chapter 17 electric potential - mad science -...

537

Chapter 17

ELECTRIC POTENTIAL

Conceptual Questions

1. (a) The electric field does positive work on –q as it moves closer to +Q.

(b) The potential increases as –q moves closer to +Q.

(c) The potential energy of –q decreases.

(d) If the fixed charge instead has a value –Q, the electric field does negative work, the potential decreases, and the potential energy increases.

2. Such a capacitor can be built by replacing the air between the capacitor plates with a dielectric material. This change not only increases the maximum possible voltage across the capacitor but also increases the amount of charge on the capacitor plates for a given potential difference.

3. While standing on a high voltage wire, the magnitude of a bird’s electric potential varies between –100 kV and +100 kV. Important for the bird is the fact that although its body is at a non-zero potential, the potential difference across its body is small. If a large potential difference existed across its body, the bird would be electrocuted.

4. A positive charge in an electric field moves toward a position of lower potential. A negative charge in this situation moves toward a position of higher potential.

5. Zero work is required to move a charge between two points at the same potential. An external force may need to be applied to move the charge but the work done to start the charge in motion will be negated by the work done to stop it.

6. If the charge of a point particle is negative, its electric potential energy decreases as it is moved towards a region of higher electric potential.

7. If all parts of a conductor in electrostatic equilibrium were not at the same potential, electric fields would exist within the conductor and charges would not remain stationary. The assumption of electrostatic equilibrium would therefore be invalid.

8. There is no physical significance to zero potential—only potential differences have physical consequences. The potential of the earth is often taken to be zero and therefore an object that is grounded has zero potential. This is only a reference value however and the potential of the earth could be taken to be any other quantity as long as other values were appropriately offset by the same potential.

9. If the electric field is zero throughout a region of space, the electric potential must be constant throughout that region.

10. The woman’s head has acquired a net charge. The microscopic charges (electrons or ions) distribute themselves so as to maximize their separation from one another, as a result of their mutual repulsion. This is why the charges move out onto the woman’s hair, which then spreads out in response to the repulsive electrical forces. The charged strands of hair orient themselves parallel to the electric field lines, which emanate radially outward from the woman’s head, as though it were a charged conducting sphere.

11. If the potential is constant throughout a region of space, the electric field must be zero throughout that region.

Chapter 17: Electric Potential College Physics

538

12. If a uniform electric field exists throughout a region of space, the potential must be linearly changing in the direction parallel to the field and unchanging in the directions perpendicular to the field.

13. It doesn’t matter which points we choose because the potential on each plate is constant over the whole plate.

14. The factor of 1/2 appears because the average height of the water in the pool is (1/2)h. The work required to fill the pool is exactly equal to the potential energy of the water in the pool, (1/2)Mgh. This is analogous to the charging of a capacitor. The charge Q on the capacitor is like the mass M of water in the pool, and the electrical potential energy Q∆V of the charge is like the gravitational potential energy Mgh. Thus, by analogy, the total potential energy of a charged capacitor is (1/2)Q∆V, which is correct.

15. Nothing happens to the capacitance, which depends only on the geometry and electrical properties of the materials in the capacitor. Since C = Q/V, if the charge doubles then the voltage doubles as well.

16. Cow A is more likely to be killed because the potential difference between its front and hind legs will be greater than that for cow B.

17. We can’t say anything about the electric field if all we know is the potential at a single point. The electric field tells us how the potential changes if we move from one point to another.

18. As long as the person touching the dome is isolated from the ground, there is no complete circuit for current to flow through, so she is safe.

19. The electric field points from regions of higher potential to regions of lower potential. Therefore, the upper atmosphere is at a higher potential than the Earth.

20. Since the capacitor is connected to a battery the whole time, its voltage ∆V remains constant. The capacitance is proportional to κ, so when the dielectric is removed the capacitance decreases by a factor of 3. Since Q = C∆V, and ∆V remains constant, the charge on the capacitor Q decreases by a factor of 3 as well. The electric field remains constant, since ∆V is constant. The energy stored decreases by a factor of 3.

21. The plates are isolated so the charge remains constant. The capacitance is given by 0 / .C A dκ= � As the plates are moved closer, all these quantities remain constant except d, which decreases. Therefore, the capacitance increases. Since Q = C∆V, and Q is constant, ∆V must decrease. The electric field remains constant. The energy stored in the capacitor decreases as well.

22. The potential close to the positive charge must be positive, while close to the negative charge it must be negative. Therefore, there is a point in region B where the potential is zero. If we move very far away from the two charges, they will look like a single point of charge –3 µC, so the potential very far away must be negative. Thus, there must be a point in region A with a potential of zero as well. The electric field can only be zero in region A, and this does not occur at the same point where the potential is zero.

Multiple-Choice Questions

1. (f) 2. (a) 3. (e) 4. (c) 5. (d) 6. (b) 7. (f) 8. (e) 9. (b) 10. (b) 11. (b) 12. (d)

College Physics Chapter 17: Electric Potential

539

Problems

1. Strategy Use Eq. (17-1). Ignore units and the constant k for simplicity. Solution The electric potential energy is given by E 1 2/ .U kq q r=

1 2 2 1 2 4 2 2 4 2(a) 2; (b) 2; (c) 4; (d) 2; (e) 21 1 2 2 4× × − × − − × − × −= = − = − = = −

Ranking these situations in order of electric potential energy, from highest to lowest, we have (a) = (d), (b) = (e), (c).

2. Strategy Use Eq. (17-1). Solution Compute the electric potential energy.

9 2 2 6 61 2

E(8.988 10 N m C )(5.0 10 C)( 2.0 10 C) 18 mJ

5.0 mq qU k

r

− −× ⋅ × − ×= = = −

3. (a) Strategy Use Eq. (17-1). Solution Compute the electric potential energy.

9 2 2 19 19181 2

E 9(8.988 10 N m C )(1.602 10 C)( 1.602 10 C) 4.36 10 J

0.0529 10 mq qU k

r

− −−

−× ⋅ × − ×= = = − ×

×

(b) Strategy and Solution The negative sign signifies that the force between the two charges is attractive; the potential energy is lower than if the two were separated by a larger distance.

4. Strategy The work done by the applied force is positive, since the direction of the applied force was in the direction of motion. (The force between the two charges is repulsive.) The potential energy of the charges is positive, so the work done on the charges is equal to their potential energy. Use Eq. (17-1). Solution Compute the work done on the charges.

9 2 2 6 21 2

E(8.988 10 N m C )(6.5 10 C) 8.4 J

0.045 mq q

W U kr

−× ⋅ ×= = = =

5. Strategy The work done by the external agent is positive since the potential energy increases. Use Eq. (17-1). Solution Find the work done by the external agent.

2 9 2 2 19 213

f i 15f

(8.988 10 N m C )(1.602 10 C) 0 2.3 10 J1.0 10 m

keW U U U Ur

−−

∞ −× ⋅ ×= ∆ = − = − = − = ×

×

6. Strategy The work done on the charges is equal to their potential energy. Let the upper charge by 1, the lower left-hand charge be 2, and the right-hand charge be 3. Also, let 0.16 ma = and 0.12 m.b = Use Eq. (17-2). Solution Compute the work done on the charges.

1 3 2 31 2E 2 2

6 6 6 69 2 2

2 2

6 6

(5.5 10 C)( 6.5 10 C) (5.5 10 C)(2.5 10 C)(8.988 10 N m C )0.12 m (0.16 m) (0.12 m)

( 6.5 10 C)(2.5 10 C) 3.0 J0.16 m

q q q qq qW U k

b aa b− − − −

− −

⎛ ⎞= = + +⎜ ⎟⎜ ⎟+⎝ ⎠

⎡ × − × × ×⎢= × ⋅ +⎢ +⎣

⎤− × ×+ = −⎥⎥⎦


540

7. Strategy The energy required to break the hydrogen bond must be great enough to overcome the electric potential energy of the hydrogen bond. Use Eq. (17-2). Solution Compute the electric potential energy of the hydrogen bond.

N O N C H O H CE

NO NC HO HC NO NC HO HC9 2 2 19 2

209

( 0.3 )( 0.4 ) ( 0.3 )(0.4 ) (0.3 )( 0.4 ) (0.3 )(0.4 )

(8.988 10 N m C )(0.12)(1.602 10 C) 1 1 1 1 4 10 J0.30 0.42 0.18 0.3010 m

kq q kq q kq q kq q e e e e e e e eU kr r r r r r r r

−−

−

⎡ ⎤− − − −= + + + = + + +⎢ ⎥⎣ ⎦

× ⋅ × ⎡ ⎤= − − + = − ×⎢ ⎥⎣ ⎦

The energy the must be supplied to break the hydrogen bond is 4 × 10–20 J.

