Chapter 15 Electric Forces and Electric Fields · Chapter 15 Electric Forces and Electric Fields...

Chapter 15

Electric Forces and Electric Fields

Quick Quizzes

1. (b). Object A must have a net charge because two neutral objects do not attract each other. Since object A is attracted to positively-charged object B, the net charge on A must be negative.

2. (b). By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other.

3. (c). The electric field at point P is due to charges other than the test charge. Thus, it is unchanged when the test charge is altered. However, the direction of the force this field exerts on the test change is reversed when the sign of the test charge is changed.

4. (a). If a test charge is at the center of the ring, the force exerted on the test charge by charge on any small segment of the ring will be balanced by the force exerted by charge on the diametrically opposite segment of the ring. The net force on the test charge, and hence the electric field at this location, must then be zero.

5. (c) and (d). The electron and the proton have equal magnitude charges of opposite signs. The forces exerted on these particles by the electric field have equal magnitude and opposite directions. The electron experiences an acceleration of greater magnitude than does the proton because the electron’s mass is much smaller than that of the proton.

6. (a). The field is greatest at point A because this is where the field lines are closest together. The absence of lines at point C indicates that the electric field there is zero.

7. (c). When a plane area A is in a uniform electric field E, the flux through that area is cosE EA θΦ = where θ is the angle the electric field makes with the line normal to the

plane of A. If A lies in the xy-plane and E is in the z-direction, then 0θ = ° and

( )( )2 2 N C .00 m 20.0 N m C= ⋅5.00E EAΦ = = 4 .

8. (b). If 60θ = in Quick Quiz 15.7 above, then

°

( )( ) ( )2 2cos 5.00 N C 4.00 m cos 60 10.0 N m CEA θ = ° = ⋅EΦ =

9. (d). Gauss’s law states that the electric flux through any closed surface is equal to the net enclosed charge divided by the permittivity of free space. For the surface shown in Figure 15.28, the net enclosed charge is Q 6 C= − which gives ( )0 06 CE QΦ = ∈ = − ∈ .

1

2 CHAPTER 15

10. (b) and (d). Since the net flux through the surface is zero, Gauss’s law says that the net change enclosed by that surface must be zero as stated in (b). Statement (d) must be true because there would be a net flux through the surface if more lines entered the surface than left it (or vise-versa).

Electric Forces and Electric Fields 3

Answers to Even Numbered Conceptual Questions

2. Conducting shoes are worn to avoid the build up of a static charge on them as the wearer walks. Rubber-soled shoes acquire a charge by friction with the floor and could discharge with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen.

4. Electrons are more mobile than protons and are more easily freed from atoms than are protons.

6. No. Object A might have a charge opposite in sign to that of B, but it also might be neutral. In this latter case, object B causes object A to be polarized, pulling charge of one sign to the near face of A and pushing an equal amount of charge of the opposite sign to the far face. Then the force of attraction exerted by B on the induced charge on the near side of A is slightly larger than the force of repulsion exerted by B on the induced charge on the far side of A. Therefore, the net force on A is toward B.

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8. If the test charge was large, its presence would tend to move the charges creating the field you are investigating and, thus, alter the field you wish to investigate.

10. She is not shocked. She becomes part of the dome of the Van de Graaff, and charges flow onto her body. They do not jump to her body via a spark, however, so she is not shocked.

12. An electric field once established by a positive or negative charge extends in all directions from the charge. Thus, it can exist in empty space if that is what surrounds the charge.

14. No. Life would be no different if electrons were positively charged and protons were negatively charged. Opposite charges would still attract, and like charges would still repel. The designation of charges as positive and negative is merely a definition.

16. The antenna is similar to a lightning rod and can induce a bolt to strike it. A wire from the antenna to the ground provides a pathway for the charges to move away from the house in case a lightning strike does occur.

18. (a) If the charge is tripled, the flux through the surface is also tripled, because the net flux is proportional to the charge inside the surface. (b) The flux remains constant when the volume changes because the surface surrounds the same amount of charge, regardless of its volume. (c) The flux does not change when the shape of the closed surface changes. (d) The flux through the closed surface remains unchanged as the charge inside the surface is moved to another location inside that surface. (e) The flux is zero because the charge inside the surface is zero. All of these conclusions are arrived at through an understanding of Gauss’s law.

20. (a) –Q (b) +Q (c) 0 (d) 0 (e) +Q (See the discussion of Faraday’s ice-pail experiment in the textbook.)

4 CHAPTER 15

22. The magnitude of the electric force on the electron of charge e due to a uniform electric field is . Thus, the force is constant. Compare this to the force on a projectile of mass m moving in the gravitational field of the Earth. The magnitude of the gravitational force is mg. In both cases, the particle is subject to a constant force in the vertical direction and has an initial velocity in the horizontal direction. Thus, the path will be the same in each case—the electron will move as a projectile with an acceleration in the vertical direction and constant velocity in the horizontal direction. Once the electron leaves the region between the plates, the electric field disappears, and the electron continues moving in a straight line according to Newton’s first law.

