Electric Potential 17-27

17 ELECTRIC POTENTIAL

Responses to Questions

1. If two points are at the same potential, then no NET work is done in moving a test charge from one point to the other. Along some segments of the path, some positive work might be done, but along other segments of the path, negative work would be done. And if the object is moved strictly along an equipotential line, then no work would be done along any segment of the path. Along any segment of the path where positive or negative work is done, a force would have to be exerted. If the object is moved along an equipotential line, then no force would be exerted along that segment of the path. This is analogous to climbing up and then back down a flight of stairs to get from one point to another point on the same floor of a building. Gravitational potential increased while going up the stairs, and decreased while going down the stairs. A force was required both to go up the stairs and to go down the stairs. If instead you walked on the level from one point to another, then the gravitational potential was constant, and no force was needed to change gravitational potential.

2. A negative charge will move toward a region of higher potential. A positive charge will move toward a region of lower potential. The potential energy of both charges decreases as they move, because they gain kinetic energy.

3. (a) Electric potential, a scalar, is the electric potential energy per unit charge at a point in space. Electric field, a vector, is the electric force per unit charge at a point in space. (b) Electric potential energy is the work done against the electric force in moving a charge from a specified location of zero potential energy to some other location. Electric potential is the electric potential energy per unit charge.

4. The potential energy of the electron is proportional to the voltage used to accelerate it. Thus, if the voltage is multiplied by a factor of 4, then the potential energy is increased by a factor of 4 also. Then, by energy conservation, we assume that all of the potential energy is converted to kinetic energy during the acceleration process. Thus the kinetic energy has increased by a factor of 4 also. Finally, since the speed is proportional to the square root of kinetic energy, the speed must increase by a factor of 2.

5. The electric field is zero at the midpoint of the line segment joining the two equal positive charges. The electric field due to each charge is of the same magnitude at that location, because the location is equidistant from both charges, but the two fields are in the opposite direction. Thus the net electric

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field is zero there. The electric potential is never zero along that line, except at infinity. The electric potential due to each charge is positive, so the total potential, which is the algebraic sum of the two potentials, is always positive.

6. A negative particle will have its electric potential energy decrease if it moves from a region of low electric potential to one of high potential. By Eq. 17–3, if the charge is negative and the potential difference is positive, then the change in potential energy will be negative and therefore decrease.  7. There is no general relationship between the value of V and the value of E. Instead, the magnitude of  E is equal to the rate at which V decreases over a short distance. Consider the point midway between  two positive charges. E is 0 there, but V is high. Or consider the point midway between two negative  charges. E is also 0 there, but V is low, because it is negative. Finally, consider the point midway  between positive and negative charges of equal magnitude. There E is not 0, because it points toward the negative charge, but V is zero.

8. Two equipotential lines cannot cross. That would indicate that a region in space had two different values for the potential. For example, if a 40-V line and a 50-V line crossed, then the potential at the point of crossing would be both 40 V and 50 V, which is impossible. As an analogy, imagine contour lines on a topographic map. They also never cross because one point on the surface of the Earth cannot have two different values for elevation above sea level. Likewise, the electric field is perpendicular to the equipotential lines. If two lines crossed, then the electric field at that point would point in two different directions simultaneously, which is not possible.

9. The equipotential lines (in black) are perpendicular to the electric field lines (in red).

10. Any imbalance of charge that exists would quickly be resolved. Suppose the positive plate, connected to the positive terminal of the battery, had more charge than the negative plate. Then negative charges from the negative battery terminal would be attracted to the negative plate by the more charged positive plate. This would continue only until the negative plate had the same magnitude of charge as the positive plate. If the negative plate became “overcharged,” then the opposite transfer of charge would take place, again until equilibrium was reached. Another way to explain the balance of charge is that neither the battery nor the capacitor can create or destroy charge. Since they were neutral before they were connected, they must be neutral after they are connected. The charge removed from one plate appears as excess charge on the other plate. This is true regardless of the conductor size or shape.

11. (a) Once the two spheres are placed in contact with each other, they effectively become one larger conductor. They will have the same potential because the potential everywhere on a conducting surface is constant. (b) Because the spheres are identical in size, an amount of charge Q/2 will flow from the initially charged sphere to the initially neutral sphere so that they will have equal charges. 12. A force is required to increase the separation of the plates of an isolated capacitor because you are pulling a positive plate away from a negative plate. Since unlike charges attract, it takes a force to move the oppositely charged plates apart. The work done in increasing the separation goes into

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increasing the electric potential energy stored between the plates. The capacitance decreases, and the potential between the plates increases since the charge has to remain the same.