8. Strategy Let 1 2 10.0 nCq q q= − = = and d = 4.00 cm. Use Eq. (17-1). Solution Find the total electric potential energy for the two charges.

2 9 2 2 9 21 2

E(8.988 10 N m C )(10.0 10 C) 11.2 J

2 2(0.0400 m)kq q kqU

r d

−× ⋅ ×= = − = − = − µ

9. Strategy Let 1 2 10.0 nC,q q q= − = = d = 4.00 cm, and 3 4.2 nC.q = − Use Eq. (17-2). Solution Find the total electric potential energy of the three charges at point a.

21 3 2 3 3 31 2

E 312 13 23

9 2 2 9 9 9

112 3 2 3

(8.988 10 N m C )(10.0 10 C) 10.0 10 C 2( 4.2 10 C) 17.5 J0.0400 m 2 3

q q q q qq qqq q q kq qU k k qr r r d d d d

− − −

⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞= + + = − + − = − + −⎢ ⎥⎜ ⎟ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦⎢ ⎥⎝ ⎠ ⎣ ⎦

⎡ ⎤× ⋅ × × − ×= − + = − µ⎢ ⎥⎢ ⎥⎣ ⎦

10. Strategy Let 1 2 10.0 nC,q q q= − = = d = 4.00 cm, and 3 4.2 nC.q = − Use Eq. (17-2). Solution Find the total electric potential energy of the three charges at point b.

2 21 3 2 3 3 31 2

E12 13 23

9 2 2 9 2

2 2

(8.988 10 N m C )(10.0 10 C) 11.2 J2(0.0400 m)

q q q q qq qqq q q kqU k kr r r d d d d

−

⎛ ⎞⎛ ⎞= + + = − + − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

× ⋅ ×= − = − µ

11. Strategy Let 1 2 10.0 nC,q q q= − = = d = 4.00 cm, and 3 4.2 nC.q = − Use Eq. (17-2). Solution Find the total electric potential energy of the three charges at point c.

2 21 3 2 3 3 31 2

E12 13 23

9 2 2 9 2

2 2 2 2

(8.988 10 N m C )(10.0 10 C) 11.2 J2(0.0400 m)

q q q q qq qqq q q kqU k kr r r d d d d

−

⎛ ⎞⎛ ⎞= + + = − + − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

× ⋅ ×= − = − µ


541

12. Strategy Use Eq. (17-2).

Solution Find the electric potential energy.

1 3 2 31 2E

12 13 23

6 69 2 2 6

2 2

6 6

3.0 10 C 1.0 10 C (8.988 10 N m C ) (4.0 10 C)3.0 m(4.0 m) (3.0 m)

(3.0 10 C)( 1.0 10 C) 2.8 mJ4.0 m

q q q qq qU kr r r

− −−

− −

⎛ ⎞= + +⎜ ⎟

⎝ ⎠⎡ ⎛ ⎞× − ×⎢ ⎜ ⎟= × ⋅ × +⎢ ⎜ ⎟+⎝ ⎠⎣

⎤× − ×+ =⎥⎥⎦

x (m)0.0 4.02.0

y (m)

2.0

4.0

0.0

3 2

1

13. Strategy Use Eqs. (6-8) and (17-1).

Solution Compute the work done by the electric field.

1 3 2 3field i f 12 12 13 23 13 23

13 239 9

9 2 2 9

( )

8.00 10 C 8.00 10 C(8.988 10 N m C )(2.00 10 C) 2.70 J0.0400 m 0.0400 m 0.1200 m

kq q kq qW U U U U U U U U Ur r

− −−

= −∆ = − = − + + = − − = − −

⎛ ⎞× − ×= − × ⋅ × + = − µ⎜ ⎟⎜ ⎟+⎝ ⎠



1 3 2 3field i f 12 12 13 23 13 23

13 23

9 2 2 9 9

( )

1 1(8.988 10 N m C )(2.00 10 C)(8.00 10 C) 1.80 J0.0800 m 0.0400 m

kq q kq qW U U U U U U U U Ur r

− −

= −∆ = − = − + + = − − = − −

⎛ ⎞= − × ⋅ × × − = µ⎜ ⎟⎝ ⎠



field i f 12 13i 23i 12 13f 23f 13i 13f 23i 23f

1 3 2 313i 13f 23i 23f

9 2 2 9 9

( )

1 1 1 1

1 1(8.988 10 N m C ) (8.00 10 C)(2.00 10 C)0.0400 m 0.0800 m

( 8.00

W U U U U U U U U U U U U U

k q q q qr r r r

− −

= −∆ = − = + + − + + = − + −⎡ ⎤⎛ ⎞ ⎛ ⎞

= − + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎛ ⎞= × ⋅ × × −⎢ ⎜ ⎟⎝ ⎠⎣

+ − × 9 9 1 110 C)(2.00 10 C) 4.49 J0.0400 m 0.1200 m 0.0400 m

− − ⎤⎛ ⎞× − = µ⎥⎜ ⎟+⎝ ⎠⎦


542



field i f 12 13i 23i 12 13f 23f 13i 13f 23i 23f

1 3 2 313i 13f 23i 23f

9 2 2 9 9

( )

1 1 1 1

1 1 1 1(8.988 10 N m C )(8.00 10 C)(2.00 10 C)0.0800 m 0.120 m 0.0400 m 0.120

W U U U U U U U U U U U U U

k q q q qr r r r

− −

= −∆ = − = + + − + + = − + −⎡ ⎤⎛ ⎞ ⎛ ⎞

= − + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= × ⋅ × × − − +m

1.80 J

⎛ ⎞⎜ ⎟⎝ ⎠

= − µ

17. Strategy The potential difference is the change in electric potential energy per unit charge. Use E .U q V∆ = ∆ Solution Find the change in the electric potential energy.

E (3.0 nC)(25 V) 75 nJU q V∆ = ∆ = =

18. Strategy The potential difference is the change in electric potential energy per unit charge. Use E .U q V∆ = ∆ Solution Find the change in the electric potential energy.

19 17E f i( ) ( ) ( 1.602 10 C)[ 360 V ( 240 V)] 1.92 10 JB AU q V q V V e V V − −∆ = ∆ = − = − = − × − − − = ×

19. Strategy Use the principle of superposition and Eq. (17-9). Solution Sum the electric fields at the center due to each charge.

0a b c d a b a b= + + + = + − − =E E E E E E E E E Do the same for the potential at the center.

2 2

9 2 2 67

(0.020 m) (0.020 m)2

4 4(8.988 10 N m C )(9.0 10 C) 2.3 10 Vi

i

kQ kQVr r

−

+

× ⋅ ×= Σ = = = ×

20. Strategy Use the principle of superposition and Eq. (17-9). Solution Sum the electric fields at the center due to each charge.

2 2

2 2

9 2 26 6 8

2(0.020 m) (0.020 m)

2

toward

8.988 10 N m C (9.0 10 C 3.0 10 C) toward 5.4 10 N C toward

a ca b c d a b c b a c

k q k qc

r r

c c− −

+

⎛ ⎞= + + + = + + − = + = +⎜ ⎟⎜ ⎟

⎝ ⎠× ⋅= × + × = ×

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

E E E E E E E E E E E

Do the same for the potential at the center.

(3 ) 0i

i

kQ kV q q q qr r

= Σ = − − − =


543

21. Strategy Use Eqs. (17-7), (17-8), (17-9), and (6-8).

Solution Find the potentials, potential difference, change in electric potential energy, and work done by the electric field.

(a) 9 2 2 9(8.988 10 N m C )( 50.0 10 C) 1.5 kV

0.30 mkQVr

−× ⋅ − ×= = = −

(b) 9 2 2 9(8.988 10 N m C )( 50.0 10 C) 900 V

0.50 mV

−× ⋅ − ×= = −

(c) 9 2 2 91 1 1 1(8.988 10 N m C )( 50.0 10 C) 600 V0.50 m 0.30 mB A

V kQr r

−⎛ ⎞ ⎛ ⎞∆ = − = × ⋅ − × − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

0,V∆ > so the potential increases .

(d) 9 2 7E ( 1.0 10 C)(6.0 10 V) 6.0 10 JU q V − −∆ = ∆ = − × × = − ×

E 0,U∆ < so the potential energy decreases .

