

e e e=−

−− −

=−a f a fc ha f

e j2 2 2 2 2

4.

When x is much, much greater than a, we find Ea k q

xe≅

43

b g.

P23.23 (a) One of the charges creates at P a field E =+

k Q nR x

e2 2 at an angle θ to

the x-axis as shown.

When all the charges produce field, for n > 1 , the componentsperpendicular to the x-axis add to zero.

The total field is nk Q n

R xk Qx

R x

e eb ge j

cosi i

2 2 2 2 3 2+=

+θ .

FIG. P23.23

(b) A circle of charge corresponds to letting n grow beyond all bounds, but the result does notdepend on n. Smearing the charge around the circle does not change its amount or itsdistance from the field point, so it does not change the field .

P23.24 E r i i ii

i= = − + − + − + =−

+ + +FHG

IKJ = −∑

k qr

k qa

k q

a

k q

a

k qa

k qa

e e e e e e2 2 2 2 2 2 2

2

22 31

12

13 6

e j a f e j a f e j … …π

Chapter 23 11

Section 23.5 Electric Field of a Continuous Charge Distribution

P23.25 Ek

d d

k Q

d dk Q

d de e e=+

=+

=+

=× ×

+

−λa f

b ga f a f

e je ja fa f8 99 10 22 0 10

0 290 0 140 0 290

9 6. .

. . .

E = ×1 59 106. , N C directed toward the rod.FIG. P23.25

P23.26 Ek dqxe= z 2 , where dq dx= λ 0

E kdxx

kx

kxe

xe

x

e= = −FHGIKJ =

∞ ∞

zλ λλ

0 2 00

00 0

1The direction is or left for − >i λ 0 0

P23.27 Ek xQ

x a

x

x

x

x

e=+

=× ×

+=

×

+

−

2 2 3 2

9 6

2 2 3 2

5

2 3 2

8 99 10 75 0 10

0 100

6 74 10

0 010 0e je je je j e j

. .

.

.

.

(a) At x = 0 010 0. m , E i i= × =6 64 10 6 646. . N C MN C

(b) At x = 0 050 0. m , E i i= × =2 41 10 24 17. . N C MN C

(c) At x = 0 300. m , E i i= × =6 40 10 6 406. . N C MN C

(d) At x = 1 00. m , E i i= × =6 64 10 0 6645. . N C MN C

P23.28 E Ei

i i i= =−L

NMMM

O

QPPP= − = − −

FHG

IKJ = −z z z∞

−∞ ∞

dk x dx

xk x x dx k x

xk

xe

xe

xe

x

eλ

λ λλ0 0

3 0 03

0 0 20

00 0 0

12 2

e j e j

P23.29 Ek Qx

x a

e=+2 2 3 2e j

For a maximum, dEdx

Qkx a

x

x ae=

+−

+

L

NMMM

O

QPPP=

1 30

2 2 3 2

2

2 2 5 2e j e jx a x2 2 23 0+ − = or x

a=

2.

Substituting into the expression for E gives

Ek Qa

a

k Q

a

k Qa

Qa

e e e= = = =∈2 3

23 3 6 33

22 3 2 3

22 2

02e j π

.

12 Electric Fields

P23.30 E kx

x Re= −

+

FHG

IKJ2 1

2 2π σ

Ex

x

x

x= × × −

+

F

HGG

I

KJJ = × −

+

FHG

IKJ

−2 8 99 10 7 90 10 10 350

4 46 10 10 123

9 3

2 2

8

2π . .

..

.e je j a f

(a) At x = 0 050 0. m , E = × =3 83 10 3838. N C MN C

(b) At x = 0 100. m , E = × =3 24 10 3248. N C MN C

(c) At x = 0 500. m , E = × =8 07 10 80 77. . N C MN C

(d) At x = 2 00. m , E = × =6 68 10 6 688. . N C MN C

P23.31 (a) From Example 23.9: E kx

x Re= −

+

FHG

IKJ2 1

2 2π σ

σπ

= = ×

= × = × =

−QR

E

23

8 7

1 84 10

1 04 10 0 900 9 36 10 93 6

.

. . . .

C m

N C N C MN C

2

e ja fappx: E ke= =2 104π σ MN C about 11% highb g

(b) E = × −+

FHG

IKJ = × =1 04 10 1

30 0

3 001 04 10 0 004 96 0 5168

2

8..

.. . . N C

cm

30.0 cm N C MN C

2e j e jb g

appx: E kQre= = ×

×=

−

29

6

28 99 105 20 10

0 300 519.

.