13. If the electric field in a region of space is uniform, then you can infer that the electric potential is increasing or decreasing uniformly in that region. For example, if the electric field is 10 V/m in a region of space, then you can infer that the potential difference between two points 1 meter apart (measured parallel to the direction of the field) is 10 V. If the electric potential in a region of space is uniform, then you can infer that the electric field there is zero.

14. The electric potential energy of two unlike charges is negative if we take the 0 location for potential energy to be when the charges are infinitely far apart. The electric potential energy of two like charges is positive. In the case of unlike charges, work must be done to separate the charges. In the case of like charges, work must be done to move the charges together.

15. (c) If the voltage across a capacitor is doubled, then the amount of energy it can store is quadrupled: 2 PE 1 CV . = 2 Q2 Q2 16. (a) If the capacitor is isolated, then Q remains constant, and PE = 1 becomes PE = 1 and 2 C 2 KC the stored energy decreases. 2 PE 1 CV (b) If the capacitor remains connected to a battery so V does not change, then = 2 becomes 2 PE 1 KCV = 2 and the stored energy increases.

17. (a) When the dielectric is removed, the capacitance decreases by a factor of K. (b) The charge decreases since Q= CV and the capacitance decreases while the potential difference remains constant. The “lost” charge returns to the battery. (c) The potential difference stays the same because it is equal to the battery voltage. (d) If the potential difference remains the same and the capacitance decreases, then the energy stored 2 PE 1 CV . in the capacitor must also decrease, since = 2 (e) The electric field between the plates will stay the same because the potential difference across the plates and the distance between the plates remain constant.

18. We meant that the capacitance did not depend on the amount of charge stored or on the potential difference between the capacitor plates. Changing the amount of charge stored or the potential difference will not change the capacitance.

Solutions to Problems

1. The work done by the electric field can be found from Eq. 17–2b. W V= -ba � W - = qV - -( . 7 � 7 = 10-6 C)( . 65 V) 5 0 10 - 4 J baq ba ba

2. The work done by the electric field can be found from Eq. 17–2b.

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W V= -ba � W - = qV - .(1 � 60 - 10-19 C)( = 45 V . 125 V) 2 72 10 - 17 J baq ba ba = - (1e )( - 170 V) = 1 . 70 102 eV

3. Energy is conserved, so the change in potential energy is the opposite of the change in kinetic energy. The change in potential energy is related to the change in potential. DPE =q D V = -D KE -DKE KE- KE mu 2(9 . 11� 10- 31 kg)(6 0 10 5 m/s) 2 DV = =initial final = = = -1 . 025 V� . 1 0 V q q2 q 2(- 1 . 60 10-19 C)

The final potential should be lower than the initial potential in order to stop the electron.

4. The kinetic energy gained is equal to the work done on the electron by the electric field. The potential difference must be positive for the electron to gain potential energy. Use Eq. 17–2b. W V= -ba � W - = qV - -( . 1 � 60 � 10-19 C)(1 . 85 10 4 V) 2 96 10 - 15 J baq ba ba = - ( - 1e )(1 . 85� 104 V) . 1 85 10 4 eV

5. The kinetic energy gained by the electron is the work done by the electric force. Use Eq. 17–2b to calculate the potential difference. -16 Wba 6. 45 10 J Vba = - = - = 4030 V q (- 1 . 60 10-19 C)

The electron moves from low potential to high potential, so plate B is at the higher potential.

6. We assume that the electric field is uniform. Use Eq. 17–4b, using the magnitude of the electric field. V 220 V E =ba = =3 . 2 104 V/m d 6. 8 10-3 m

7. The magnitude of the voltage can be found from Eq. 17–4b, using the magnitude of the electric field. V E=ba � V = Ed .(525 V/m)(11 � 0 . 10-3 m) 5 78 V d ba

8. The distance between the plates is found from Eq. 17–4b, using the magnitude of the electric field. V V 45 V E=ba� d = ba = . 2 4 10-2 m d E 1900 V/m

9. The gain of kinetic energy comes from a loss of potential energy due to conservation of energy, and the magnitude of the potential difference is the energy per unit charge. The helium nucleus has a charge of 2e. DPE D KE 85 . 0 103 eV DV = = - = - = -4 . 25 104 V q q2 e

The negative sign indicates that the helium nucleus had to go from a higher potential to a lower potential.