(e) 7field E 6.0 10 JW U −= −∆ = ×


Solution Find the electric potential at point P due to the charges. 3 3

9 2 22 2

2.0 10 C 4.0 10 C(8.988 10 N m C )4.0 m (4.0 m) (3.0 m)

2.7 MV

i

i

kQVr

− −⎛ ⎞× − ×⎜ ⎟= Σ = × ⋅ +⎜ ⎟+⎝ ⎠

= −

x (m)0.0 4.02.0

y (m)

2.0

4.0

0.0P2.0 mC

− 4.0 mC

23. Strategy and Solution

(a) Since V is positive, q is positive .

(b) 1 ,Vr

∝ so since the potential is doubled, the distance is halved or 10.0 cm .

24. Strategy Just outside the surface of the sphere, the electric potential is given by ,V Er= where r is the radius of the sphere. Solution Find the electric potential.

5 5(8.40 10 V m)(0.750 m) 6.30 10 VV Er= = × = ×


544

25. (a) Strategy and Solution Since ,i

i

kqV

r= Σ the minimum or most negative value of the potential is the case

where the two negative charges are closer to 0x = than the two positive charges.

x

y

+_+ _

(b) Strategy Let d = 1.0 m and q = 1.0 µC. Use Eq. (17-9). Solution Find the potential at the origin.

9 2 2 6

3 32 2 2 2

2 2 4(8.988 10 N m C )(1.0 10 C)2 23 3 1.0 m

36 kV

id d

i

kQ q q q q kqV kr dd d

−⎛ ⎞ × ⋅ ×⎛ ⎞⎜ ⎟= Σ = + + − = + + − =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠=

26. (a) Strategy Use Eq. (16-5). Solution Find E.R

0 00 E 02 2

0 E

(3 ) , so 3 .kQ k QE E R RR R

= = = =

(b) Strategy Use Eq. (17-8). Solution Find V.R

0 00 V 0

0 V

(3 ) , so 3 .kQ k QV V R RR R

= = = =

27. Strategy Use Eq. (17-9). Solution Find the electric potential at the third corner, B.

9 2 29 98.988 10 N m C( ) (2.0 10 C 1.0 10 C) 9.0 V

1.0 mi

A Bi

kQ kV Q Qr r

− −× ⋅= Σ = + = × − × =

28. (a) Strategy Use Eq. (17-9). Solution Find the electric potential at each point.

9 99 2 2 4.2 10 C 6.4 10 C(8.988 10 N m C ) 270 V

0.060 m 0.159 makq kqVr r

− −+ −

+ −

⎛ ⎞× − ×= + = × ⋅ + =⎜ ⎟⎜ ⎟⎝ ⎠

9 99 2 2 4.2 10 C 6.4 10 C(8.988 10 N m C ) 160 V

0.120 m 0.120 mbV− −⎛ ⎞× − ×= × ⋅ + = −⎜ ⎟⎜ ⎟

⎝ ⎠

(b) Strategy Use Eq. (17-6). Solution Compute the potential difference for the trip from a to b.

160 V 270 V 430 Vb aV V V∆ = − = − − = −


545

(c) Strategy The work done by an external agent is equal to the change in electric potential energy of the point charge when moved from a to b. Use Eq. (17-7). Solution Compute the work done.

9 7(1.50 10 C)( 430 V) 6.5 10 JW q V − −= ∆ = × − = − ×

29. (a) Strategy Use Eq. (17-9). Solution Find the potential at the points.

9 99 2 21 2

1 2

2.50 10 C 2.50 10 C(8.988 10 N m C ) 300 V0.050 m 0.150 ma

kQ kQVr r

− −⎛ ⎞× − ×= + = × ⋅ + =⎜ ⎟⎜ ⎟⎝ ⎠

If 92.50 10 C,Q −= × 1 2

1 20 .b

kQ kQ kQ kQVr r r r

= + = − =

(b) Strategy The work done by an external agent is equal to the change in electric potential energy of the point charge when moved from infinity to b. Use Eq. (17-7). Solution Compute the work done.

( ) (0 0) 0bW q V q V V q∞= ∆ = − = − =

30. (a) Strategy Let 4.00 cm, 12.0 cm, and 8.00 nC.d r q= = = Use Eq. (17-9). Let V = 0 at infinity. Solution Find the potentials at points a and b.

9 2 2 91 2

1 2

( ) 3 3(8.988 10 N m C )(8.00 10 C) 1350 V4 4 4(0.0400 m)a

kq kq kq k q kqVr r d d d

−− × ⋅ ×= + = + = = =

9 2 2 9( ) (8.988 10 N m C )(8.00 10 C) 899 V2 2 2(0.0400 m)bkq k q kqVd d d

−− × ⋅ ×= + = − = − = −

(b) Strategy The potential difference is the change in electric potential energy per unit charge. Use E .U q V∆ = ∆ Solution Compute the change in electric potential energy.

9E (2.00 10 C)( 899 V 1348 V) 4.49 JU q V −∆ = ∆ = × − − = − µ

31. (a) Strategy Let 4.00 cm, 12.0 cm, and 8.00 nC.d r q= = = Use Eq. (17-9). Let V = 0 at infinity. Solution Find the potentials at points b and c.

9 2 2 91 2

1 2

( ) (8.988 10 N m C )(8.00 10 C) 899 V2 2 2(0.0400 m)b

kq kq kq k q kqVr r d d d

−− × ⋅ ×= + = + = − = − = −

( ) 0ckq k qVr r

−= + =

(b) Strategy The potential difference is the change in electric potential energy per unit charge. Use E .U q V∆ = ∆ Solution Compute the change in electric potential energy.

9E (2.00 10 C)(0 899 V) 1.80 JU q V −∆ = ∆ = × + = µ


546

32. Strategy Rewrite each unit in terms of kg, m, s, and C. Solution Show that 1 N C 1 V m.=

2kg m s1 N CC

⋅= and 2 2 2J kg m s kg m s1 V m ,

m C m C C⋅ ⋅= = =

⋅ ⋅ therefore 1 N C 1 V m.=

33. Strategy Since the electric field points in the direction of maximum potential decrease, points farther down the electric field lines are lower in potential. The ratio of the field lines is 12/8 = 1.5, which means that the magnitude of the positive charge is 1.5 times that of the negative charge; use this to find where the potential is zero—changes from positive to negative, moving from left to right. Solution Find where the potential is zero.

(1.5 ) ( ) 1.5 10, so and 0.6 .kq kq k q k qV x dr r x d x x d x

+ −

+ −

−= + = + = = =− −

This is 0.6 times the distance d from the

positive charge to the negative charge. For points to the left of this vertical line, the potential is positive. For points to the right of this line, the potential is negative. Only A and B are to the left of the line, so the potential is positive. Since A is closer to the positive charge, the potential is higher at A than at B. The potential is negative at the other three points. The magnitude of the potential is higher the closer a point is to the negative charge; thus, the potential is most negative at C and least negative at E. Ranking these points in order of the potential, from highest to lowest, we have A, B, E, D, C.

34. (a) Strategy Use Eq. (16-4b). Solution Find the electric force that acts on the particle.

9(4.2 10 C)(240 N C to the right) 1.0 to the rightq −= = × = µΝ F E

(b) Strategy The work done on the particle is equal to the electric force times the displacement of the particle. Solution Compute the work done on the particle.

(1.0 N)(0.25 m) 0.25 JW Fd= = µ = µ

(c) Strategy The electric field points in the direction of decreasing potential, so a bV V> and 0.a bV V− > Use Eq. (17-10). Solution Compute the potential difference.

(240 N C)(0.25 m) 60 Va bV V Ed− = = =

35. (a) Strategy and Solution Positive work is required to move an electron (negative charge) from low potential to high potential, so Y is at the higher potential.

(b) Strategy Use Eqs. (6-8) and (17-5). Solution Find the potential difference.

19field

E field 198.0 10 J, so 5.0 V .

1.602 10 JY XWq V e V U W V V V

e

−

−×∆ = − ∆ = ∆ = − ∆ = − = = =×


547

36. Strategy V Ed∆ = for a uniform electric field. Solution Find the distance between the equipotential surfaces.

1.0 V 1.0 cm100.0 N C

VdE

∆= = =

37. Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together where the field is stronger. The electric field always points in the direction of maximum potential decrease. Solution Outside the sphere, E is radially directed (toward the sphere), and 1.V r−∝ The equipotential surfaces are perpendicular to E at any point, so they are spheres . Inside the sphere, 0=E and V is constant.

38. Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together where the field is stronger. The electric field always points in the direction of maximum potential decrease. Solution Outside the cylinder, E is radially directed away from the axis of the cylinder. The equipotential surfaces are perpendicular to E at any point, so they are cylinders .

39. Strategy The rate at which work is done by the electric organs is equal to the rate of change of the electric potential energy. Use Eq. (17-7). The total amount of work done in one pulse is equal to the rate times the duration of the pulse.