..e j a f b g MN C about 0.6% high

P23.32 The electric field at a distance x is E kx

x Rx e= −

+

LNMM

OQPP2 1

2 2π σ

This is equivalent to E kR x

x e= −+

L

NMM

O

QPP2 1

1

1 2 2π σ

For large x, Rx

2

2 1<< and 1 12

2

2

2

2+ ≈ +Rx

Rx

so E kR x

kR x

R xx e e= −

+

F

HGG

I

KJJ =

+ −

+2 1

1

1 22

1 2 1

1 22 2

2 2

2 2π σ π σ

e je je je j

Substitute σπ

=QR2 , E

k Q x

R xk Q x

Rx

ee=

+= +FHG

IKJ

1

1 2 2

2

2 22

2e je j

But for x R>> , 1

21

2 2 2x R x+≈ , so E

k Qxxe≈ 2 for a disk at large distances

Chapter 23 13

P23.33 Due to symmetry E dEy y= =z 0 , and E dE kdq

rx e= =z zsinsin

θθ

2

where dq ds rd= =λ λ θ ,

so that, Ekr

dkr

krx

e e e= = − =zλ θ θλ

θλπ

πsin cos

00

2a f

where λ =qL

and rL

=π

. FIG. P23.33

Thus, Ek qLxe= =

× ⋅ × −2 2 8 99 10 7 50 10

0 1402

9 6

2

π π. .

.

N m C C

m

2 2e je ja f .

Solving, Ex = ×2 16 107. N C .

Since the rod has a negative charge, E i i= − × = −2 16 10 21 67. .e j N C MN C .

P23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has

charge Qdx

h and produces, at the chosen point, a field

dk x

x R

Qdxh

eE i=+2 2 3 2e j

.

The total field is

E E ii

Ei i

= =+

= +

=+

−=

+−

+ +

L

NMMM

O

QPPP

z z z+ −

=

+

−

=

+

dk Qxdx

h x R

k Qh

x R xdx

k Qh

x R k Qh d R d h R

e

d

d he

x d

d h

e

x d

d h

e

all charge2 2 3 2

2 2 3 2

2 2 1 2

2 2 1 2 2 2 1 2

22

2 1 21 1

e je j

e jb g e j a fe j

(b) Think of the cylinder as a stack of disks, each with thickness dx, charge Qdx

h, and charge-

per-area σπ

=QdxR h2 . One disk produces a field

dk QdxR h

x

x R

eE i= −+

F

HGG

I

KJJ

212 2 2 1 2

ππ e j

.

So, E E i= = −+

F

HGG

I

KJJz z

=

+

dk QdxR h

x

x R

e

x d

d h

all charge

212 2 2 1 2e j

Ei i

Ei

Ei

= − +LNMM

OQPP = −

+L

N

MMM

O

Q

PPP

= + − − + + + +LNM

OQP

= + + − + +LNM

OQP

+ −

=

++

+

z z2 12

22 1

2 1 2

2

2

22 2 1 2

2

2 2 1 2

22 2 1 2 2 2 1 2

22 2 1 2 2 2 1 2

k QR h

dx x R xdxk QR h

xx R

k QR h

d h d d h R d R

k QR h

h d R d h R

e

d

d h

x d

d he

dd h

d

d h

e

e

e j e j

a fe j e j

e j a fe j

14 Electric Fields

P23.35 (a) The electric field at point P due to each element of length dx, is

dEk dq

x ye=+2 2 and is directed along the line joining the element to

point P. By symmetry,E dEx x= =z 0 and since dq dx= λ ,

E E dE dEy y= = =z z cosθ where cosθ =+

y

x y2 2.

Therefore, E k ydx

x y

kye

e=+

=z22

2 2 3 20

20λ

λ θ

e jsin

.FIG. P23.35

(b) For a bar of infinite length, θ 0 90= ° and Ekyye=

2 λ.

P23.36 (a) The whole surface area of the cylinder is A r rL r r L= + = +2 2 22π π π a f .Q A= = × + = ×− −σ π15 0 10 2 0 025 0 0 025 0 0 060 0 2 00 109 10. . . . . C m m m m C2e j b g

(b) For the curved lateral surface only, A rL= 2π .

Q A= = × = ×− −σ π15 0 10 2 0 025 0 0 060 1 41 109 10. . . . C m m 0 m C2e j b ga f

(c) Q V r L= = = × = ×− −ρ ρπ π2 9 2 11500 10 0 025 0 0 060 0 5 89 10 C m m m C3e j b g b g. . .

P23.37 (a) Every object has the same volume, V = = × −8 0 030 2 16 103 4. . 0 m m3a f .

For each, Q V= = × × = ×− − −ρ 400 10 2 16 10 8 64 109 4 11 C m m C3 3e je j. .