10. Find the distance corresponding to the maximum electric field, using Eq. 17–4b for the magnitude of the electric field. V V 45 V E=ba� d = ba = .1 椿 5 10-5 m 2 10 - 5 m d E 3 106 V/m

11. By the work-energy theorem, the total work done, by the external force and the electric field together, is the change in kinetic energy. The work done by the electric field is given by Eq. 17–2b.

Wexternal+ W electric =KE final - KE initial� W external - q( V B = V A ) KE final -4 - 4 Wexternal- KE final 15. 0� 10 J . 4 82 10 J (VB- V A ) = = = - 157 V q -6 . 50 10-6 C

Since the potential difference is negative, we see that Va> V b.

12. The kinetic energy of the electron is given in each case. Use the kinetic energy to find the speed.

2KE 2(850 eV)(1. 60 10-19 J/eV) (a) 1 mu2=KE � u = = . 1 7 10 7 m/s 2 m 9. 11 10-31 kg

2KE 2(0. 50� 103 eV)(1 60 10- 19 J/eV) (b) 1 muv2=KE � u = = . 1 3 10 7 m/s 2 m 9. 11 10-31 kg

13. The kinetic energy of the proton is given. Use the kinetic energy to find the speed.

2KE 2(4. 2� 103 eV)(1 60 10- 19 J/eV) 1 muv2=KE � u = = . 9 0 10 5 m/s 2 m 1. 67 10-27 kg

14. The kinetic energy of the alpha particle is given. Use the kinetic energy to find the speed. 6- 19 22KE 2(5. 53� 10 eV)(1 60 10 J/eV) 7 1 muv =KE � u = = . 1 63 10 m/s 2 m 6. 64 10-27 kg

15. Use Eq. 17–4b without the negative sign in order to find the magnitude of the voltage difference. DV E=瓺 V = E D x =(3 创 106 V/m)(1 10- 3 m) = 3000 V Dx No harm is done because very little charge is transferred between you and the doorknob.

16. (a) The electron was accelerated through a potential difference of 4.8 kV (moving from low potential to high potential) in gaining 4.8 keV of kinetic energy. The proton is accelerated through the opposite potential difference as the electron and has the exact opposite charge. Thus the proton gains the same kinetic energy, 4.8 keV . (b) Both the proton and the electron have the same kinetic energy. Use that fact to find the ratio of the speeds. -27 u mp 1. 67 10 kg 1mu2= 1 m u 2 � e = = . 42 8 2p p 2 e e -31 upm e 9. 11 10 kg The electron will be moving 42.8 times as fast as the proton.

18. Use Eq. 17–5 to find the potential. 1Q 3. 00 10-6 C V = =(8 . 99醋 109 N m 2 /C 2 ) = 1 . 80 10 5 V -1 4pe0 r 1. 50 10 m

19. Use Eq. 17–5 to find the charge. 1Q 骣 1 V=� Q =(4pe ) rV .琪 (0 = 15 m)(165 V) . 2 8 10-9 C 0 琪 9 2 2 4pe0 r 桫8. 99醋 10 N m /C

20. The work required is the difference in potential energy between the two locations. The test charge has Q Q potential energy due to each of the other charges, given in Conceptual Example 17–7 as PE = k 1 2 . r So to find the work, calculate the difference in potential energy between the two locations. Let Q represent the 35m C charge, let q represent the 0. 50m C test charge, and let d represent the 46-cm distance. kQq kQq kQq kQq PE= + PE = + initiald/2 d /2 final [ d /2- 0 . 12 m] [ d /2 + 0 . 12 m]

kQq kQq骣 kQq Work=PE - PE = + - 2琪 final initial [d /2- 0.12 m] [ d /2 + 0.12 m]桫 d /2 轾 1 1 2 =kQq 犏 + - 臌[0.23 m- 0.12 m] [0.23 m + 0.12 m] 0.23 m =(8.99醋 109 N m 2 /C 2 )(35 创 10- 6 C)(0.50 10 - 6 C)(3.25 m - 1 ) = 0.5117 J 0.51 J

An external force needs to do positive work to move the charge.