Solution

(a) Compute the rate at which work is done. 3(0.20 10 V)(18 C s) 3.6 kWW q V qV

t t t∆ ∆ ⎛ ⎞= = ∆ = × =⎜ ⎟∆ ∆ ∆⎝ ⎠

(b) Compute the total amount of work done. 3(3.6 10 W)(0.0015 s) 5.4 JWW t

t∆= ∆ = × =∆


548

40. (a) Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together where the field is stronger. The electric field always points in the direction of maximum potential decrease. Solution The electric field lines are radial. They begin on the point charge and end on the inner surface of the shell. Then they begin again on the surface and extend to infinity.

+

(b) Strategy Use Eqs. (16-5) and (17-8), and the principle of superposition. Solution For 1,r r< E is that due to the point charge, 2 .E kq r= For 1 2,r r r< < 0,E = since this is inside a

conductor. For 2,r r> E once again is that due to the point charge, 2 .kq r For 1,r r V kq r< = (point charge). For 1 2 1, ,r r r V kq r< < = since V is continuous, and it is constant in a conductor. For 2,r r>

1 2

kq kq kqVr r r

⎛ ⎞= + −⎜ ⎟

⎝ ⎠ (to preserve continuity). The graphs of the electric field magnitude and potential:

0 r1 r2r

E

kq

r

kq

r

2

2

1

2

0 r1 r2r

V

kqr1

41. Strategy Since the electric field is uniform, we can use Eq. (17-10). Solution Find the magnitude of the charge on the drop in terms of the elementary charge e.

16

19(0.16 m)(9.6 10 N) and , so 2 .

( ) (1.602 10 C)(480 V)F Fe Fe dFeF qE V Ed q e eE Ee V d e e V

−

−×= ∆ = − = = = = = =

−∆ ∆ ×

42. Strategy Since the electric field is uniform, we can use Eq. (17-10). Use Newton’s second law.

Solution Find the magnitude of the charge on the drop.

15 219

3

0, so (1.0 10 kg)(9.80 m s )(0.16 m) 1.6 10 C .

9.76 10 V

F qE mgmg mg mgdq eE V d V

−−

Σ = − =×= = = = = × =

−∆ ∆ ×

qE

mg


549

43. Strategy Use conservation of energy and Eq. (17-7). Solution Find the change in kinetic energy.

19 3 3 142(1.602 10 C)(200.0 10 V 500.0 10 V) 9.612 10 JK U q V − −∆ = −∆ = − ∆ = − × × − × = ×

44. Strategy Use conservation of energy and Eq. (17-7). Solution Find the potential difference.

2 31 6 22

191 (9.109 10 kg)(7.26 10 m s), so 150 V .2 2 2(1.602 10 C)

mvU e V K mv Ve

−

−× ×∆ = − ∆ = −∆ = − ∆ = = =

×

45. (a) Strategy The electric field always points in the direction of maximum potential decrease. Electrons, being negatively charged, move in the direction opposite the direction of the electric field; that is, in the direction of potential increase. Solution Since the speed of the electron decreased, it must have traveled in the direction of the electric field, so it moved in the direction of potential decrease; that is, to a lower potential .

(b) Strategy The kinetic energy of the electron decreased, so its potential energy increased. Use conservation of energy and Eq. (17-7). Solution Compute the potential difference the electron moved through.

2 2 31 6 2 6 2f i

19( ) (9.109 10 kg)[(2.50 10 m s) (8.50 10 m s) ] 188 V

2 2(1.602 10 C)m v vU KV

q e e

−

−−∆ −∆ × × − ×∆ = = = = = −

− ×

46. Strategy According to Example 17.8, the speed of the electrons at the anode is proportional to the square root of the potential difference. Use a proportion. Solution Find the speed of the electrons.

7 72 2 22 1

1 1 1

6.0 kV, so and (6.5 10 m s) 4.6 10 m s .12 kV

v V Vv V v vv V V

∆ ∆∝ ∆ = = = × = ×∆ ∆

47. Strategy According to Example 17.8, the speed of the electrons at the anode is proportional to the square root of the potential difference. Use a proportion. Solution Find the potential difference.

22 2 72 2 2

2 1 71 1 1

3.0 10 m s, so and (12 kV) 2.6 kV .6.5 10 m s

V v vv V V VV v v

⎛ ⎞⎛ ⎞ ⎛ ⎞∆ ×∝ ∆ = ∆ = ∆ = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟∆ ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

48. Strategy The field is uniform. Use conservation of energy and Eq. (17-10). Solution Find the kinetic energy increase.

19 19(1.602 10 C)(500.0 N C)(0.0030 m) 2.4 10 J .K U q V eEd − −∆ = −∆ = − ∆ = = × = ×

49. Strategy Use conservation of energy and Eq. (17-7). Solution Find the final kinetic energy.

f i16 19 3 16

f i

2 , so

2 1.20 10 J 2(1.602 10 C)( 0.50 10 V) 2.8 10 J .

K K K U q V e V

K K e V − − −

∆ = − = −∆ = − ∆ = − ∆

= − ∆ = × − × − × = ×


550

50. Strategy The force between the nuclei is repulsive, since they both have positive charge. Use conservation of energy and Eq. (17-1). Solution Find the closest distance that a helium nucleus approaches the gold nucleus.

1 2 2f i He i

9 2 2 19 2Au He 14

2 2 27 7 2He i He i

1 , so2

2 2 (79 )(2 ) 316(8.988 10 N m C )(1.602 10 C) 4.85 10 m .(6.68 10 kg)(1.50 10 m s)

kq qU K m v

rkq q k e erm v m v

−−

−

= = =

× ⋅ ×= = = = ×× ×

Note that the radius of the gold nucleus is about 157 10 m−× and the radius of the gold atom is about 101 10 m.−×

51. Strategy The electron must have enough kinetic energy at point A to overcome the potential decrease between A and C. Use conservation of energy and Eq. (17-7). Solution Find the required kinetic energy.

19 17(1.602 10 C)( 60.0 V 100.0 V) 2.56 10 JAK U e V − −= ∆ = − ∆ = − × − − = ×

52. Strategy and Solution Since positive charges move through decreases in potential, and since the potential and potential energy are greatest at A, the proton will spontaneously travel from point A to point E. So, 0 .AK =

53. Strategy Use conservation of energy and Eq. (17-7). Ignore units for simplicity. Solution The change in kinetic energy is given by .K U q V∆ = −∆ = − ∆ (a) ( 5) ( 50 100) 750; (b) ( 5) [50 ( 50)] 500; (c) 25 (20 50) 750;(d) ( 1) ( 100 400) 500; (e) 1 [ 250 ( 100)] 150; (f) 5 (250 100) 750

− − × − − = − − − × − − = − × − =− − × − − = − − × − − − = − × − = −

Ranking the changes in kinetic energy, from largest to smallest, we have (c), (b), (e), (d), (a) = (f).

54. Strategy Use conservation of energy, field ,W U= −∆ and the fact that the field is uniform. Solution Find the kinetic energy of each electron when it leaves the space between the plates.

f i field

19 315 15

f i

, so

(1.602 10 C)(100.0 10 V)(0.0030 m)2.0 10 J 6.0 10 J .0.0120 m

VK K K U W eE y e yd

VK K e yd

−− −

∆⎛ ⎞∆ = − = −∆ = = ∆ = ∆⎜ ⎟⎝ ⎠

∆ × ×⎛ ⎞= + ∆ = × + = ×⎜ ⎟⎝ ⎠

55. Strategy Use the definition of capacitance, Eq. (17-14). Solution Find the magnitude of the charge on each plate.

(2.0 µF)(9.0 V) 18 µCQ C V= ∆ = =

56. (a) Strategy Use the definition of capacitance, Eq. (17-14). Solution Find the potential difference between the plates.

0.75 µC, so 50 mV .15.0 µF

QQ C V VC

= ∆ ∆ = = =

(b) Strategy and Solution The plate with the positive charge is at the higher potential, so the +0.75-µC plate.


551

57. Strategy Use the definition of capacitance, Eq. (17-14). Solution

6 4(10.2 10 F)( 60.0 V) 6.12 10 CQ C V − −= ∆ = × − = − ×

612 Cµ of charge must be removed from each plate.

58. (a) Strategy Use Eq. (17-10). Solution Compute the maximum potential difference across the capacitor.

6max max (3 10 V m)(0.0010 m) 3 kVV E d∆ = = × =

(b) Strategy Use the definition of capacitance, Eq. (17-14). Solution Compute the magnitude of the greatest charge.