(b) We must count the 9 00. cm2 squares painted with charge:

(i) 6 4 24× = squares

Q A= = × × = ×− − −σ 15 0 10 24 0 9 00 10 3 24 109 4 10. . . . C m m C2 2e j e j(ii) 34 squares exposed

Q A= = × × = ×− − −σ 15 0 10 34 0 9 00 10 4 59 109 4 10. . . . C m m C2 2e j e j(iii) 34 squares

Q A= = × × = ×− − −σ 15 0 10 34 0 9 00 10 4 59 109 4 10. . . . C m m C2 2e j e j(iv) 32 squares

Q A= = × × = ×− − −σ 15 0 10 32 0 9 00 10 4 32 109 4 10. . . . C m m C2 2e j e j(c) (i) total edge length: = ×24 0 030 0. mb g

Q = = × × = ×− −λ 80 0 10 24 0 030 0 5 76 1012 11. . . C m m Ce j b g

(ii) Q = = × × = ×− −λ 80 0 10 44 0 030 0 1 06 1012 10. . . C m m Ce j b gcontinued on next page

Chapter 23 15

(iii) Q = = × × = ×− −λ 80 0 10 64 0 030 0 1 54 1012 10. . . C m m Ce j b g

(iv) Q = = × × = ×− −λ 80 0 10 40 0 030 0 0 960 1012 10. . . C m m Ce j b g

Section 23.6 Electric Field Lines

P23.38

FIG. P23.38

P23.39

FIG. P23.39

P23.40 (a)qq

1

2

618

13

=−

= −

(b) q q1 2 is negative, is positive

P23.41 (a) The electric field has the general appearance shown. It is zeroat the center , where (by symmetry) one can see that the three

charges individually produce fields that cancel out.

In addition to the center of the triangle, the electric field lines in thesecond figure to the right indicate three other points near themiddle of each leg of the triangle where E = 0 , but they are moredifficult to find mathematically.

(b) You may need to review vector addition in Chapter Three. Theelectric field at point P can be found by adding the electric fieldvectors due to each of the two lower point charges: E E E= +1 2 .

The electric field from a point charge is E r= kq

re 2 .

As shown in the solution figure at right,

E1 2= kq

ae to the right and upward at 60°

E2 2= kq

ae to the left and upward at 60°FIG. P23.41

E E E i j i j j

j

= + = ° + ° + − ° + ° = °

=

1 2 2 2

2

60 60 60 60 2 60

1 73

kq

ak

qa

kq

a

e e

e

cos sin cos sin sin

.

e j e j e j

16 Electric Fields

Section 23.7 Motion of Charged Particles in a Uniform Electric Field

P23.42 F qE ma= = aqEm

=

v v atf i= + vqEtmf =

electron: ve =× ×

×= ×

− −

−

1 602 10 520 48 0 10

9 11 104 39 10

19 9

316

. .

..

e ja fe j m s

in a direction opposite to the field

proton: vp =× ×

×= ×

− −

−

1 602 10 520 48 0 10

1 67 102 39 10

19 9

273

. .

..

e ja fe j m s

in the same direction as the field

P23.43 (a) aqEm

= =×

×= ×

−

−

1 602 10 640

1 67 106 14 10

19

2710.

..

a f m s2

(b) v v atf i= + 1 20 10 6 14 106 10. .× = ×e jt t = × −1 95 10 5. s

(c) x x v v tf i i f− = +12d i x f = × × =−1

21 20 10 1 95 10 11 76 5. . .e je j m

(d) K mv= = × × = ×− −12

12

1 67 10 1 20 10 1 20 102 27 6 2 15. . . kg m s Je je j

P23.44 (a) aqEm

= =× ×

×= ×

−

−

1 602 10 6 00 10

1 67 105 76 10

19 5

2713

. .

..

e je je j

m s so a i= − ×5 76 1013. m s2

(b) v v a x xf i f i= + −2 d i0 2 5 76 10 0 070 02 13= + − ×vi . .e jb g v ii = ×2 84 106. m s

(c) v v atf i= +

0 2 84 10 5 76 106 13= × + − ×. .e jt t = × −4 93 10 8. s

P23.45 The required electric field will be in the direction of motion .

Work done = ∆K

so, − = −Fd mvi12

2 (since the final velocity = 0 )

which becomes eEd K=

and EKed

= .

Chapter 23 17

P23.46 The acceleration is given by

v v a x xf i f i2 2 2= + −d i or v a hf

2 0 2= + −a f .

Solving av

hf= −2

2.

Now F a∑ = m : − + = −mg qmv

hfj Ej2

2.

Therefore qmv

hmgfE j= − +

FHG

IKJ

2

2.

(a) Gravity alone would give the bead downward impact velocity

2 9 80 5 00 9 90. . . m s m m s2e ja f = .

To change this to 21.0 m/s down, a downward electric field must exert a downward

electric force.

(b) qmE

v

hgf= −

FHG

IKJ=

×

×⋅⋅

FHGIKJ −LNMM

OQPP =

−2 3 2

21 00 10 21 0

2 5 009 80 3 43

. .

.. .

kg1.00 10 N C

N skg m

m s

m m s C4

22b g

a f µ

P23.47 (a) tx

vx= =

×= × =−0 050 0

4 50 101 11 10 1115

7..

. s ns

(b) aqEmy = =

× ×

×= ×

−

−

1 602 10 9 60 10

1 67 109 21 10

19 3

2711

. .