21. (a) The electric potential is given by Eq. 17–5. 1Q 1. 60 10-19 C V = =(8 . 99醋 109 N m 2 /C 2 ) = 5 . 754 椿 10 5 V 5 . 8 10 5 V -15 4pe0 r 2. 5 10 m (b) The potential energy of a pair of charges is derived in Conceptual Example 17.7. Q Q (1. 60 10-19 C) 2 PE =k 1 2 =(8 . 99醋 109 N m 2 /C 2 ) = 9 . 2 � 10- 14 J . 0 58 MeV r 2. 5 10-15 m

22. The potential at the corner is the sum of the potentials due to each of the charges, found using Eq. 17–5. k(3 Q ) kQ k (- 2 Q ) kQ骣 1 kQ V = + + =琪1 + = (2 + 2) ℓ2ℓ ℓ ℓ桫 2 2 ℓ

23. By energy conservation, all of the initial potential energy will change to kinetic energy of the electron when the electron is far away. The other charge is fixed and has no kinetic energy. When the electron is far away, there is no potential energy. k(- e )( Q ) E= E�PE � KE 1 mu 2 initial final initial final r 2 2k (- e )( Q ) 2(8 . 99醋 109 N m 2 /C 2 )( - 1 . 60 � 10- 19 C)( . 6 50 10 - 9 C) u = = mr (9. 11� 10-31 kg)(0 245 m)

= 9 . 15 106 m/s

24. By energy conservation, all of the initial potential energy of the charges will change to kinetic energy when the charges are very far away from each other. By momentum conservation, since the initial momentum is zero and the charges have identical masses, the charges will have equal speeds in opposite directions from each other as they move. Thus each charge will have the same kinetic energy. kQ2 E= E�PE � KE 2(1 muv2 ) initial final initial final r 2 kQ2(8. 99醋 10 9 N m 2 /C 2 )(9 . 5 10- 6 C) 2 u = = =3 . 9 103 m/s mr (1. 0� 10-6 kg)(0 053 m)

25. (a) Because of the inverse square nature of the electric x d field, any location where the field is zero must be

closer to the weaker charge (q2 ). Also, in between q2 < 0 q1 > 0 the two charges, the fields due to the two charges are parallel to each other (both to the left) and cannot cancel. Thus the only places where the field can be zero are closer to the weaker charge, but not between them. In the diagram, they are labeled x. Take the positive direction to the right.

q2 q1 2 2 E= k - k =0� q2 ( d = x ) q 1 x x2( d+ x ) 2 -6 q2 2. 0 10 C x= d =(4 . 0 cm) = 18 cm left of q2 -6 - 6 q1- q 2 3. 0� 10 C . 2 0 10 C (b) The potential due to the positive charge is positive d x x everywhere, and the potential due to the negative 2 1 charge is negative everywhere. Since the negative q < 0 charge is smaller in magnitude than the positive 2 q1 > 0 charge, any point where the potential is zero must be closer to the negative charge. So consider locations between the charges (position x1) and to the left of the negative charge (position x2 ) as shown in the diagram. kq kq q d (- 2.0 10-6 C)(4.0 cm) V=1 + 2 =0 � x = 2 location 1 1 -6 (d- x1 ) x 1 ( q 2 - q 1 ) (- 5.0 10 C) kq kq q d (- 2.0 10-6 C)(4.0 cm) V=1 + 2 =0� x - = 2 - = 8.0 cm location 2 2 -6 (d+ x2 ) x 2 ( q 1 + q 2 ) (1.0 10 C) So the two locations where the potential is zero are 1.6 cm from the negative charge toward the positive charge and 8.0 cm from the negative charge away from the positive charge.