6 3(2.0 10 F)(3 10 V) 6 mCQ C V −= ∆ = × × =


(a) Since E does not depend upon the separation of the plates 0( ),E σ= � it stays the same .

(b) Since , increasesV d V∆ ∝ ∆ if d increases.

60. Strategy Use the definition of capacitance, Eq. (17-14), and the fact that the potential difference depends upon the plate separation.

Solution

(a) Compute the potential difference between the plates. 6

90.800 10 C 667 V1.20 10 F

QVC

−

−×∆ = = =×

(b) Since , doublesV d V∆ ∝ ∆ if d doubles.


(a) The battery maintains a constant potential difference between the plates, so ∆V stays the same.

(b) The electric field magnitude increases because the same potential difference occurs over a shorter distance. (E = ∆V/d for a uniform field.)

(c) To maintain a constant potential difference while the plate spacing changes, the battery must change the charge on the plates. A larger electric field means that the charge increases. Check: Q = C∆V; C increases and ∆V doesn’t change, so Q increases.


552

62. Strategy Use the definition of capacitance, Eq. (17-14), and the fact that the potential difference depends upon the plate separation.

Solution

(a) Find the magnitude of the charge on each plate. (1.20 nF)(12 V) 14 nCQ C V= ∆ = =

(b) The capacitor remains connected to the battery, so the potential difference stays the same. From E = ∆V/d, increasing d means that the electric field decreases. The electric field is proportional to the charge per unit area on the plates, so the charge decreases.

63. Strategy The capacitance of a parallel plate capacitor is directly proportional to its area. Form a proportion.

Solution Find the capacitance for each situation.

(a) 1

12 2 2 22 1 1

1 1 1 1

1, so (0.694 pF) 0.347 pF .2

AC A AC C C

C A A A= = = = =

(b) 2

132 2 22 1 1

1 1 1 1

2, so (0.694 pF) 0.463 pF .3

AC A AC C C

C A A A= = = = =

64. Strategy Use Eq. (17-10). Solution Compute the plate separation.

61.5 V 1500 km

1.0 10 V mVdE −

∆= = =×

65. Strategy Use the definition of capacitance, Eq. (17-14). Solution Find the capacitance of the spheres.

143.2 10 C, so 8.0 pF .0.0040 V

QQ C V CV

−×= ∆ = = =∆

66. Strategy Use the definition of capacitance and the definition of potential for a spherical conductor. Solution Find the capacitance of the Moon if wrapped in aluminum foil.

64

9 2 21.737 10 m, so 1.933 10 F .

8.988 10 N m CkQr

Q Q rQ C V CV k

−×= ∆ = = = = = ×∆ × ⋅


553

67. Strategy The electrons are accelerated by the electric field between the plates of the capacitor. When they emerge from the positive plate, their speed will be greater than their speed when they entered the capacitor. Solution Find the acceleration of the electrons while they are inside the capacitor.

, so .F ma e VV Ed d d ae e md

∆∆ = = = =

Find the speed of the electrons as they emerge from the capacitor. 2 2

f i

192 6 2 6

f i 31

22 2 2 , so

2 2(1.602 10 C)(40.0 V)(2.50 10 m s) 4.51 10 m s .9.109 10 kg

x x x

x x

e V e Vv v a x ad dmd m

e Vv vm

−

−

∆ ∆⎛ ⎞− = ∆ = = =⎜ ⎟⎝ ⎠

∆ ×= + = × + = ××

68. Strategy Use the definition of electric flux and Gauss’s law.

Solution

(a) The Gaussian surface is a cylinder whose axis is parallel to a radius vector (of the sphere) through it. One end is just within the conductor and the other is just outside it. The ends have area A, with a radius much smaller than that of the sphere, so the electric field is approximately uniform. Find E just outside the conductor.

E0

cos QEA θΦ = =�

Cylindrical surface: E cos90 (0) 0EA EAΦ = ° = = End inside conductor: E cos (0) cos 0,EA Aθ θΦ = = = since the electric field is zero inside a conductor.

End just outside the conductor: E0

cos0 ,QEA EAΦ = ° = =�

so 0 0

.QEA

σ= =� �

(b) Consider an area A of the surface of an arbitrary conductor. If A is small enough such that its surface is approximately flat, then the electric field will be nearly uniform just outside the surface. Comparing an area of the same size on a spherical conductor with the same charge density to that of the arbitrary conductor, we see that the electric field just outside either conductor should be 0 ;σ � as long as A is small enough that it is approximately flat, then this holds for any conductor.

69. (a) Strategy Use Eq. (16-6). Solution The electric field between the plates is

113

12 2 20

4.0 10 C 3.3 10 V m .[8.854 10 C (N m )](0.062 m)(0.022 m)

QEA

−

−×= = = ×

× ⋅�

(b) Strategy Use the definition of the dielectric constant, Eq. (17-17). Solution Find the electric field between the plates of the capacitor with the dielectric.

30 0 23.3 10 V m, so 6.0 10 V m .

5.5E E

EE

κκ

×= = = = ×

70. Strategy Assume the field is uniform. Use Eq. (17-10). Solution Find the maximum possible height for the bottom of the thundercloud.

8

51.00 10 V 300 m .

3.33 10 V mVdE

∆ ×= = =×


554

71. (a) Strategy Use Eq. (17-10). Solution Compute the magnitude of the average electric field between the cow’s front and hind legs.

35(400 200) 10 V 10 V m

1.8 mVEd

∆ − ×= = = 1.1×

Since the electric field always points in the direction of decreasing potential, the average electric field is 510 V m toward the hind legs .1.1×

(b) Strategy and Solution The front and hind legs of Cow B are nearly at the same potential, whereas those for Cow A span a potential difference of approximately 200 kV, thus Cow A is more likely to be killed.

72. Strategy Use the definition of capacitance, Eq. (17-14). Solution Compute the capacitance of the capacitor.

60.020 10 C 83 pF .240 V

QCV

−×= = =∆

73. Strategy The spark flies between the spheres when the electric field between them exceeds the dielectric strength. The magnitude of the electric field is given by ,V d∆ where d is the distance between the spheres. Solution Find d.

6900 V, so 0.30 mm .

3.0 10 V mV VE dd E

∆ ∆= = = =×


Solution

(a) Find the greatest ,dκ since A and 0� are constant. 3.5 7.035, 3.5,

0.10 2.0= = and 2.0 0.2.

10.0= Since 35 > 3.5 > 0.2, the paper is the best choice.

12 2 2 4 20

33.5[8.854 10 C (N m )](120 10 m ) 3.7 nF

0.10 10 mAC

dκ

− −

−× ⋅ ×= = =

×�

(b) Compute the smallest capacitance. 12 2 2 4 2

32.0[8.854 10 C (N m )](120 10 m ) 21 pF

10.0 10 mC

− −

−× ⋅ ×= =

×

75. Strategy Use Eq. (17-16). Solution Compute the capacitance of the capacitor.

12 2 20

32.5[8.854 10 C (N m )](0.30 m)(0.40 m) 89 nF

0.030 10 mAC

dκ

−

−× ⋅= = =

×�

76. Strategy Use Eq. (17-16). Solution Find the average dielectric constant of the tissue in the limb.

120

12 2 2 4 20

(0.030 m)(0.59 10 F), so 5.0 .[8.854 10 C (N m )](4.0 10 m )

A dCCd A

κ κ−

− −×= = = =

× ⋅ ×�

�


555

77. (a) Strategy Use Eq. (17-18c). Solution Compute the capacitance.

2 2 2 2(8.0 10 C), so 7.1 F .2 2 2(450 J)Q QU C

C U

−×= = = = µ

(b) Strategy Use Eq. (17-18a). Solution Compute the potential difference.

42

1 2 2(450 J), so 1.1 10 V .2 8.0 10 C

UU Q V VQ −= ∆ ∆ = = = ×

×

78. Strategy Use Eq. (17-19). The dielectric strength of air is 3 kV mm, which is equal to the maximum electric field. Solution Compute the maximum electric energy density in dry air.

2 12 2 2 6 2 30

1 1 (1.00054)[8.854 10 C (N m )](3 10 V m) 40 J m2 2

u Eκ −= = × ⋅ × =�

79. Strategy The capacitance of a capacitor is inversely proportional to the distance between the plates and 2 (2 ).U Q C=

Solution Form a proportion to find the ratio of the new capacitance to the old.

2 1 1

1 2 1

11.50 1.50

C d dC d d

= = =

Form a proportion to find the energy stored in the capacitor in terms of the old. 2

2 2 12 12

1 21

(2 )1.50, so 1.50 .