E F e= E


Answers to Even Numbered Problems

2. 135.71 10 C×

4. ( )2 21.91 eF k q= a along the diagonal toward the negative charge

6. 92.25 10 N m−×

8. 5.08 m

10. ( )6 46.7 N leftF = , ( )1.5 157 N rightF = , ( )2 111 N leftF− =

12. 73.89 10 N at 11.3 below axisx−× ° +

14. , stable if third bead has positive charge 0.634x = d

16. 1.45 m beyond the –3.00 nC charge

18. (a) 72.00 10 N C to the right× (b) 40.0 N to the left

20. (a) 13 25.27 10 m s× (b) 55.27 10 m s×

22. 41.63 10 N C× directed opposite to the proton’s velocity

24. 31.88 10 N C at 4.40 below axisx× ° +

26. at 0.85 my = +

28. (a) 1 2 1 3q q =− (b) 2 10, 0q q> <

34. (a) zero (b) 61.8 10 N C× (c) 51.1 10 N C×

36. ~ 1 mµ

38. (a) 6 22.0 10 N m C× ⋅ (b) 0

40. (a) –55.7 nC (b) negative, with a spherically symmetric distribution

42. 5 25.65 10 N m C× ⋅

48. (a) (b) 88.2 10 N−× 62.2 10 m s×

50. 5.25 Cµ

6 CHAPTER 15

52. (a) downward (b) 3.43 Cµ

54. 102.51 10−×

56. (a) 0.307 s (b) Yes, the absence of gravity produces a 2.28% difference.

60. (d) 0 at 9.47 m, 0.x y= = + =E

62. 2.0 Cµ

64. (a) 37.0° or 53.0° (b) 7 71.66 10 s and 2.21 10 s − −× ×


Problem Solutions

15.1 Since the charges have opposite signs, the force is one of attraction .

Its magnitude is

( )( )

( )

9 921 2 9 8

22 2

4.5 10 C 2.8 10 CN m8.99 10 1.1 10 N

C 3.2 mek q q

Fr

− −−

× × ⋅= = × = ×

15.2 The electrical force would need to have the same magnitude as the current gravitational force, or

2

2 2 giving E moon E moone

e

q M m GM mk G q

r r k= =

This yields

( )( )( )11 2 2 24 22

139 2 2

6.67 10 N m kg 5.98 10 kg 7.36 10 kg5.71 10 C

8.99 10 N m Cq

−× ⋅ × ×= =

× ⋅×

15.3 ( )( )

2

2 79ek e eF

r=

( )( )

( )( )

21929

22 14

158 1.60 10 CN m8.99 10 91 repulsion

C 2.0 10 m

−

−

× ⋅= × = ×

N

8 CHAPTER 15

15.4 The attractive forces exerted on the positive charge by the negative charges are shown in the sketch and have magnitudes

( )1

2 2

2 3 22 2 and 22

e ek q k q k qF F F

a aa= = = =

2e

( )2 2

2 3 2 2 2cos 45 0.707 1.352

e e ex

k q k q k qF F F

a a a

Σ = + ° = + =

2

and ( )2 2 2

1 3 2 2 2sin 45 0.707 1.352

e e ey

k q k q k qF F F

a a a

Σ = + ° = + =

( ) ( )

2

22

21.91 eR x y

k qF F F

a= Σ + Σ = and ( )1 1tan 1 45− −tan y

x

F

Fθ

Σ = = = ° Σ

so 2

21.91 along the diagonal toward the negaeR

k qa

=

F tive charge

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15.5 (a) ( ) ( )

( )

2192

922 15

2 4 1.60 102 N m8.99 10 36.8 N

2 5.00 10 m

k eeCr

−

−

× ⋅ = = × = ×

F

(b) The mass of an alpha particle is 4.002 6 um = , where is the unified mass unit. The acceleration of either alpha particle is then

27u 1.66 10 kg−= ×

( )275.54 10= × 2

27

36.8 N m s

4.002 6 1.66 10 kgF

am −

= =×


15.6 The attractive force between the charged ends tends to compress the molecule. Its magnitude is

( ) ( )

( )

2192 29 1

22 2 6

1.60 10 C1 N m8.99 10 4.89 10 N

C 2.17 10 mek e

Fr

−−

−

× ⋅= = × = ×

×7 .

The compression of the “spring” is ( ) ( )( )6 80.010 0 0.010 0 2.17 10 m 2.17 10 mx r − −= = × = × ,

so the spring constant is 17

98

4.89 10 N2.25 10 N m

2.17 10 mFx

−−

−

×= = = ×

×k

15.7 1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton and one electron. Thus, each charge has magnitude Aq N e= . The distance separating

these charges is , where is Earth’s radius. Thus,

2 Er R=

( )( )

ER

( )( )( )

2

2

910

e A

E

e

= ×

223 1925

22 6

2

10 1.60 10 CN m8.99 5.12 10 N

C 4 6.38 10 m

k NF

R

−

=

× × ⋅ = × ×

6.02

15.8 The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus, 2 2

e ek e r m g=

or ( )( )

( )( )

29 2 2 192

31 2

8.99 10 N m C 1.60 10 C5.08 m

9.11 10 kg 9.80 m se

e

k er

m g

−

−

× ⋅ ×= = =

×

15.9 (a) The spherically symmetric charge distributions behave as if all charge was located at the centers of the spheres. Therefore, the magnitude of the attractive force is

( )( )

( )

9 921 2 9 5

22 2

12 10 C 18 10 CN m8.99 10 2.2 10 N

C 0.30 mek q q

Fr

− −−

× × ⋅= = × = ×

10 CHAPTER 15

(b) When the spheres are connected by a conducting wire, the net charge will divide equally between the two identical spheres.

Thus, the force is now

91 2 6.0 10 Cnetq q q −= + = − ×

( ) ( )( )

22 929

22 2

6.0 10 C2 N m8.99 10

C 4 0.30 me netk q

Fr

−− × ⋅= = ×

or 79.0 10 N (repulsion)F −= ×

15.10 The forces are as shown in the sketch at the right.

( )( )

( )

6 6291 2

1 22 2 -212

6.00 10 C 1.50 10 CN m8.99 10 89.9 N

C 3.00 10 mek q q

Fr

− −× × ⋅= = × =

×

( )( )

( )

6 621 3 9

2 22 2 -213

6.00 10 C 2.00 10 CN m8.99 10 43.2

C 5.00 10 m

ek q qF

r

− −× × ⋅= = × =

× N

( )( )

( )

6 622 3 9

3 22 2 -223

1.50 10 C 2.00 10 CN m8.99 10 67.4

C 2.00 10 m

ek q qF

r

− −× × ⋅= = × =

×

6 C

N

The net force on the µ charge is 6 1 2 46.7 N (to the left)F= − =

5 C

F F

The net force on the 1. µ charge is 1.5 1 3 157 N (to the right)F= + =

2 C

F F

The net force on the µ− charge is 2 2 3 111 N (to the left)F− = + =F F

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15.11 In the sketch at the right, is the resultant of the forces that are exerted on the charge at the origin by the 6.00 nC and the –3.00 nC charges respectively.