26. Let the side length of the equilateral triangle be ℓ. Imagine bringing the electrons in from infinity one at a time. It takes no work to bring the first -e ℓ -e electron to its final location, because there are no other charges present. Thus W1 = 0. The work done in bringing in the second electron to its ℓ ℓ final location is equal to the charge on the electron times the potential (due to the first electron) at the final location of the second electron. -e 骣 1e 1 e2 Thus W2 =( - e )琪 - = . The work done in bringing the 桫 4pe0ℓ 4 pe 0 L third electron to its final location is equal to the charge on the electron times the potential (due to the 骣 1e 1 e 1 2 e2 first two electrons). Thus W3 =( - e )琪 - - = . The total work done is the sum 桫 4pe0ℓ 4 pe 0 ℓ 4 pe 0 ℓ

W1+ W 2 + W 3. 1e2 1 2 e 2 1 3 e 2 3(8. 99醋 10 9 N m 2 /C 2 )(1 . 60 10- 19 C) 2 W= W + W + W =0 + + = = 1 2 3 -10 4pe0ℓ 4 pe 0 ℓ 4 pe 0 ℓ (1. 0 10 m) 骣 1 eV =6 . 9� 10-18 J . 6 � 9 10 - 18 J琪 43 eV 琪 -19 桫1. 60 10 J

27. (a) The potential due to a point charge is given by Eq. 17–5.

     Eb Ea Eb- E a Ea q  Eb

kq kq 骣1 1 Vba= V b - V a = - = kq琪 - rb r a桫 r b r a

9 2 2- 6骣 1 1 4 =(8 . 99醋 10 N m /C )( - 3 . 8 � 10 C)琪 = . 1 6 10 V 桫0. 88 m 0 . 62 m (b) The magnitude of the electric field due to a point charge is given by Eq. 16–4a. The direction of the electric field due to a negative charge is toward the charge, so the field at point a will point downward, and the field at point b will point to the right. See the vector diagram.

 k q (8.99醋 109 N m 2 /C 2 )(3.8 10- 6 C) E =right = right = 4.4114 104 V/m, right b 2 2 rb (0.88 m)  k q (8.99醋 109 N m 2 /C 2 )(3.8 10- 6 C) E =down = down = 8.8871 104 V/m, down a 2 2 rb (0.62 m)   2 2 4 2 4 2 4 Eb- E a =E a + E b =(4.4114创 10 V/M) (8.8871 10 V/m) = 9.9 10 V/m E 88871 q =tan-1a = tan - 1 = 64 Eb 44114

28. The potential energy of the two-charge configuration (assuming they are both point charges) is given in Conceptual Example 17–7. Note that the charges have equal but opposite charge. Q Q e2 PE =k1 2 = - k r r

2 骣 1 1 DPE = PEfinal - PE initial =ke 琪 - 桫rinitial r final 骣1 1 骣 1 eV =(8 . 99醋 109 N m 2 /C 2 )(1 . 60 � 10- 19 C) 2 琪琪琪-9 - 9 琪 - 19 桫0. 110� 10 m 0 � 100 10 m 桫 1 60 10 J = -1 . 31 eV

Thus 1.3 eV of potential energy was lost.

29. We assume that all of the energy the proton gains in being accelerated by the voltage is changed to potential energy just as the proton’s outer edge reaches the outer radius of the silicon nucleus. r e(14 e ) PE= PE � eV k initial final initial r -19 14e 9 2 2 (14)(1. 60 10 C) 6 Vinitial = k =(8 . 99醋 10 N m /C ) = 4 . 2 10 V r (1. 2 + 3 . 6) 10-15 m

30. Use Eq. 17–5. 骣kq k(- q ) 骣 kq k ( - q ) 骣 1 1 1 1 V= V - V =琪 + - 琪 + = kq 琪 - - + BA B A 桫d- b b 桫 b d - b 桫 d - b b b d - b 骣 1 1 2kq (2 b- d ) = 2kq琪 - = 桫d- b b b( d - b )

31. (a) The electric potential is found from Eq. 17–5. q (1. 60 10-19 C) V= k p =(8 . 99醋 109 N m 2 /C 2 ) 27 V initial -10 ratom 0. 53 10 m (b) The kinetic energy can be found from the fact that the magnitude of the net force on the electron, which is the attraction by the proton, is causing circular motion.

2 2 2 meu qp q e 2e 2 e F= = k� mu � kKE = 1 m u 1 k net2 e2 e 2 ratomratom r atom r aton (1.60 10-19 C) 2 =1 (8.99醋 109 N m 2 /C 2 ) = 2.171 椿 10- 18 J 2.2 10 - 18 J 2 (0.53 10-10 m) 1 eV = 2.171创 10-18 J 13.57 eV 14 eV 1.60 10-19 J (c) The total energy of the electron is the sum of its KE and PE. The PE is found from Eq. 17–2a and is negative since the electron’s charge is negative. e2 e 2 e 2 EPE KE eV1 m2 k 1 k 1 k total= + = - +2 eu = - + 2 = - 2 ratom r atom r atom = -2 . 171椿 10-18 J - 2 . 2 椿 10 - 18 J - 14 eV

(d) If the electron is taken to infinity at rest, then both its PE and KE would be 0. The amount of energy needed by the electron to have a total energy of 0 is just the opposite of the answer to part (c), 2. 2 10-18 J or 14 eV .