(2 )U Q C C

U UU CQ C

= = = =

Thus, the energy increases by 50% .

80. Strategy The energy stored in the capacitor is given by 212 ( ) .U C V= ∆ When the plate separation is increased,

the capacitance changes but the potential difference stays the same, so the energy in the capacitor changes as well. The work done on the capacitor in separating the plates is negative the change in energy. Solution Form a proportion. The capacitance is inversely proportional to the plate separation.

2f f f i

2i i fi

( ) 1.00 cm 0.500.2.00 cm( )

U C V C dU C dC V

∆= = = = =

∆ So, the energy is reduced by half.

Find the work done on the capacitor. 02 2

i f i i i ii

12 2 2 4 22

1 0.5000.500 0.500 0.500 ( ) ( )2 2

0.500[8.854 10 C (N m )](314 10 m ) (20.0 V) 2.78 nJ2(0.0100 m)

AW U U U U U C V V

d− −

= − = − = = ∆ = ∆

× ⋅ ×= =

�


556

81. (a) Strategy Use Eq. (17-15). Solution Find the capacitance for the thundercloud.

12 2 20 [8.854 10 C (N m )](4500 m)(2500 m) 0.18 F

550 mAC

d

−× ⋅= = = µ�

(b) Strategy Use Eq. (17-18c). Solution Find the energy stored in the capacitor.

2 28

6(18 C) 10 J

2 2(0.1811 10 F)QU

C −= = = 8.9××

82. (a) Strategy The capacitance after the slab is removed is equal to the capacitance with the slab divided by the dielectric constant. Solution Compute the capacitance.

06.0 F 2.0 F

3.0CCκ

µ= = = µ

(b) Strategy Use Eqs. (17-10) and (17-17). Solution Find the potential difference across the capacitor.

00

0 3.0(1.5 V) 4.5 V

E EE d Ed

V V

κκκ

==

∆ = ∆ = =

(c) Strategy Use the definition of capacitance, Eq. (17-14). Solution Compute the charge on the plates.

(2.0 F)(4.5 V) 9.0 CQ C V= ∆ = µ = µ

(d) Strategy Use Eq. (17-18c). Solution Compute the energy stored in the capacitor.

2 6 2

6(9.0 10 C) 20 J

2 2(2.0 10 F)QU

C

−

−×= = = µ

×

83. (a) Strategy Use the definition of capacitance, Eq. (17-14), and Eq. (17-15). Solution Find the charge on the capacitor.

12 2 2 20

3[8.854 10 C (N m )](0.100 m) (150 V) 18 nC

0.75 10 mAQ C V V

d

−

−× ⋅= ∆ = ∆ = =

×�

(b) Strategy Use Eq. (17-18a). Solution Compute the energy stored in the capacitor.

91 1 (17.7 10 C)(150 V) 1.3 J2 2

U Q V −= ∆ = × = µ


557

84. (a) Strategy Use Eq. (17-15). Solution Compute the capacitance.

12 2 2 20

3 3[8.854 10 C (N m )](0.100 m) 59.0 pF

0.75 10 m 0.750 10 mAC

d

−

− −× ⋅= = =

× + ×�

(b) Strategy From Problem 83, 18 nC.Q = Use Eq. (17-18c) and conservation of energy. Solution Compute the new energy stored in the capacitor.

2 9 2

12(18 10 C) 2.7 J

2 2(59.0 10 F)QU

C

−

−×= = = µ

×

Work was done on the capacitor when the plates were separated; that work has been stored in the capacitoras potential energy.

85. (a) Strategy where 10.0 kW and 2.0 ms.U P t P t= ∆ = ∆ = Use Eq. (17-18b). Solution Find the initial potential difference.

26

1 2 2(10.0 kW)(2.0 ms)( ) , so 630 V .2 100.0 10 F

UU C V VC −= ∆ ∆ = = =

×

(b) Strategy Use Eq. (17-18c). Solution Find the initial charge.

26, so 2 2(100.0 10 F)(10.0 kW)(2.0 ms) 0.063 C .

2QU Q CU

C−= = = × =

86. Strategy Use Eq. (17-18a). Solution Compute the energy stored in the capacitor.

61 1 (0.020 10 C)(240 V) 2.4 J2 2

U Q V −= ∆ = × = µ

87. Strategy The work done by the external agent is equal to the change in potential energy of the capacitor. Use Eq. (17-18c) and the fact that the capacitance is inversely proportional to the plate separation. Solution Find the work required to double the plate separation.

2 2 2 2 6 2i f

9f i i f i i

(0.80 10 C)1 1 (2 1) 0.27 mJ2 2 2 2 2(1.20 10 F)Q Q Q C Q dW UC C C C C d

−

−⎛ ⎞ ⎛ ⎞ ×= ∆ = − = − = − = − =⎜ ⎟ ⎜ ⎟

×⎝ ⎠ ⎝ ⎠

88. Strategy Use Eq. (17-18b). Solution Find the required potential difference.

26

1 2 2(300 J)( ) , so 8 kV .2 9 10 F

UU C V VC −= ∆ ∆ = = =

×


558

89. (a) Strategy Use the definition of capacitance, Eq. (17-14). Solution Compute the charge that passes through the body tissues.

6 3(15 10 F)(9.0 10 V) 0.14 CQ C V −= ∆ = × × =

(b) Strategy Use Eq. (17-18b) and the definition of average power. Solution Find the average power delivered to the tissues.

2 6 3 2

av 3( ) (15 10 F)(9.0 10 V) 0.30 MW2 2(2.0 10 s)

E U C VPt t t

−

−∆ ∆ × ×= = = = =∆ ∆ ∆ ×

90. Strategy Assume the that thundercloud and Earth system behaves like a capacitor. Use Eq. (17-18a). Solution Find the electric potential energy released by the lightning strike.

81 1 (25.0 C)(1.00 10 V) 1.25 GJ .2 2

U Q V= ∆ = × =


(a) Electrons travel opposite the direction of the electric field, so E is directed upward .

(b) For a uniform electric field, ,y ye VF eE ma

d∆Σ = = = so .y

e Vamd∆= Thus, .y y

y

v v mdt

a e V∆ = =

∆

(c) Since the electron gains kinetic energy, its potential energy decreases .

92. (a) Strategy Assume that the thundercloud and Earth system acts like a capacitor. Use Eq. (17-18a). Solution Find the electric potential energy released by the lightning strike.

91 1 (20.0 C)(1.00 10 V) 10.0 GJ2 2

U Q V= ∆ = × =

(b) Strategy Use the definition of latent heat. Solution Find the mass of sap that is vaporized.

99

V1.00 10 Jenergy absorbed 0.100(10.0 10 J), so 443 kg .

2,256,000 J kgQ mL m ×= = = × = =

(c) Strategy Divide 10.0% of the total energy released from the lightning strike by the homeowner’s monthly energy use. Solution

9

30.100(10.0 10 J) 0.694 month

(400.0 10 W h month)(3600 s h)t ×∆ = =

× ⋅


559

93. (a) Strategy Let L R 10.0 nC.q q= − = Use Eqs. (17-5) and (17-9). Solution Find the electric potential energy of the point charge at each location.

L RL

L R L R

9 2 2 9 9

1 1

1 1(8.988 10 N m C )( 4.2 10 C)(10.0 10 C) 00.0400 m 0.0400 m

b bkq kqU qV q kqqr r r r

− −

⎛ ⎞ ⎛ ⎞= = + = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞= × ⋅ − × × − =⎜ ⎟⎝ ⎠

9 2 2 9 9 1 1(8.988 10 N m C )( 4.2 10 C)(10.0 10 C) 00.0800 m 0.0800 mc cU qV − − ⎛ ⎞= = × ⋅ − × × − =⎜ ⎟

⎝ ⎠

The change in electric potential energy is ∆U = 0 – 0 = 0 .

(b) Strategy Compute the electric potential energy at point a as in part (a). The work done by the external force is negative the work done by the field. Use Eq. (6-8). Solution Find the electric potential energy of the point charge at point a.

L RL

L R L R

9 2 2 9 9

1 1

1 1(8.988 10 N m C )( 4.2 10 C)(10.0 10 C) 6.3 J0.0400 m 0.1200 m

a akq kqU qV q kqqr r r r

− −

⎛ ⎞ ⎛ ⎞= = + = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞= × ⋅ − × × − = − µ⎜ ⎟⎝ ⎠

Find the work required to move the point charge.

field 6.3 J 0 6.3 Ja bW W U U U= − = ∆ = − = − µ − = − µ

94. (a) Strategy Use Eq. (17-10). Solution Compute the minimum thickness of the titanium dioxide.