RF

6 and F 3F

( )( )( )

( )( )( )

9 929

2

6

9 929 5

2

6.00 10 C 5.00 10 C10

0.300 m

10 N

3.00 10 C 5.00 10 C10 1.35 10 N

0.100 m

− −

−

− −−

× × = ×

× × = × = ×

( )

6

3

8.99

3.00

8.99

F

F

= ×

2

2

N mC

N mC

⋅

⋅

( )

The resultant is 2 2 5

6 3 1.38 10 NRF F F −= ×= + at 1 3

6

tan 77.5FF

θ − = = °

or 510 N 77.5 below axisx−= × ° −1.38RF at

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15.12 Consider the arrangement of charges shown in the sketch at the right. The distance r is

( ) ( )2 20.500 m 0.500 m 0.707 mr = + =

The forces exerted on the 6.00 nC charge are

( )( )( )

9 929

2 22

7

6.00 10 C 2.00 10 CN m8.99 10

C 0.707 m

2.16 10 N

F− −

−

× × ⋅= ×

= ×

and ( )( )

( )

9 929

3 22

6.00 10 C 3.00 10 CN m8.99 10 3.24

C 0.707 mF

− −× × ⋅= ×

710 N−= ×

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12 CHAPTER 15

Thus, Σ = and

The resultant force on the 6.00 nC charge is then

( ) 72 3 cos 45.0 3.81 10 NxF F F −+ ° = ×

( ) 82 3 sin 45.0 7.63 10 NyF F F −Σ = − ° = − ×

( ) ( )22 73.89 10 NR x yF F F −= Σ + Σ = × at 1tan 11.3y

x

F

Fθ − Σ = = − ° Σ

or 73.89 10 N at 11.3 below + axR x−= × °F is

15.13 The forces on the 7.00 µC charge are shown at the right.

( )( )( )

( )(( )

6 629

1 22

6 629

2 22

7.00 10 C 2.00 10 CN m8.99 10

C 0.500 m

0.503 N

7.00 10 C 4.00 10 CN m8.99 10

C 0.500 m

1.01 N

F

F

− −

− −

× × ⋅= ×

=

× × ⋅= ×

=

(

Thus, )1 2 cos60.0 0.755 NxF F FΣ = + ° =

( )1 2 sin60.0 0.436 NyF F FΣ = − ° = −

( ) ( )

and

The resultant force on the 7.00 µC charge is

22

0.872 NR x yF F F= Σ + Σ = at 1tan 3y

x

F

Fθ − Σ

0.0= = − Σ °

or 0.872 N at 30.0 below the + axisR x= °F

)

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15.14 Assume that the third bead has charge Q and is located at 0 x d< < . Then the forces exerted on it by the +3q charge and by the +1q charge have magnitudes

( )

3 2

3ek Q qF

x= and

( )( )1 2

ek Q qF

d x=

− respectively

These forces are in opposite directions, so charge Q is in equilibrium if . This gives 3F F= 1

( )2 23 d x x− = , and solving for x, the equilibrium position is seen to be

0.6341 1 3

dx d= =

+

This is a position of stable equilibrium if 0Q > . In that case, a small displacement from

the equilibrium position produces a net force directed so as to move Q back toward the equilibrium position.

15.15 Consider the free-body diagram of one of the spheres given at the right. Here, T is the tension in the string and eF is the repulsive electrical force exerted by the other sphere.

, or 0 cos5.0yF TΣ = ⇒ ° = mg cos5.0

mgT =

°

tan 5.0mg °

At equilibrium, the distance separating the two spheres is

0 sin 5.0x eF F TΣ = ⇒ = ° =

2 sin 5.0r L= ° .

Thus, becomes tan 5.0eF mg= °( )

2

22 sin 5.0

ek q

Ltan 5.0mg= °

° and yields

( )

( )( )( )3 2

9 2

0.20 10 kg 9. m s nC

8.99 10 m

−

°

=2

tan 5.07.2

C

× °

tan 5.02 sin 5.0

2 0.300 m sin 5.0

e

mgq L

k= °

= ° 80

N× ⋅

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14 CHAPTER 15

15.16 The required position is shown in the sketch at the right. Note that this places q closer to the smaller charge, which will allow the two forces to cancel. Requiring that gives

6F F= 3

( )( )

( )2 2

nC 3.00 nC

00 me ek q k

x=

6.00

0.6x +

q ( )222 0.600 mx x= +, or

Solving for x gives the equilibrium position as

0.6

x00 m

1.45 m beyond the 2 1

= =−

3.00 nC charge−

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15.17 For the object to “float” it is necessary that the electrical force support the weight, or

( )( )6

32

24 10 C 610 N C or 1.5 10 kg

9.8 m sqE

qE mg mg

−−

×= = = = ×

15.18 (a) Taking to the right as positive, the resultant electric field at point P is given by

( ) ( ) ( )