32. The dipole moment is the product of the magnitude of one of the charges and the separation distance. p= Qℓ =(1 . 60� 10-19 C)(0 � 53 . 10 - 10 m) 醋 8 5 10 - 30 C m r 33. The potential due to the dipole is given by Eq. 17–6b. kp cosq (8. 99醋 109 N m 2 /C 2 )(4 . 2 醋 10- 30 C m) cos 0 -Q Q (a) V = = r2(2. 4 10- 9 m) 2

=6 . 6 10-3 V

kp cosq (8. 99醋 109 N m 2 /C 2 )(4 . 2 醋 10- 30 C m) cos 45 (b) V = = r2(2. 4 10- 9 m) 2 r q =4 . 6 10-3 V -Q r Q kp cosq (8. 99醋 109 N m 2 /C 2 )(4 . 2 醋 10- 30 C m) cos 135 (c) V = = q r2(2. 4 10- 9 m) 2 -Q Q = -4 . 6 10-3 V

   34. We assume that p1 and p2 are equal in magnitude and that each makes a 52 angle with p. The  magnitude of p1 is also given by p1 = qd, where q is the charge on the hydrogen atom and d is the distance between the H and the O. p p=2 p cos 52爱 p = = qd 1 1 2 cos 52 p 6. 1醋 10-30 C m q = = =5 . 2� 10-20 C , about 3 2 electron’s charge 2d cos 52 2(0. 96窗 10-10 m) cos 52

35. The capacitance is found from Eq. 17–7.

Q 2. 5 10-3 C Q= CV� C = = . 2 6 10-6 F V 960 V

36. The voltage is found from Eq. 17–7. Q 16. 5 10-8 C Q= CV� V = = 19 V C 8. 5 10-9 F

37. We assume the capacitor is fully charged, according to Eq. 17–7.

Q= CV =(5 . 00� 10-6 F)(12 = 0 V) . 6 00 10 - 5 C

38. The area can be found from Eq. 17–8. e A Cd (0. 20 F)(3 . 2 10-3 m) C=0 � A = = . 7 2 107 m 2 -12 2 2 d e0 (8. 85醋 10 C /N m ) Note that this is not very realistic—if it were square, each side would be about 8500 m.

39. Let Q1 and V1 be the initial charge and voltage on the capacitor, and let Q2 and V2 be the final charge and voltage on the capacitor. Use Eq. 17–7 to relate the charges and voltages to the capacitance.

Q1= CV 1 Q 2 = CV 2 Q 2 - Q 1 = CV 2 - CV 1 = C( V 2 - V 1 ) Q- Q 15 10-6 C C =2 1 = =6 . 3 10-7 F V2- V 1 24 V

40. The desired electric field is the value of V/ d for the capacitor. Combine Eq. 17–7 and Eq. 17–8 to find the charge. e AV Q= CV =0 =e AE =(8 . 85醋 10-12 C 2 /N m 2 )(45 . 0 � 10 - 4 m 2 )(8 50 10 5 V/m) d 0 =3 . 39 10-8 C

41. Combine Eq. 17–7 and Eq. 17–8 to find the area. -6 e0 AV Q (4. 2 10 C) 2 Q= CV = =e0 AE� A = = . 0 24 m de0 E 骣2000 V (8. 85醋 10-12 C 2 /N m 2 )琪琪 -3 桫10 m

42. The work to move the charge between the capacitor plates is W= qV , where V is the voltage difference between the plates, assuming that q<< Q so that the charge on the capacitor does not change appreciably. The charge is then found from Eq. 17–7. -6 骣Q CW (15 10 F)(18 J) W= qV = q琪 � Q = = . 0 90 C 桫C q 0. 30 10-3 C

The assumption that q<< Q is justified.