65.00 V 1.25 m

10 V mVdE

∆= = = µ4.00×

(b) Strategy Use Eq. (17-16). Solution Find the area of the plates.

620

12 2 20

(1.25 10 m)(1.0 F), so 1600 m .90.0[8.854 10 C (N m )]

A dCC Ad

κκ

−

−×= = = =× ⋅

��

95. (a) Strategy The energy required to break the hydrogen bond must be great enough to overcome the electric potential energy of the hydrogen bond. Use Eq. (17-2). Solution Compute the electric potential energy of the hydrogen bond.

OH OH OH H H OH H HE

OHOH OHH HOH HH

OHOH OHH HOH HH9 2 2 19 2

9

( 0.35 )( 0.35 ) ( 0.35 )( 0.35 ) ( 0.35 )( 0.35 ) ( 0.35 )( 0.35 )

(8.988 10 N m C )(0.1225)(1.602 10 C) 1 1 10.27 0.37 010 m

kq q kq q kq q kq qUr r r r

e e e e e e e ekr r r r

−

−

= + + +

⎡ ⎤− − − + + − + += + + +⎢ ⎥⎣ ⎦

× ⋅ ×= − − 201 3 10 J.17 0.27

−⎡ ⎤+ = − ×⎢ ⎥⎣ ⎦

The energy the must be supplied to break the hydrogen bond is 3 × 10–20 J.


560

(b) Strategy Each hydrogen in a water molecule can form one hydrogen bond with an oxygen in another water molecule (H2O); thus, there are two bonds per molecule. The molar mass of water is 0.018 kg/mol. The heat of vaporization of water is 2.256 MJ/kg. Solution Estimate the energy required to break the hydrogen bonds in 1 kg of water.

20 233.3 10 J 2 bonds/molecule 6 10 molecules/mol 0.018 kg/mol 2.2 MJ/kg 2 MJ/kg−× × × × ÷ = = It is not a coincidence that these two quantities are similar, since hydrogen bonds in the liquid phase must be broken to form a gas.

96. Strategy Use Eq. (17-9). Solution Find the potential midway between the charges.

9 2 29 91 2

1 2 0.700 m2

8.988 10 N m C( ) ( 12.0 10 C 22.0 10 C) 873 Vkq kq kV q qr r r

− −× ⋅= + = + = − × − × = −

97. Strategy Use Eqs. (6-8) and (17-7). Solution Compute the work done by the electric field.

19 17field ( ) (1.602 10 C)[100.0 V ( 100.0 V)] 3.204 10 JW U e V − −= −∆ = − − ∆ = × − − = ×

98. Strategy The potential at the surface of a conducting sphere is equal to the magnitude of the electric field times the radius of the sphere. Solution Compute the potential.

6(3.0 10 N C)(0.15 m) 450 kVV Er= = × =

99. Strategy Let 212.0 10 Cq −= × and r = 1.0 nm. Use Eq. (17-9). Solution Find the potential at the sodium ions due to the other three ions.

9 2 2 2131 2

91 2 3

2 (8.988 10 N m C )(2.0 10 C) 9.0 mV2 2 2(1.0 10 m)

kqkq kq q q q kqV kr r r r r r r

−

−× ⋅ ×⎛ ⎞= + + = − + − = = =⎜ ⎟ ×⎝ ⎠

100. (a) Strategy Electric field lines begin on positive charges and end on negative charges. The same number of field lines begins on the plate and ends on the negative point charge. Field lines never cross. Use the principles of superposition and symmetry. Solution The electric field lines for the cylinder and sheet:

(b) Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together where the field is stronger. The electric field always points in the direction of maximum potential decrease. Solution The equipotential surfaces for the cylinder and sheet:


561

101. Strategy Use Newton’s second law and Eq. (4-9). Solution Find the acceleration.

, so .y y yeEF eE ma am

Σ = = =

Find the time to reach the lower plate. 31

219 4

1 2 2(9.109 10 kg)(0.040 m)( ) , so 3.0 ns .2 (1.602 10 C)(5.0 10 N C)y

m yy a t teE

−

−∆ ×∆ = ∆ ∆ = = =

× ×

102. Strategy Assume the field is uniform. Use Eq. (17-10). Solution Compute the magnitude of the electric field in the membrane.

36

990 10 V 9 10 V m .10 10 m

VEd

−

−∆ ×= = = ×

×

103. (a) Strategy Treat the nerve cell as a capacitor. Use Eqs. (17-14) and (17-16) to determine the magnitude of the charge on each surface of the membrane. Solution Find the charge on each surface.

12 2 2 7 2 3110

95.2[8.854 10 C (N m )](1.1 10 m )(70 10 V) 4.9 10 C

7.2 10 mAQ C V V

dκ − − −

−−

× ⋅ × ×= ∆ = ∆ = = ××

�

(b) Strategy Divide the total charge by the charge of one ion. Solution Find the number of ions on each surface of the membrane.

118

194.9 10 C 3.1 10 ions

1.602 10 C ionQe

−

−×= = ×

×

104. (a) Strategy Use Eqs. (17-16) and (17-18b). Solution Find the energy stored in the capacitor.

12 2 2 7 2 3 22 20

9

12

1 1 5.2[8.854 10 C (N m )](1.0 10 m )(90.0 10 V)( ) ( )2 2 2(7.5 10 m)

2.5 10 J

AU C V Vd

κ − − −

−

−

× ⋅ × ×⎛ ⎞= ∆ = ∆ =⎜ ⎟ ×⎝ ⎠

= ×

�

(b) Strategy Divide the total charge by the charge of one ion. Use the definition of capacitance, Eq. (17-14), and Eq. (17-16). Solution Find the number of ions outside of the membrane.

12 2 2 7 2 380

19 95.2[8.854 10 C (N m )](1.0 10 m )(90.0 10 V) 3.4 10 ions

(1.602 10 C ion)(7.5 10 m)A VQ C V

e e edκ − − −

− −∆∆ × ⋅ × ×= = = = ×

× ×�

105. Strategy Use Eqs. (6-8) and (17-7). Solution Find the work done by the electric field.

19 3 20field (1.602 10 C)( 90.0 10 V) 1.44 10 JW U q V e V − − −= −∆ = − ∆ = − ∆ = − × − × = ×


562

106. Strategy Use the definition of capacitance, Eq. (17-14), to find the charge that moves through the membrane. Then divide the charge by e.

Solution

(a) 6 2 2 3 3(1 10 F cm )(0.05 cm )[20 10 V ( 90 10 V)] 6 nCCQ C V A VA

− − −⎛ ⎞= ∆ = ∆ = × × − − × =⎜ ⎟⎝ ⎠

(b) 9

1019

5.5 10 C 3 10 ions1.602 10 C ion

Qe

−

−×= = ×

×

107. Strategy Use Eq. (17-16). Solution Find the capacitance of the axon.

12 2 2 12 2140

95[8.854 10 C (N m )](5 10 m ) 5 10 F

4.4 10 mAC

dκ

− −−

−× ⋅ ×= = = ×

×�

108. (a) Strategy Treat the axon as a parallel plate capacitor. Use Eq. (17-16) and the fact that the area of the curved surface of a cylinder is (2πr)L, where r and L are the radius and length of a cylinder, respectively. Solution Calculate the capacitance per unit length of the axon.

0 0

12 2 2 670

9

[(2 ) ] , so

2 2 (7.0)[8.854 10 C (N m )](5.0 10 m) 3.2 10 F m .6.0 10 m

A r LCd d

rCL d

κ κ π

πκ π − −−

−

= =

× ⋅ ×= = = ××

� �

�

(b) Strategy Use Eq. (17-10) and the fact that the magnitude of the electric field inside a parallel plate capacitor is given by 0 .σ � Solution The outside of the membrane has the positive charge, since the potential is higher outside than inside.

12 2 24 20 0

90

7.0[8.854 10 C (N m )](0.085 V), so 8.8 10 C m .6.0 10 m

E VV Ed d dd

κσ σκ κ

−−

−∆ × ⋅∆ = = = = = = ×

×�

�

109. Strategy Use Newton’s second law, ,xx v t∆ = ∆ and Eqs. (4-7) and (4-9).

Solution

(a) Since E points downward, the negatively charged electron’s change in velocity is directed upward. Find the acceleration.