..

e je je j

m s2

y y v t a tf i yi y− = +12

2: y f = × × = × =− −12

9 21 10 1 11 10 5 68 10 5 6811 7 2 3. . . .e je j m mm

(c) vx = ×4 50 105. m s v v a tyf yi y= + = × × = ×−9 21 10 1 11 10 1 02 1011 7 5. . .e je j m s

*P23.48 The particle feels a constant force: F E j j= = × − = × −− −q 1 10 2 000 2 106 3 C N C Ne jb ge j e j

and moves with acceleration: aF j

j= =× ⋅ −

×= × −∑

−

−m

2 10

2 101 10

3

1613

kg m s

kg m s

22e je j

e je j.

Its x-component of velocity is constant at 1 00 10 37 7 99 105 4. cos .× °= × m s m se j . Thus it moves in a

parabola opening downward. The maximum height it attains above the bottom plate is described by

v v a y yyf yi y f i2 2 2= + −d i: 0 6 02 10 2 10 04 2 13= × − × −. m s m s2e j e jd iy f

y f = × −1 81 10 4. m.

continued on next page

18 Electric Fields

Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetricparabola and strikes the bottom plate after a time given by

y y v t a tf i yi y= + +12

2 0 0 6 02 1012

1 104 13 2= + × + − ×. m s m s2e j e jt t

since t > 0 , t = × −1 20 10 8. s .

The particle’s range is x x v tf i x= + = + × × = ×− −0 7 99 10 1 20 10 9 61 104 8 4. . . m s s me je j .

In sum,

The particle strikes the negative plate after moving in a parabola with a height of 0.181 mmand a width of 0.961 mm.

P23.49 vi = ×9 55 103. m s

(a) aeEmy = =

×

×= ×

−

−

1 60 10 720

1 67 106 90 10

19

2710

.

..

e ja fe j

m s2

Rv

ai

y= = × −

232

1 27 10sin

.θ

m so that

9 55 10 2

6 90 101 27 10

3 2

103

. sin

..

×

×= × −e j θ

sin .2 0 961θ = θ = °36 9. 90 0 53 1. .°− = °θ

^

FIG. P23.49

(b) tR

vR

vix i= =

cosθIf θ = °36 9. , t = 167 ns . If θ = °53 1. , t = 221 ns .

Additional Problems

*P23.50 The two given charges exert equal-size forces of attraction on eachother. If a third charge, positive or negative, were placed betweenthem they could not be in equilibrium. If the third charge were at apoint x > 15 cm , it would exert a stronger force on the 45 Cµ thanon the −12 Cµ , and could not produce equilibrium for both. Thusthe third charge must be at x d= − < 0 . Its equilibrium requires

d x15 cm

q

x = 0+–

–12 Cµ 45 Cµ

FIG. P23.50

k q

d

k q

de e12 45

152 2

C C

cm

µ µb g b ga f=

+

15 4512

3 752 cm+F

HGIKJ = =

dd

.

15 1 94 cm+ =d d. d = 16 0. cm .

The third charge is at x = −16 0. cm . The equilibrium of the −12 Cµ requires

k q ke e12

16 0

45 122 2

C

cm

C C

15 cm

µ µ µb ga f

b ga f.

= q = 51 3. Cµ .

All six individual forces are now equal in magnitude, so we have equilibrium as required, and this isthe only solution.

Chapter 23 19

P23.51 The proton moves with acceleration aqEmp = =

×

×= ×

−

−

1 60 10 640

1 673 106 13 10

19

2710

.

..

C N C

kg m s2e jb g

while the e− has acceleration a ae p=×

×= × =

−

−

1 60 10 640

9 110 101 12 10 1 836

19

3114

.

..

C N C

kg m s2e jb g

.

(a) We want to find the distance traveled by the proton (i.e., d a tp=12

2 ), knowing:

4 0012

12

1 83712

2 2 2. cm = + = FHGIKJa t a t a tp e p .

Thus, d a tp= = =12

4 0021 82 .

. cm

1 837 mµ .

(b) The distance from the positive plate to where the meeting occurs equals the distance the

sodium ion travels (i.e., d a tNa Na=12

2 ). This is found from:

4 0012

12

2 2. cm Na Cl= +a t a t : 4 0012 22 99

12 35 45

2 2.. .

cm u u

= FHGIKJ + FHG

IKJ

eEt

eEt .

This may be written as 4 0012

12

0 649 1 6512

2 2 2. . . cm Na Na Na= + = FHG

IKJa t a t a tb g

so d a tNa Na cm

1.65 cm= = =

12

4 002 432 .. .

P23.52 (a) The field, E1 , due to the 4 00 10 9. × − C charge is in the –xdirection.

E r i

i

1 2

9 9

2

8 99 10 4 00 10

2 50

5 75

= =× ⋅ − ×

= −

−k qr

e. .

.

.

N m C C

m

N C

2 2e je ja f

FIG. P23.52(a)

Likewise, E2 and E3 , due to the 5 00 10 9. × − C charge and the 3 00 10 9. × − C charge are

E r i i2 2

9 9

2

8 99 10 5 00 10

2 0011 2= =

× ⋅ ×=

−k qr

e. .

..

N m C C

m N C

2 2e je ja f

E i i3

9 9

2

8 99 10 3 00 10

1 2018 7=

× ⋅ ×=

−. .