43. Use Eq. 17–8 for the capacitance. eA e A (8. 85醋 10-12 C 2 /N m 2 )(1 . 0 10 - 4 m 2 ) C=0� d = 0 = 9 10-16 m d C (1 F)

No, this is not practically achievable. The gap would have to be smaller than the radius of a proton.

44. From Eq. 17–4a, the voltage across the capacitor is the magnitude of the electric field times the separation distance of the plates. Use that with Eq. 17–7. Q (62 10-6 C) Q= CV = CEd� E = = . 3 9 104 V/m Cd (0. 80� 10-6 F)(2 0 10 - 3 m)

45. The total charge on the combination of capacitors is the sum of the charges on the two individual capacitors, since there is no battery connected to them to supply additional charge, and there is no neutralization of charge by combining positive and negative charges. The voltage across each capacitor must be the same after they are connected, since each capacitor plate is connected to a corresponding plate on the other capacitor by a constant-potential connecting wire. Use the total charge and the fact of equal potentials to find the charge on each capacitor and the common potential difference.

Q1= C 1 V 1 Q 2 = C 2 V 2 Q 1 = C 1 V final Q 2 = C 2 V final initial initial initial initial final final

QTotal= Q 1 + Q 2 = Q 1 + Q 2 = C 1 V 1 + C 2 V 2 = C 1 V final + C 2 V final initial initial final final initial initial C V+ C V 1 1 2 2 -6 - 6 initial initial (2.50�10 F)(746 V) (6.80 10 F)(562 V) V = = final -6 C1+ C 2 (9.30 10 F)

=611.46 V� 611 V = V1 V 2

-6 - 3 Q1= C 1 V final =(2 . 50� 10 F)(611 = 46) . 1 53 10 C final -6 - 3 Q2= C 2 V final =(6 . 80� 10 F)(611 = 46) . 4 16 10 C

46. After the first capacitor is disconnected from the battery, the total charge must remain constant. The voltage across each capacitor must be the same when they are connected, since each capacitor plate is connected to a corresponding plate on the other capacitor by a constant-potential connecting wire. Use the total charge and the final potential difference to find the value of the second capacitor.

QTotal= C 1 V 1 Q 1 = C 1 V final Q 2 = C 2 V final initial final final

QTotal= Q 1 + Q 2 =( C 1 + C 2 ) V final� C 1 V 1 + ( C 1 C 2 ) V final final final initial

骣V1 琪initial -6骣165 V - 5 C2= C 1琪 - 1 = (7.7� 10 F)琪 = 1 7.7 10 F 琪Vfinal 桫15 V 桫

47. Use Eq. 17–9 to calculate the capacitance with a dielectric. -2 2 A -12 2 2(6. 6 10 m) - 11 C= Ke0 =(2 . 2)(8 . 85醋 10 C /N m ) = 4 . 7 10 F d (1. 8 10-3 m)

48. Use Eq. 17–9 to calculate the capacitance with a dielectric. -2 2 A -12 2 2p (5. 0 10 m) - 10 C= Ke0 =(7)(8 . 85醋 10 C /N m ) = 1 . 7 10 F d (2. 8 10-3 m)

49. Since the capacitor is disconnected from the battery, the charge on it cannot change. The capacitance of the capacitor is increased by a factor of K, the dielectric constant.

Cinitial C initial 1 Q= Cinitial V initial = C final V final� V final = V initial = V initial .(21 = 0 V) . 9 5 V Cfinal KC initial 2. 2

50. The initial charge on the capacitor is Qinitial= C initial V. When the mica is inserted, the capacitance changes to Cfinal= KC initial , and the voltage is unchanged since the capacitor is connected to the same battery. The final charge on the capacitor is Qfinal= C final V. -9 DQ = Qfinal - Q initial = C final V - C initial V =( K - 1) C initial V = (7 - 1)(3 . 5 10 F)(32 V) =6 . 7 10-7 C

51. The capacitance is found from Eq. 17–7, with the voltage given by Eq. 17–4 (ignoring the sign). Q 0. 675 10-6 C Q= CV = C( Ed )� C = = . 4 20 10-9 F Ed (8. 24� 104 V/m)(1 95 10- 3 m)

The plate area is found from Eq. 17–9. A Cd (4. 20� 10-9 F)(1 95 10 - 3 m) C= Ke � A = = . 0 247 m2 0 -12 2 2 d Ke0 (3. 75)(8 . 85醋 10 C /N m )

52. The stored energy is given by Eq. 17–10.

PE 1CV 2 1 (2 8 10- 9 F)(650 V) 2 5 9 10 - 4 J =2 = 2 .� .