, so .y y yeEF eE ma am

Σ = = =

Find the time interval.

x

xtv∆∆ =

Find the change in velocity. 19 4

631 7

(1.602 10 C)(2.0 10 N C)(0.060 m) 7.0 10 m s(9.109 10 kg)(3.0 10 m s)y y

x

eE xv a tm v

−

−⎛ ⎞∆ × ×⎛ ⎞∆ = ∆ = = = ×⎜ ⎟⎜ ⎟ × ×⎝ ⎠⎝ ⎠

So, 67.0 10 m s upward .∆ = ×v


563

(b) Find the deflection of the electrons. 2 19 4 2

231 7 2

1 1 (1.602 10 C)(2.0 10 N C)(0.060 m)( ) 7.0 mm2 2 2(9.109 10 kg)(3.0 10 m s)y

x

eE xy a tm v

−

−⎛ ⎞∆ × ×⎛ ⎞∆ = ∆ = = =⎜ ⎟⎜ ⎟ × ×⎝ ⎠⎝ ⎠

110. Strategy The negatively charged particle will accelerate toward the positively charged plate while it is between the plates of the capacitor. Solution The particle is between the plates for a time given by .xt x v∆ = ∆ During this time, the particle travels a vertical distance 0.00100 m.y∆ = Find the acceleration of the particle.

2 222

2 221 1 ( )( ) , so .

2 2 2 ( )x

x x

v yx a xy a t a av v x

⎛ ⎞ ∆∆ ∆∆ = ∆ = = =⎜ ⎟⎜ ⎟ ∆⎝ ⎠

The magnitude of the electrical force on the particle is ,VNeE Ned

∆= where d is the plate separation and N is the

number of excess electrons on the particle. According to Newton’s second law, the acceleration of the particle is

.VdNe Ne Va

m md

∆ ∆= = We set the two expressions for the acceleration of the particle equal and solve for N.

2 2 19 2

2 2 19 22 2 2(5.00 10 kg)(0.00200 m)(35.0 m s) (0.00100 m), so 51 .( ) ( ) (1.602 10 C)(3.00 V)(0.0100 m)

x xv y md v yNe V Nmd x e V x

−

−∆ ∆∆ ×= = = =

∆ ∆ ∆ ×

111. (a) Strategy Compute the electrical and gravitational forces on the particle and compare. Refer to Problem 110.

Solution The gravitational force on the particle is 19 2 18(5.00 10 kg)(9.80 m s ) 4.90 10 N .mg − −= × = ×

The electrical force on the particle is 19

1451(1.602 10 C)(3.00 V) 1.23 10 N .0.00200 m

Ne Vd

−−∆ ×= = ×

Compare the forces. 14

318

1.226 10 N 2.50 104.90 10 N

−

−× = ××

3The electrical force is 2.50 10 times larger than the gravitational force.×

(b) Strategy Use the results from Problem 83. The horizontal component of the velocity doesn’t change. Solution The horizontal component of the velocity is 35.0 m s .xv = Compute the y-component of the particle’s velocity.

2

22 2 2(35.0 m s)(0.00100 m) 7.00 m s

0.0100 m( )x x

yx

v y v yxv a tv xx

⎛ ⎞∆ ∆∆= ∆ = = = =⎜ ⎟⎜ ⎟ ∆∆ ⎝ ⎠

112. Strategy Find the charge on the capacitor. Use Eqs. (17-14) and (17-15). Solution The charge on the capacitor is ,Q Ne= where N is the number of excess electrons.

12 2 2 20 0 6

19[8.854 10 C (N m )](0.0100 m) (3.00 V), so 8.29 10 .

(0.00200 m)(1.602 10 C)A A V

Q Ne C V V Nd de

−

−∆ × ⋅= = ∆ = ∆ = = = ×

×

� �


564

113. (a) Strategy Use the definition of capacitance, Eq. (17-14). Solution Compute the capacitance.

60.020 10 C 83 pF240 V

QCV

−×= = =∆

(b) Strategy Use Eq. (17-15). Solution Find the area of a single plate.

3 113 20

12 2 20

(0.40 10 m)(8.33 10 F), so 3.8 10 m .8.854 10 C (N m )

A dCC Ad

− −−

−× ×= = = = ×

× ⋅�

�

(c) Strategy Use Eq. (17-10). Solution Compute the voltage required to ionize the air between the plates.

3(3.0 10 V mm)(0.40 mm) 1.2 kVV Ed∆ = = × =

114. Strategy The energy in the capacitor is converted into heat in the water. Use Eqs. (14-4) and (17-18b). Solution Find the temperature change of the water.

2 6 22

3 31 ( ) (200.0 10 F)(12.0 V)( ) , so 3.44 mK .2 2 2(1.00 cm )(1.00 g cm )[4.186 J (g K)]

C VQ mc T U C V Tmc

−∆ ×= ∆ = = ∆ ∆ = = =⋅

115. Strategy Treat the nerve cell as a capacitor. Use Eqs. (17-16) and (17-18) to determine the energy stored in a single nerve cell. Multiply the energy per cell by the number of cells to estimate the total amount of stored electrical energy. Solution

2 20total

11 12 2 2 7 2 3 2

9

1 ( ) ( )2 2

10 (5)[8.854 10 C (N m )](1 10 m )(70 10 V) 0.1 J2(8 10 m)

ANU N C V Vd

κ

− − −

−

⎡ ⎤= ∆ = ∆⎢ ⎥⎣ ⎦× ⋅ × ×= =

×

�

116. (a) Strategy For a parallel plate capacitor, 0E σ= � and .V Ed∆ = Solution Find the potential difference between the plates.

6 2

12 2 20

(4.0 10 C m )(0.0060 m) 2.7 kV8.854 10 C (N m )

dV Ed σ −

−×∆ = = = =

× ⋅�

(b) Strategy Use conservation of energy and the fact that .U q V∆ = ∆ Solution Find the kinetic energy of each point charge just before it hits the positive plate.

9f i f i, so 0 ( 2.5 10 C)(2711 V) 6.8 J .K K K U q V K K q V −∆ = − = −∆ = − ∆ = − ∆ = − − × = µ

117. Strategy Use conservation of energy, field ,W U= −∆ and the fact that the field is uniform. Solution Find the final kinetic energy of the alpha particle.

f i field19 3 17

f i

2 , so

2 0 2(1.602 10 C)(10.0 10 V m)(0.010 m) 3.2 10 J .

K K K U W eEd

K K eEd − −

∆ = − = −∆ = =

= + = + × × = ×


565

118. Strategy The energy stored in the capacitor is directly proportional to the capacitance. Use Eq. (17-16) and the fact that V∆ is constant. Form a proportion. Solution Determine what happens to the energy stored in the capacitor.

( )f i

0

i

1 10 i

0 0 f

11 1 0.200,1.25

d dA

d

AU C dU C dκ

κ −∆ ∆= = = − = − = −�

� so the energy is reduced by 20.0%.

119. Strategy Use Eq. (17-18b) and form a proportion. V∆ is constant. Solution Find the energy stored in the capacitor after the dielectric is inserted.

2102

0 0210 0 002

( ), so 3.0 .

( )

C V CU C U U UU C CC V

κ κ κ∆

= = = = = =∆

120. (a) Strategy Use Eq. (17-18b), the definition of capacitance, Eq. (17-14), and the relationships between the quantities (energy, potential difference, capacitance) before and after the dielectric is inserted. Solution Calculate the initial energy stored in the capacitor (without the dielectric).

2 6 2i i i

1 1( ) (4.00 10 F)(100.0 V) 20.0 mJ2 2

U C V −= ∆ = × =

f iC Cκ= and i i f f .Q C V C V= ∆ = ∆ So, i if i

f.C VV V

C κ∆∆ = ∆ =

Calculate the final energy. 2

2 2if f f i i i i

1 1 1 1 1 20.0 mJ( ) ( ) 3.3 mJ2 2 2 6.0

VU C V C C V Uκκ κ κ

∆⎛ ⎞ ⎡ ⎤= ∆ = = ∆ = = =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

(b) Strategy and Solution Since the energy of the capacitor increases when the dielectric is removed, an external agent has to do positive work to remove the dielectric .

121. Strategy Let 0 120.0 V m, E E= be the field outside of the dielectric after it is inserted, and 2E be the field inside the dielectric. Use the principle of superposition for the potential after the dielectric is inserted and the fact that 2 1 .E E κ= Solution Find the electric field inside the dielectric. Initially: 0V E d∆ =

Finally: 1 11 2 1

112 2 2 2 2d d d E d E dV E E E

κ κ⎛ ⎞ ⎛ ⎞∆ = + = + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Solve for 1E in terms of 0.E

010 1 1

211 , so .2 1

EE d E d Eκκ

⎛ ⎞+ = =⎜ ⎟ +⎝ ⎠

Calculate 2.E

012

2 2(20.0 V m) 8.0 V m1 4.0 1

EEEκ κ

= = = =+ +

chapter 17 electric potential - mad science -...

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