1 3 2

1 3 22 2 2

1 3 2

62 69

2 22

2.00 10 CN m 6.00 10 C 1.50 10 C8.99 10

C 0.020 0 m 0.030 0 m 0.010 0 m

R

e e e

E E E E

k q k q k q

r r r

−− −

= + −

= + −

6

2

×⋅ × × = × + −

This gives 72.00 10 N CR = + ×E

or 72.00 10 N C to the rightR = ×E

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(b) ( )( )6 72.00 10 C 2.00 10 N C 40.0 NRq −= = − × × = −F E

or 40.0 N to the left=F


15.19 We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions, one due to each concentration. The contribution due to the positive charge at 3 000 m altitude is

( )( )

( )2

9 522 2

40.0 CN m8.99 10 3.60 10 N C downward

C 1 000 me

qE k

r+

⋅= = × = ×

The contribution due to the negative charge at 1 000 m altitude is

( )( )

( )2

9 522 2

40.0 CN m8.99 10 3.60 10 N C downward

C 1 000 me

qE k

r−

⋅= = × = ×

( )

The resultant field is then

57.20 10 N C downward+ −= + = ×E E E

15.20 (a) The magnitude of the force on the electron is F q E eE= = , and the acceleration is

( )( )19

13 231

1.60 10 C 300 N C5.27 10 m s

9.11 10 kge e

F eEa

m m

−

−

×= = = = ×

×

(b) ( )( )13 2 8 50 0 5.27 10 m s 1.00 10 s 5.27 10 m sat −= + = + × × = ×v v

15.21 If the electric force counterbalances the weight of the ball, then

( )( )3 2

46

5.0 10 kg 9.8 m s or 1.2 10 N C

4.0 10 Cmg

qE mg Eq

−

−

×= = = = ×

×

15.22 The force an electric field exerts on a positive change is in the direction of the field. Since this force must serve as a retarding force and bring the proton to rest, the force and hence the field must be in the direction opposite to the proton’s velocity

net

.

The work-energy theorem, f iW KE KE= −

( )

, gives the magnitude of the field as

( ) ( )( )

154

-19

3.25 10 J0 1.63 10 N C

1.60 10 C 1.25 miqE x KE−×

− ∆ = − = = ××

or iKEE

q x=

∆

16 CHAPTER 15

15.23 (a) ( )( )19

10 2-27

1.60 10 C 640 N C6.12 10 m s

1.673 10 kgp

qEFa

m m

−×= = = = ×

×

(b) 6

510 2

1.20 10 m s1.96 10 s 19.6 s

6.12 10 m sv

at µ−×∆

= × =×

= =

(c) ( )( )

262 20

10 2

1.20 10 m s 011.8 m

2 2 6.12 10 m sfv v

xa

× −−= =

×∆ =

(d) ( )( )22 27 61 11.673 10 kg 1.20 10 m s 1.20 10 J

2 2f p fm v − −× × = × 15KE = =

15.24 The altitude of the triangle is ( )0.500 m sin60.0 0.433 mh = ° = and the magnitudes of the fields due to each of the charges are

( )(( )

9 2 21

22

8.99 10 N m C 3.00

0.433 m

N C

k q × ⋅= =

( )( )( )

9 2 2 932

2 222

8.99 10 N m C 8.00 10 C1.15 10 N C

0.250 mek q

Er

−× ⋅ ×= = = ×

and ( )( )

( )

9 2 2 93

3 223

8.99 10 N m C 5.00 10 C719 N C

0.250 mek q

Er

−× ⋅ ×= = =

)9

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1

10 C

144

eEh

−×

=


Thus, 32 3 1.87 10 N CxE E E+ = ×Σ = and 1 144 N CyE EΣ = − = −

giving

( ) ( )22 31.88 10 N CR x yE E E= Σ + Σ = ×

and

( ) ( )1 1tan tan 0.0769y xE Eθ − −= Σ Σ = − 4.40= − °

Hence 31.88 10 N C at 4.40° below the + axR x= ×E is

15.25 From the symmetry of the charge distribution, students should recognize that the resultant electric field at the center is

0R =E

R = +E E

If one does not recognize this intuitively, consider: , so

1 2 +E E3

1 2x x xE E E= − = 2 2cos30 cos30 0e ek q k q

r r° − ° =

and

1 2y y yE E E= + − 3 2 2sin 30 sin 30 0e e ek q k q k qE

r r r= ° + ° − =2

Thus, 2 2 0R x yE E E= + =

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18 CHAPTER 15

15.26 If the resultant field is to be zero, the contributions of the two charges must be equal in magnitude and must have opposite directions. This is only possible at a point on the line between the two negative charges. Assume the point of interest is located on the y-axis at . Then, for equal magnitudes,

4.0 m 6.0 my− < <

1 22 2

1 2

e ek q k q

r r= or

( ) ( )2 2

C 8.0 C

m 4.0 my y

9.0

6.0

µ µ=

− +

Solving for y gives ( )86.0 m

9y y+ = −4.0 m , or 0.85 my = +

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15.27 If the resultant field is zero, the contributions from the two charges must be in opposite directions and also have equal magnitudes. Choose the line connecting the charges as the x-axis, with the origin at the –2.5 µC charge. Then, the two contributions will have opposite directions only in the regions 0x < and

. For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the point of zero resultant field is on the x-axis at

1.0 mx >0x < .

Requiring equal magnitudes gives 1 22 2

1 2

e ek q k q

r r= or

( )2

2.5 6.0 C

1.0 md d2

Cµ µ+

=

Thus, ( ) 2.5 m

6.0d d+ =

1.8 m

1.0

d =

Solving for d yields , or 1.8 m to the left of the charge− 2.5 Cµ

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15.28 The magnitude of is three times the magnitude of because 3 times as many lines

emerge from as enter q . 2q 1q

2q 1 2 13q q=

(a) Then, 1 2 1 3= −q q


(b) 2 0q > because lines emerge from it,

and 1 0<q because lines terminate on it.