..

N m C C

m N C

2 2e je ja f

E E E ER = + + =1 2 3 24 2. N C in +x direction.

(b) E r i j1 2 8 46 0 243 0 970= = − +k qr

e . . . N Cb ge jE r j

E r i j

i j

2 2

3 2

1 3 1 2 3

11 2

5 81 0 371

4 21 8 43

= = +

= = −

= + = − = + + =

k qrk qr

E E E E E E E

e

e

x x x y y y y

.

. .

. .

N C

N C +0.928

N C N C

b ge j

b ge j

ER = 9 42. N C θ = ° −63 4. above axisxFIG. P23.52(b)

20 Electric Fields

*P23.53 (a) Each ion moves in a quarter circle. The electric force causes the centripetal acceleration.

F ma∑ = qEmv

R=

2

EmvqR

=2

(b) For the x-motion, v v a x xxf xi x f i2 2 2= + −d i

0 22= +v a Rx avR

Fm

qEmx

x x= − = =2

2

Emv

qRx = −2

2. Similarly for the y-motion,

v a Ry2 0 2= + a

vR

qE

myy

= + =2

2E

mvqRy =

2

2

The magnitude of the field is

E Emv

qRxx y

2 22

2+ = ° at 135 counterclockwise from the -axis .

P23.54 From the free-body diagram shown,

Fy∑ = 0 : T cos . .15 0 1 96 10 2°= × − N .

So T = × −2 03 10 2. N .

From Fx∑ = 0 , we have qE T= °sin .15 0

or qT

E=

°=

× °

×= × =

−−sin . . sin .

.. .

15 0 2 03 10 15 0

1 00 105 25 10 5 25

2

36

N

N C C C

e jµ .

FIG. P23.54

P23.55 (a) Let us sum force components to find

F qE Tx x∑ = − =sinθ 0 , and F qE T mgy y∑ = + − =cosθ 0 .

Combining these two equations, we get

qmg

E Ex y

=+

=×

°+ ×= ×

=

−−

cot

. .

. cot . ..

.

θe je ja f

a f1 00 10 9 80

3 00 37 0 5 00 101 09 10

10 9

3

58 C

nC

(b) From the two equations for Fx∑ and Fy∑ we also find

TqEx

=°= × =−

sin .. .

37 05 44 10 5 443 N mN .

Free Body Diagram

FIG. P23.55

Chapter 23 21

P23.56 This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ. Theunknowns are q and T.

The approach to this problem should be the same as for the last problem, but withoutnumbers to substitute for the variables. Likewise, we can use the free body diagram given in thesolution to problem 55.

Again, Newton’s second law: F T qAx∑ = − + =sinθ 0 (1)

and F T qB mgy∑ = + + − =cosθ 0 (2)

(a) Substituting TqA

=sinθ

, into Eq. (2),qA

qB mgcos

sinθ

θ+ = .

Isolating q on the left, qmg

A B=

+cotθa f .

(b) Substituting this value into Eq. (1), TmgA

A B=

+cos sinθ θa f .

If we had solved this general problem first, we would only need to substitute theappropriate values in the equations for q and T to find the numerical results needed forproblem 55. If you find this problem more difficult than problem 55, the little list at the firststep is useful. It shows what symbols to think of as known data, and what to considerunknown. The list is a guide for deciding what to solve for in the analysis step, and forrecognizing when we have an answer.

P23.57 Fk q q

re= 1 2

2 : tan..

θ =15 060 0

θ = °14 0.

F

F

F

F F FF F F

x

y

1

9 6 2

2

3

9 6 2

2

2

9 6 2

2

3 2

1 2

8 99 10 10 0 10

0 15040 0

8 99 10 10 0 10

0 6002 50

8 99 10 10 0 10

0 6192 35

14 0 2 50 2 35 14 0 4 7814 0 40 0 2 35 14 0 40 6

=× ×

=

=× ×

=

=× ×

=

= − − °= − − °= −= − − °= − − °= −

−

−

−

. .

..

. .

..

. .

..

cos . . . cos . .sin . . . sin . .

e je ja f

e je ja f

e je ja f

N

N

N

N N

FIG. P23.57

F F F

F

F

x y

y

x

net N= + = + =

= =−−

= °

2 2 2 24 78 40 6 40 9

40 64 78

263

. . .

tan.

.

a f a fφ

φ

22 Electric Fields

P23.58 From Figure A: d cos . .30 0 15 0°= cm,

or d =°

15 030 0

.cos .

cm

From Figure B: θ = FHG

IKJ

−sin.

1

50 0d cm

θ =°

FHG

IKJ = °−sin

.cos .

.1 15 030 0

20 3 cm

50.0 cma fF

mgq= tanθ

or F mgq = °tan .20 3 (1)

From Figure C: F Fq = °2 30 0cos .