15.29 Note in the sketches at the right that electric field lines originate on positive charges and terminate on negative charges. The density of lines is twice as great for the 2q− charge in (b) as it is for the charge in (a).

1q

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15.30 Rough sketches for these charge configurations are shown below.

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15.31 (a) The sketch for (a) is shown at the right. Note that four times as many lines should leave as emerge from although, for clarity, this is not shown in this sketch.

1q

2q

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(b) The field pattern looks the same here as that shown for (a) with the exception that the arrows are reversed on the field lines.

20 CHAPTER 15

15.32 (a) In the sketch for (a) at the right, note that there are no lines inside the sphere. On the outside of the sphere, the field lines are uniformly spaced and radially outward.

(b) In the sketch for (b) above, note that the lines are perpendicular to the surface at the points where they emerge. They should also be symmetrical about the symmetry axes of the cube. The field is zero inside the cube.

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15.33 (a) Zero net charge on each surface of the sphere.

(b) The negative charge lowered into the sphere repels 5 C to the outsideµ− surface,

and leaves 5 C on the insideµ+ surface of the sphere.

(c) The negative charge lowered inside the sphere neutralizes the inner surface, leaving zero charge on the inside . This leaves 5 C on the outsideµ− surface of the

sphere.

(d) When the object is removed, the sphere is left with 5.00 C on the outsideµ−

surface and zero charge on the inside .

15.34 (a) The dome is a closed conducting surface. Therefore, the electric field is everywhere inside it. At the surface and outside of this spherically symmetric charge distribution, the field is as if all the charge were concentrated at the center of the sphere.

zero

(b) At the surface,

( )( )

( )

9 2 2 46

22

8.99 10 N m C 2.0 10 C1.8 10 N C

1.0 mek q

ER

−× ⋅ ×= = = ×


(c) Outside the spherical dome, 2ek q

Er

= . Thus, at 4.0 mr = ,

( )( )

( )

9 2 2 4

2

8.99 10 N m C 2.0 10 CC

4.0 mE

−× ⋅ ×= = 51.1 10 N×

15.35 For a uniformly charged sphere, the field is strongest at the surface.

Thus, maxmax 2

ek qE

R= ,

or ( ) ( )2 62

3maxmax 9 2 2

2.0 m 3.0 10 N C1.3 10 C

8.99 10 N m Ce

R Eq

k−

×= = = ×

× ⋅

15.36 If the weight of the drop is balanced by the electric force, then mg or the

mass of the drop must be

q E eE= =

( )( )19 416

2

1.6 10 C 3 10 N C5 10 kg

9.8 m seE

mg

−−

× ×= = ≈ ×

But, 343

m V rρ ρ π = =

and the radius of the drop is 1 3

34

mπρ

r

=

( )( )

1 316

73

3 5 10 kg5.2 10 m or ~ 1 m

4 858 kg mr r µ

π

−−

× = = ×

15.37 (a) ( )( )19 4 151.60 10 C 3.0 10 N C 4.8 10 NE − −= = × × = ×F q

(b) 15

12 227

4.8 10 N2.9 10 m s

1.673 10 kgp

Fm

−

−

×= ×

×a = =

15.38 The flux through an area is cosE EA θΦ = , where θ is the angle between the direction of the field E and the line perpendicular to the area A.

(a) ( )( )5 2 6cos 6.2 10 N C 3.2 m cos0 2.0 10 N m CE EA θ = × ° = × ⋅ 2Φ =

(b) In this case, 90θ = ° and 0EΦ =

22 CHAPTER 15

15.39 The area of the rectangular plane is ( )( ) 20.350 m 0.700 m 0.245 mA = = .

(a) When the plane is parallel to the yz plane, 0θ = ° , and the flux is

( )( )3 2m cos 2cos 3.50 10 N C 0.245 0 858 N m CE EA θΦ = = × ° = ⋅

(b) When the plane is parallel to the x-axis, 90θ = ° and 0EΦ =

(c) ( )( )3 2cos 3.50 10 N C 0.245 m cos 40.0 657 N m CE EA θ = × ° = ⋅ 2Φ =

15.40 In this problem, we consider part (b) first.

(b) Since the field is radial everywhere, the charge distribution generating it must be spherically symmetric . Also, since the field is radially inward, the net charge

inside the sphere is negative charge .

(a) Outside a spherically symmetric charge distribution, the field is 2ek Q

Er

= . Thus, just

outside the surface where r R= , the magnitude of the field is 2Re Q=E k , so

( ) ( )22

82

0.750 m C5.57 10 C 55.7 nC

8.99 10 Ce

R EQ

k−= = = × =

0Q

9 2

890 N

N m× ⋅

Since we have determined that < , we now have 55.7= − nCQ

15.41 cosE EA θΦ = and Φ = when , maxE EΦ 0θ = °

Thus, ( )

( ), mE , max

2

4 5

4E E

A dπΦ Φ

= = = 2

0 N

40 m

× ⋅

5 2ax 6

.2 1 m C4.1 10 N C

0.π= ×

15.42 ( )22cos 4 cos0 4e

E e

k qEA R k q

Rθ π Φ = = ° =

π

( )( )9 2 2 6 5 24 8.99 10 N m C 5.00 10 C 5.65 10 N m CE π −Φ = × ⋅ × = × ⋅


15.43 We choose a spherical gaussian surface, concentric with the charged spherical shell and of radius r. Then, ( )2 2cos 4 cos0 4EA E r r Eθ π πΣ = ° = .

(a) For r (that is, outside the shell), the total charge enclosed by the gaussian surface is Q q . Thus, Gauss’s law gives

a>0q= + − = 24 0, or r E Eπ = = 0 .

(b) Inside the shell, r , and the enclosed charge is Q qa< = + .

Therefore, from Gauss’s law, 224 , or r E E

rπ

0 04q q

π= = =∈ ∈ 2

ek qr

The field for is r < a 2 directed radially ouek qr

=E tward .

15.44 Construct a gaussian surface just barely inside the surface of the conductor, where E 0= .

Since E inside, Gauss’ law says 0=0

= 0Q∈

inside. Thus, any excess charge residing on

the conductor must be outside our gaussian surface (that is, on the surface of the conductor).

15.45 0 at all points inside the conductor, and cE = os cos90 0θ = ° = on the cylindrical surface. Thus, the only flux through the gaussian surface is on the outside end cap and Gauss’s

law reduces to cos capo

QEA EAθΣ = =

∈.

The charge enclosed by the gaussian surface is Q Aσ= , where A is the cross-sectional area of the cylinder and also the area of the end cap, so Gauss’s law becomes

o

AEA

σ=∈

, or o

Eσ

=∈

15.46 Choose a very small cylindrical gaussian surface with one end inside the conductor. Position the other end parallel to and just outside the surface of the conductor. Since, in static conditions, E 0= at all points inside a conductor, there is no flux through the inside end cap of the gaussian surface. At all points outside, but very close to, a conductor the electric field is perpendicular to the conducting surface. Thus, it is parallel to the cylindrical side of the gaussian surface and no flux passes through this cylindrical side. The total flux through the gaussian surface is then EAΦ = , where A is the cross-sectional area of the cylinder as well as the area of the end cap.

24 CHAPTER 15

The total charge enclosed by the cylindrical gaussian surface is Q Aσ= , where σ is the charge density on the conducting surface. Hence, Gauss’s law gives

0

AEA

σ=∈

or o

Eσ

=∈

15.47 ( )( )

( )

29 2 1921 2

22 2 -15

8.99 10 N m C 1.60 10 C57.5 N

2.00 10 m

e ek q q k e

Fr r

−× ⋅ ×= = = =

×

15.48 (a)

( )( )( )

21 22 2

29 2 2 198

210

8.99 10 N m C 1.60 10 C8.2 10 N

0.53 10 m

e ek q q k e

r r

−−

−

= =

× ⋅ ×= =

×

F

×

(b) ( )2e c ea m v r= =F m , so

( )( )10 86

-31

0.53 10 m 8.2 10 N2.2 10 m s

9.11 10 kge

r Fv

m

− −× ×⋅= = = ×

×

15.49 The three contributions to the resultant electric field at the point of interest are shown in the sketch at the right. The magnitude of the resultant field is

1 2RE E E= − + + 3E

1 22 2

1 2

e ek q k

r r3 1 2

2 2 23 1 2 3

eR e

q k q q q qE k

r r r r32

+ = − + += − +

( ) ( ) ( )

2 9 9 9

2 22

m 4.0 10 C 5.0 10 C 3.0 10 CC 1.2 m2.5 m 2.0 m

− −9 N

8.99 10 RE 2

− ⋅ × × × − + + = ×

24 N CRE = + , or 24 N C in the + directionR x=E

��

�

��

��

��

��

��

��

��


15.50 Consider the free-body diagram shown at the right.

0 cos or cosy

mgF T mg Tθ

θΣ = ⇒ = =

0 sin tanx eF F T mg

θ θΣ = ⇒ = =

eF qE=

tanqE mg

Since , we have

θ= , or tanmg

qE

θ=

(

)( )3 2

3

2.00 10 kg 9.80 m s tan

1.00 10 N Cq 6

15.05.25 10 C 5.25 Cµ

−

×−

× °= = × =

��

��

��

��

15.51 (a) At a point on the x-axis, the contributions by the two charges to the resultant field have equal magnitudes

given by 1 2 2 ek qE E

r= = .

The components of the resultant field are

1 2 2 2sin sin 0e ey y y

k q k qE E E

r rθ θ = − = − =

and ( )

1 2 2 2 2

2cos cos cosee e

x x x

k qk q k qE E E

r r rθ θ θ

= + = + =

Since ( )3 22 2 3 2 2

cos b r b br r r a b

θ= = =

+

( )( )

, the resultant field is

3 22 2

2 in the + directione

R

k q bx

a b=

+E

��

��

�

�

��

��

��

��

26 CHAPTER 15

(b) Note that the result of part (a) may be written as ( )

( )3 22 2

eR

k Q bE

a b=

+ where Q 2q= is

the total charge in the charge distribution generating the field. In the case of a uniformly charged circular ring, consider the ring to consist of a very large number of pairs of charges uniformly spaced around the ring. Each pair consists of two identical charges located diametrically opposite each other on the ring. The total charge of pair number i is Q . At a point on the axis of the ring, this pair of charges generates an electric field contribution that is parallel to the axis and

has magnitude

i

( )3 22 2

e ii

k bQ

a b=

+

( )

E .

The resultant electric field of the ring is the summation of the contributions by all pairs of charges, or

( )3 22 2 2

e eR i i

k b k bQE E Q

a b a b

= Σ = Σ = + +

3 22

where Q is the total charge on the ring.

iQ= Σ

( )3 2 in the + directionR

bxE

2 2

ek Q

a b=

+

15.52 (a) ( )

( )( )

( )22 2

0 221.0 m s 044.1 m s downward

2 2 5.00 my y

y

v v

y

− −= = =

∆

ya g>

a

Since , the electrical force must be directed downward, aiding the

gravitational force in accelerating the bead. Because the bead is positively charged, the electrical force acting on it is in the direction of the electric field. Thus, the field is directed downward .

(b) Taking downward as positive, y yF qE mg maΣ = + = .