Fk q

qe=

LNMM

OQPP °2

0 30030 0

2

2.cos .

ma f (2)

Combining equations (1) and (2),

20 300

30 0 20 32

2

k qmge

.cos . tan .

ma fLNMM

OQPP °= °

qmg

k

q

q

e

22

23 2

9

14 7

0 300 20 32 30 0

2 00 10 9 80 0 300 20 3

2 8 99 10 30 0

4 20 10 2 05 10 0 205

=°

°

=× °

× ⋅ °

= × = × =

−

− −

. tan .cos .

. . . tan .

. cos .

. . .

m

kg m s m

N m C

C C C

2

2 2

2

a f

e je ja fe j

µ

Figure A

Figure B

Figure C

FIG. P23.58

P23.59 Charge Q2

resides on each block, which repel as point charges: Fk Q Q

Lk L Le

i= = −2 2

2

b gb g b g.

Solving for Q, Q Lk L L

ki

e=

−2b g

.

*P23.60 If we place one more charge q at the 29th vertex, the total force on the central charge will add up to

zero: F28 charges +k qQ

ae

2 away from vertex 29 0= F28 chargesek

toward vertex 29=qQ

a2 .

P23.61 According to the result of Example 23.7, the left-hand rod createsthis field at a distance d from its right-hand end:

Ek Q

d a d

dFk QQ

adx

d d a

Fk Q

adx

x x ak Q

a aa xx

e

e

e

x b a

be

b a

b

=+

=+

=+

= −+F

HGIKJ

= − −z

2

2 2

2 2 212

22

2

2

2

a f

a f

a f ln

FIG. P23.61

Fk Q

aa bb

bb a

k Qa

bb a b a

k Qa

bb a

e e e=+

−+

+−

FHG

IKJ = − +

=FHGIKJ −

FHG

IKJ

2

2

2

2

2 2

2

2

2 242

2 4 2 2 4 4ln ln ln lna fa f

Chapter 23 23

P23.62 At equilibrium, the distance between the charges is r = °= × −2 0 100 10 0 3 47 10 2. sin . .m mb gNow consider the forces on the sphere with charge +q , and use Fy∑ = 0 :

Fy∑ = 0 : T mgcos .10 0°= , or Tmg

=°cos .10 0

(1)

Fx∑ = 0 : F F F Tnet = − = °2 1 10 0sin . (2)

Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1).

Fmg

mgnet2kg m s N=

°°

= °= × °= ×− −sin .cos .

tan . . . tan . .10 0

10 010 0 2 00 10 9 80 10 0 3 46 103 3e je j

Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exertedby −q , and F2 is the force exerted on +q by the external electric field.

r

L

+q–q

θ θ

FIG. P23.62

F F Fnet = −2 1 or F F F2 1= +net

F19

8 8

3 228 99 10

5 00 10 5 00 10

3 47 101 87 10= × ⋅

× ×

×= ×

− −

−

−.. .

.. N m C

C C

m N2 2e j e je j

e jThus, F F F2 1= +net yields F2

3 2 23 46 10 1 87 10 2 21 10= × + × = ×− − −. . . N N N

and F qE2 = , or EFq

= =××

= × =−

−2

2

852 21 10

4 43 10 443.

. N

5.00 10 C N C kN C .

P23.63 Q d Rd R R R

Q

dFd

R

Rd

R

F

y

y

= = = = − − =

= = = =

=∈FHG

IKJ =

∈

FHGG

IKJJ

= × ⋅×

z z− °

°

− °°

−

λ λ θ θ λ θ λ λ

µ λ µ λ µ

πµ λ

θπ

µ λ θ θ

090 0

90 0

0 90 090 0

0 0

0

02

0

02

2

9

1 1 2

12 0 2 0 600 12 0 10 0

14

3 00 14

3 00

8 99 103 00 10

cos sin

. . . .

.cos

. cos

..

.

.

.. a f

b ga fb gb g b ge j

e j

C m C so C m

C C

N m C

0

2 26 6

2

90 0

90 0

3

2

2

2

2

10 0 10

0 600

8 99 30 00 600

1012

12

2

0 45012

14

2 0 707

C C m

m

N

N N Downward.

e je ja f

a f e j

a f

.

.cos

. ..

cos

. sin .

.

. ×

= +FHGIKJ

= +FHG

IKJ =

−

− °

°

−

−

−

zz

θ θ

θ θ

θ θ

π

π

π

π

d

F d

F

y

y

1

0

–1

cosθ

0° 360°

0° 360°

1

0

cos θ2

FIG. P23.63

Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0 .

P23.64 At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can belocated by determining the angle θ corresponding to equilibrium.

In terms of lengths s, 12

3a , and r, shown in Figure P23.64, the charge at the origin exerts an

attractive force

k Qq

s a

e

+ 12

23e j

continued on next page

24 Electric Fields

The other two charges exert equal repulsive forces of magnitude k Qqre

2 . The horizontal components

of the two repulsive forces add, balancing the attractive force,

F k Qqr s a

enet = −+

L

NMMM

O

QPPP=

2 1

302

12

2cosθ

e j

From Figure P23.64 ra

=12

sinθs a=

12

cotθ

The equilibrium condition, in terms of θ, is Fa

k Qqenet =FHGIKJ −

+

F

HGG

I

KJJ =

42

1

302

22cos sin

cotθ θ

θe j.

Thus the equilibrium value of θ satisfies 2 3 12 2cos sin cotθ θ θ+ =e j .

One method for solving for θ is to tabulate the left side. To three significant figures a value of θcorresponding to equilibrium is 81.7°.

The distance from the vertical side of the triangle to the equilibrium position is

s a a= °=12

81 7 0 072 9cot . . .

FIG. P23.64

θ θ θ θ2 3

6070809081

81 581 7

2 2cos sin cot

.

.

+

°°°°°°°

e j4

2.6541.226

01.0911.0240.997

A second zero-field point is on the negative side of the x-axis, where θ = − °9 16. and s a= −3 10. .

P23.65 (a) From the 2Q charge we have F Te − =2 2 0sinθ and mg T− =2 2 0cosθ .

Combining these we findFmg

TT

e = =2 2

2 22

sincos

tanθθ

θ .

From the Q charge we have F Te = =1 1 0sinθ and mg T− =1 1 0cosθ .

Combining these we findFmg

TT

e = =1 1

1 11

sincos

tanθθ

θ or θ θ2 1= .

(b) Fk QQ

rk Qre

e e= =2 2

2

2

2

FIG. P23.65

If we assume θ is small then tanθ ≈r 2

.

Substitute expressions for Fe and tanθ into either equation found in part (a) and solve for r.

Fmg

e = tanθ then 2 1

2

2

2k Qr mg

re FHGIKJ ≈ and solving for r we find r

k Qmge≈

FHG

IKJ

4 2 1 3

.

Chapter 23 25

P23.66 (a) The distance from each corner to the center of the square is

L L L2 2 2

2 2FHGIKJ + FHGIKJ = .

The distance from each positive charge to −Q is then

zL2

2

2+ . Each positive charge exerts a force directed

+q +q

+q+q z

–Q

L/2L/2

x

FIG. P23.66

along the line joining q and −Q , of magnitudek Qq

z Le

2 2 2+.

The line of force makes an angle with the z-axis whose cosine isz

z L2 2 2+

The four charges together exert forces whose x and y components add to zero, while the

z-components add to F k= −+

4

22 2 3 2

k Qqz

z L

e

e j

(b) For z L>> , the magnitude of this force is Fk Qqz

L

k Qq

Lz maz

e ez= − = −

FHG

IKJ =

4

2

4 22 3 2

3 2

3e ja f

Therefore, the object’s vertical acceleration is of the form a zz = −ω2

with ω 23 2

3 3

4 2 128= =a f k Qq

mLk Qq

mLe e .

Since the acceleration of the object is always oppositely directed to its excursion fromequilibrium and in magnitude proportional to it, the object will execute simple harmonicmotion with a period given by

TmLk Qq

mLk Qqe e

= = =2 2

128 81 4

3

1 4

3πω

π πa f a f .

P23.67 (a) The total non-contact force on the cork ball is: F qE mg m gqEm

= + = +FHGIKJ ,

which is constant and directed downward. Therefore, it behaves like a simple pendulum inthe presence of a modified uniform gravitational field with a period given by:

TL

g qE m=

+=

+ × × ×

=

− −2 2

0 500

2 00 10 1 00 10 1 00 10

0 307

6 5 3π π

.

. . .

.

m

9.80 m s C N C kg

s

2 e je j

(b) Yes . Without gravity in part (a), we get TL

qE m= 2π

T =× × ×

=− −

20 500

2 00 10 1 00 1 00 100 314

6 3π

.

. . ..

m

C 10 N C kg s

5e je j (a 2.28% difference).

26 Electric Fields

P23.68 The bowl exerts a normal force on each bead, directed along theradius line or at 60.0° above the horizontal. Consider the free-bodydiagram of the bead on the left:

F n mgy = °− =∑ sin .60 0 0 ,

or nmg

=°sin .60 0

.

Also, F F nx e∑ = − + °=cos .60 0 0 ,

ork qR

nmg mge

2

2 60 060 0 3

= °=°=cos .

tan ..

Thus, q Rmg

ke

=FHGIKJ3

1 2

.

n

mg

60.0°Fe

FIG. P23.68

P23.69 (a) There are 7 terms which contribute:

3 are s away (along sides)

3 are 2s away (face diagonals) and sin cosθ θ= =12

1 is 3s away (body diagonal) and sinφ =13

.

The component in each direction is the same by symmetry.