Therefore,

( )

( ) ( )3 2.80 m s3.43 10 6

4

1.00 10 kg 44.1 9 C 3.43 C

1.00 10 N C

ym a gq

E

µ−

−

−=

× −×

×

= = =


15.53 Because of the spherical symmetry of the charge distribution, any electric field present will be radial in direction. If a field does exist at distance R from the center, it is the same as if the net charge located within

were concentrated as a point charge at the center of the inner sphere. Charge located at does not contribute to the field at .

r R≤

r R>r R=

��

��

��

��

�� (a) At 1.00r = , cm 0E = since static

electric fields cannot exist within conducting materials.

(b) The net charge located at 3.00 cmr ≤ is Q 8.00 Cµ= + . Thus, at ,

3.00 cmr =

( )( )

( )( )

2

8.99 10 N

3.00

k Q

× ⋅9 2m N C outward

1

eEr

=

××

2

22

C 8.00 10

0 m

−

−

× 67

C7.99 10= =

(c) At 4.50r = , cm 0E = since this is located within conducting materials.

(d) The net charge located at 7.00 cmr ≤ is 4.00 CQ µ= + . Thus, at ,

7.00 cmr =

( )( )

( )( )

2

8.99 10 N

7.

ek Qr

× ⋅9 2m N C outward

00 1

E

−

=

××

×

2

22

C 4.00 10

0 m−

66

C7.34 10= =

28 CHAPTER 15

15.54 The charges on the spheres will be equal in magnitude and opposite in sign. From 2 2

eF k q r= , this charge must be

( )( )242

39 2 2

1.00 10 N 1.00 m1.05 10 C

8.99 10 N m Ce

F rq

k−

×⋅= = = ×

× ⋅

The number of electrons transferred is

3

1519

1.05 10 C6.59 10

1.60 10 Cq

ne

−

−

×= = = ×

×

The total number of electrons in 100-g of silver is

( ) 250 g 2.62 10=23electrons atoms 1 mole47 6.02 10 10

atom mole 107.87 gN

= ×

×

Thus, the fraction transferred is

15

1025

6.59 102.51 10

2.62 10nN

−×= = ×

× (that is, 2.51 out of every 10 billion).

15.55

( ) ( )( )4 5 2

cos

2.00 10 N C 6.00 m 3.00 m cos10.0 3.55 10 N m C

E EA θΦ =

= × ° = × ⋅

15.56 (a) The downward electrical force acting on the ball is ( )( )6 52.00 10 C 1.00 10 N C 0.200 NeF qE −= = × × =

The total downward force acting on the ball is then ( )( )-3 20.200 N+ 1.00 10 kg 9.80 m seF F mg= + = × 0.210 N=

Thus, the ball will behave as if it was in a modified gravitational field where the effective free-fall acceleration is

2-3

0.210 N“ ” 210 m s

1.00 10 kgF

gm

= = =×


The period of the pendulum will be

2

0.500 m2 2 0.307 s

" " 210 m sL

Tg

π π= = =

(b) Y . The force of gravity is a significant portion of the total downward force

acting on the ball. Without gravity, the effective acceleration would be

es

2-3

0.200 N“ ” 200 m s

1.00 10 kgeF

gm

= = =×

giving 2

0.500 m2 0.31

200 m sT π= = 4 s

a 2.28% difference from the correct value with gravity included.

15.57 The sketch at the right gives a free-body diagram of the

positively charged sphere. Here, 2 2

1 eF k q r= is the attractive

force exerted by the negatively charged sphere and 2F qE= is exerted by the electric field.

0 cos10 or cos10y

mgF T mg TΣ = ⇒ ° = =

°

2

2 1 20 + sin10 or 0ex

k qF F F T qE

rΣ = ⇒ = ° = °tan1mg+

At equilibrium, the distance between the two spheres is ( )2 sin10r L= ° . Thus,

( )

( )(( )

) ( )( )( )

2

9 2 2 8 3 2

2 8

tan10

4 sin10

8.99 10 N m C 5.0 10 C 2.0 g 9.80 m s tan10

5.0 10 C4 0.100 m sin10

ek q mgE

qL

−

°= +

°

× ⋅ ×= +

×°

10 k− −× °

or the needed electric field strength is 510 N= ×4.4 CE

��

��

��

��

��

30 CHAPTER 15

15.58 As shown in the sketch, the electric field at any point on the x-axis consists of two parts, one due to each of the charges in the dipole.

2 2e ek q k q

E E Er r+ −+ −

= − = −

( ) ( )

( ) ( )( ) ( ) ( )

2 2

2 2 2 2 2 2

4e ee e

k q k q x a x a axE k q k q

x a x a x a x a x a2

+ − − = − = = − + − + −

2 2x a>>

Thus, if , this gives 4 3

44 ee

k qaaxE k q

x x ≈ =

� �

�

��

��

��

��

��

��

15.59 (a) Consider the free-body diagram for the ball given in the sketch.

0 sin 37.0 or sin 37.0

xx x

qEF T qE TΣ = ⇒ ° = =

°

0 cos37.0 or y yF qE T mg qE qEΣ = ⇒ + ° = +

and

Thus,

cot 37.0y x mg° =

( )( )( )

3 2

5

8

1.00 10 kg 9.80 m s

cot 37.0 5.00 3.00 cot 37.0 10 N C

1.09 10 C 10.9 nC

y x

mgq

E E

−

−

×= =

+ ° + ° ×

= × =

��

�

��

��

��

��

(b) From Σ , we found that 0xF =sin 37.0

xqET =

°.

Hence, ( )( )8 5

31.09 10 C 3.00 10 N C

5.44 10 Nsin 37.0

−−

× ×= =

°T ×

15.60 (a) At any point on the x-axis in the range 0 1.00 mx< < , the contributions made to the resultant electric field by the two charges are both in the positive x direction. Thus, it is not possible for these contributions to cancel each other and yield a zero field.