F i j k i j k= + +LNM

OQP + + = + +

k qs

k qs

e e2

2

2

212

2 21

3 31 90.e j a fe j

FIG. P23.69

(b) F F F Fk qsx y ze= + + =2 2 2

2

23 29. away from the origin

P23.70 (a) Zero contribution from the same face due to symmetry, oppositeface contributes

4 2

k qr

e sinφFHG

IKJ where r

s ss s s= FHG

IKJ + FHGIKJ + = =

2 21 5 1 22

2 22 . .

sinφ =sr

Ek qsr

k qs

k qs

e e e= = =44

1 222 183 3 2 2..a f

(b) The direction is the direction.kFIG. P23.70

Chapter 23 27

P23.71 The field on the axis of the ring is calculated in Example 23.8, E Ek xQ

x ax

e= =+2 2 3 2e j

The force experienced by a charge −q placed along the axis of the ring is F k Qqx

x ae= −

+

L

NMMM

O

QPPP2 2 3 2e j

and when x a<< , this becomes Fk Qq

axe= −FHGIKJ3

This expression for the force is in the form of Hooke’s law, with an

effective spring constant of kk Qq

ae= 3

Since ω π= =2 fkm

, we have fk Qqma

e=1

2 3π.

P23.72 dk dq

x

x

x

k x dx

x

e eE

i j i j=

+

− +

+

F

HGG

I

KJJ =

− +

+2 2 2 2 2 2 3 20 150

0 150

0 150

0 150

0 150.

.

.

.

. m

m

m

m

ma f a fe ja f

λ

E Ei j

= =− +

+z z

=

d kx dx

xe

xall charge

m m

mλ

.

.

. 0 150

0 1502 2 3 20

0 400 e ja f

FIG. P23.72

Ei j

E i j

E i j i j

=+

++

+

L

NMMM

O

QPPP

= × ⋅ × − + −

= − + × = − +

− − −

kx

x

xeλ

.

.

. .

. . . . .

. . . .

. .

2 2

0

0 400

2 2 2

0

0 400

9 9 1 1

3

0 150

0 150

0 150 0 150

8 99 10 35 0 10 2 34 6 67 6 24 0

1 36 1 96 10 1 36 1 96

m

m

m m

N m C C m m m

N C kN C

m m

2 2

a fa f

a f a f

e je j a f a fe j e j

P23.73 The electrostatic forces exerted on the two charges result in a net torqueτ θ θ= − = −2 2Fa Eqasin sin .

For small θ, sinθ θ≈ and using p qa= 2 , we have τ θ= −Ep .

The torque produces an angular acceleration given by τ αθ

= =I Iddt

2

2 .

Combining these two expressions for torque, we haveddt

EpI

2

2 0θ

θ+ FHGIKJ = . FIG. P23.73

This equation can be written in the formddt

2

22θ

ω θ= − where ω 2 =EpI

.

This is the same form as Equation 15.5 and the frequency of oscillation is found by comparison with

Equation 15.11, or fpEI

qaEI

= =1

21

22

π π.

28 Electric Fields

ANSWERS TO EVEN PROBLEMS

P23.2 (a) 2 62 1024. × ; (b) 2.38 electrons for every109 present

P23.36 (a) 200 pC; (b) 141 pC; (c) 58.9 pC

P23.38 see the solution

P23.4 57 5. NP23.40 (a) −

13

; (b) q1 is negative and q2 is positiveP23.6 2 51 10 9. × −

P23.42 electron: 4 39. Mm s ; proton: 2 39. km sP23.8 514 kN

P23.10 x d= 0 634. . The equilibrium is stable if thethird bead has positive charge.

P23.44 (a) −57 6. i Tm s2 ; (b) 2 84. i Mm s; (c) 49.3 ns

P23.46 (a) down; (b) 3 43. Cµ

P23.12 (a) period =π2

3mdk qQe

where m is the mass

of the object with charge −Q ; (b) 4 3ak qQmd

e

P23.48 The particle strikes the negative plate aftermoving in a parabola 0.181 mm high and0.961 mm.

P23.50 Possible only with +51 3. Cµ atx = −16 0. cmP23.14 1 49. g

P23.16 720 kN C down P23.52 (a) 24 2. N C at 0°; (b) 9 42. N C at 117°

P23.18 (a) 18 0 218. i j− kN C; P23.54 5 25. Cµ

(b) 36 0 436. i j−e jmNP23.56 (a)

mgA Bcotθ +

; (b) mgA

A Bcos sinθ θ+P23.20 (a) 12 9. j kN C ; (b) −38 6. j mN

P23.58 0 205. CµP23.22 see the solution

P23.60k qQ

ae

2 toward the 29th vertexP23.24 −

π 2

26k qa

e i

P23.62 443 kN CiP23.26

kxeλ 0

0−ie j

P23.64 0 072 9. a

P23.28k

xeλ 0

02−ie j P23.66 see the solution; the period is

π81 4

3mLk Qqe

P23.30 (a) 383 MN C away; (b) 324 MN C away;(c) 80 7. MN C away; (d) 6 68. MN C away

P23.68 Rmg

ke 3

1 2FHGIKJP23.32 see the solution

P23.70 (a) see the solution; (b) kP23.34 (a)

k Qh

d R d h Re i 2 2 1 2 2 2 1 2+ − + +L

NMOQP

− −e j a fe j ;

P23.72 − +1 36 1 96. .i je j kN C(b)

22

k QR h

e ih d R d h R+ + − + +LNM

OQP

2 2 1 2 2 2 1 2e j a fe j