(b) Any point on the x-axis in the range 0x < is located closer to the larger magnitude charge ( )5.00 Cq µ= than the smaller magnitude charge ( )4.00 Cq µ= . Thus, the

contribution to the resultant electric field by the larger charge will always have a greater magnitude than the contribution made by the smaller charge. It is not possible for these contributions to cancel to give a zero resultant field.

(c) If a point is on the x-axis in the region , the contributions made by the two charges are in opposite directions. Also, a point in this region is closer to the smaller magnitude charge than it is to the larger charge. Thus, there is a location in this region where the contributions of these charges to the total field will have equal magnitudes and cancel each other.

1.00 mx >

(d) When the contributions by the two charges cancel each other, their magnitudes must be equal. That is,

( ) ( )

( )22

5.00 C 4.00

1.00 me e

Ck k

x x

µ µ=

− or

41.00 m

5x x− = +

Thus, the resultant field is zero at 1.00 m

9.47 m1 4 5

x = = +−

15.61 We assume that the two spheres have equal charges, so the repulsive force that one exerts on the other has magnitude

2 2e eF k q r=

3.0r

. From Figure P15.61 in the textbook, observe that the distance separating the two spheres is

From the free-body diagram of one sphere given above, observe that

( ) cm 2 5.0 cm sin10 4.7 cm 0.047 m= + ° = =

0 cos10 or cos10°yF T mg T mgΣ = ⇒ ° = =

and 0 sin10 sin10 tan1cos10x e

mgF F T m Σ = ⇒ = ° = ° = °

0g °

��

��

��

��

32 CHAPTER 15

Thus, 2 2 tan10e r mg= °k q

or ( )( )( )222

9 2 2

0.015 kg 9.8 m s 0.047 m tan10tan108.99 10 N m Ce

mgrq

k

°°= =

× ⋅

88.0 10 Cq −= ×

giving or 7~ 10 Cq −

15.62 Consider the free-body diagram of the rightmost charge given below. 0 cos or cosyF T mg T mgθ θΣ = ⇒ = =

and ( )0 = sin cos sin tx eF F T mg anmgθ θ θΣ = ⇒ = = θ

But, ( ) ( )

2 2 2 2

2 22 2 21 2

54 sinsin 2 sin

e e e e ee

k q k q k q k q k qF

r r LL L

2

2θθ θ= + = + =

Thus, 2

2 2

5tan

4 sinek q

mgL

θθ= or

2 24 sin tan5 e

L mgq

kθ θ

=

5 , 0.10 kg, and 0.300 mm L

If 4θ = ° = =

( ) ( )

then

( ) ( ) ( )

( )24 0.300 m 0

q =2 2

9 2 2

kg 9.80 m s sin 45

.99 10 N m C× ⋅

.10

5 8

tan 45° °

or 62.0 10 C 2.0q Cµ−= × =

�

��

��

��

15.63 (a) When an electron (negative charge) moves distance x∆ in the direction of an electric field, the work done on it is ( ) ( ) ( )cos cos180eW F x eE x eE xθ= ∆ = ∆ ° = − ∆

From the work-energy theorem ( )net f iKEW KE= − with 0fKE = , we have

( ) ieE x KE− ∆ = −( )

, or ( )( )

17

19

1.60 10 J1.60 10 C 0.100

−

−

××

310 N CiKEE

e x= = ×

∆1.00=

m


(b) The magnitude of the retarding force acting on the electron is , and Newton’s second law gives the acceleration as

eF e= E

ea F m eE m= − = − . Thus, the time required to bring the electron to rest is

( ) ( )

00 2 2i iKE m m KEv v

ta eE m eE

−−= = =

−

( )(

or

)

( )( )31 17

819 3

2 9.11 10 kg 1.60 10 J3.37 10 s 33.7 n

1.60 10 C 1.00 10 N Ct

− −

−−

× ×= =

× ×s× =

(c) After bringing the electron to rest, the electric force continues to act on it causing the electron to accelerate in the direction opposite to the field at a rate of

( )( )19 3

14 2-31

1.60 10 C 1.00 10 N C1.76 10 m s

9.11 10 kgeE

am

−× ×= = = ×

×

15.64 (a) The acceleration of the protons is downward (in the direction of the field) and

( )( )19

10 227

1.60 10 C 720 N C6.90 10 m s

1.67 10 kge

y

F eEa

m m

−

−

×= = = = ×

×

The time of flight for the proton is twice the time required to reach the peak of the arc, or

0 02 sin2 2 y

peak

y y

v vt t

a a

θ = = =

( )

The horizontal distance traveled in this time is

2

0 00 0

2 sin sin 2cosx

y y

v vR v t v

a a

θ θθ = = =

�

��

��

34 CHAPTER 15

Thus, if R = × , we must have

31.27 10 m−

( )( )( )

10 2 3

220

6.90 10 m s 1.27 10 msin 2 0.961

9 550 m s

ya R

vθ

−× ×= = =

2 73.9 or 2 180 73.9 106.1°

giving θ θ= ° = ° − ° = . Hence, 37.0 or 53.0°θ = °

(b) The time of flight for each possible angle of projection is:

For 37.0θ = ° : ( ) 70

10 2

2 9 550 m s sin 37.02 sin1.66 10 s

6.90 10 m sy

vt

a

θ −°= = = ×

×

For 53.0θ = ° : ( ) 70

10 2

2 9 550 m s sin 53.02 sin2.21 10 s

6.90 10 m sy

vt

a

θ −°= = = ×

×

Chapter 15 Electric Forces and Electric Fields · Chapter 15 Electric Forces and Electric Fields...

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Transcript of Chapter 15 Electric Forces and Electric Fields · Chapter 15 Electric Forces and Electric Fields...