53. The capacitance can be found from the stored energy using Eq. 17–10. 2(PE ) 2(1200 J) PE =1 CV2 � = C = = . 9 6 10- 5 F 2 V 2(5. 0 10 3 V) 2

54. The two charged plates form a capacitor. Use Eq. 17–8 to calculate the capacitance and Eq. 17–10 for the energy stored in the capacitor. e A Q2 Q 2 d (3. 7� 10- 4 C) 2 (1 5 10 - 3 m) C =0 PE =1 = 1 = 1 = 1 . 8 103 J 2 2 2 -12 2 2 - 2 2 d Ce0 A (8. 85醋 10 C /N m )(8 . 0 10 m)

55. (a) Use Eq. 17–8 to estimate the capacitance. e A (8. 85醋 10-12 C 2 /N m 2 )p (4 . 5 in .� 0 . 0254 m/in ) 2 C =0 = =9 . 081椿 10-12 F 9 10 - 12 F d (4 10-2 m) (b) Use Eq. 17–7 to estimate the charge on each plate.

Q= CV =(9 . 081� 10-12 F)(9 V) . 8 椿 173 10 - 11 C 8 10 - 11 C

(c) Use Eq. 17–4b to estimate the (assumed uniform) electric field between the plates. The actual location of the field measurement is not critical in this approximation. V 9 V E = = = 225 V/m 200 V/m d 4 10-2 m (d) By energy conservation, the work done by the battery to charge the plates is the potential energy stored in the capacitor, given by Eq. 17–10.

PE 1QV 1 (8 173 10-11 C)(9 V) 3 678 10 - 10 J 4 10 - 10 J =2 = 2 .� . 椿 (e) If a dielectric is inserted, then the capacitance changes. Thus the charge on the capacitor and the work done by the battery also change. The electric field does not change because it only depends on the battery voltage and the plate separation.

A 56. (a) From Eq. 17–8 for the capacitance, C = e , since the separation of the plates is halved, the 0 d Q2 capacitance is doubled. Then from Eq. 17–10 for the energy, PE = 1 , since the charge is 2 C constant and the capacitance is doubled, the energy is halved. So the energy stored changes by a 1 . factor of 2 (b) The work done to move the plates together will be negative work, since the plates will naturally attract each other. The work is the change in the stored energy. 骣 Q2 Q 2 Q 2 d PE- PE =1 PE - PE = - 1 PE = - 1琪 1 =- 1 = - final initial2 initial initial 2 initial 2琪 2 4 A 桫 Ce 4e0 A 0 d

57. The energy density is given by Eq. 17–11.

Energy density1E2 1 (8 85醋 10- 12 C 2 /N m 2 )(150 V/m) 2 1 0 10 - 7 J/m 3 =2e0 = 2 . = .

58. (a) Before the two capacitors are connected, all of the stored energy is in the first capacitor. Use Eq. 17–10.

PE 1CV 2 1 (3 70 10- 6 F)(12 0 V) 2 2 664 10 - 4 J 2 66 10 - 4 J =2 = 2 .� = . 椿 . (b) After the first capacitor is disconnected from the battery, the total charge must remain constant, and the voltage across each capacitor must be the same, since each capacitor plate is connected to a corresponding plate on the other capacitor by a connecting wire which always has a constant potential. Use the total charge and the fact of equal potentials to find the charge on each capacitor and then calculate the stored energy. -6 - 5 Qtotal= C 1 V initial =(3 . 70� 10 F)(12 = 0 V) . 4 44 10 C

Q1 Q 2 Qtotal- Q 1 Q1= C 1 V Q 2 = C 2 V = = C1 C 2 C 2 C (3. 70 10-6 F) Q= Q 1 =(4 . 44� 10-5 C) . 1 888 10 - 5 C 1 total -6 C1+ C 2 (8. 70 10 F) -5 - 5 - 5 Q2= Q total - Q 1 =4 . 44� 10 C . 1 � 888 . 10 C 2 552 10 C

Q2 Q 2 轾(1. 888� 10-5 C) 2 (2 552 10 - 5 C) 2 PE= PE + PE =11 + 1 2 = 1 + total 1 2 2 2 2 犏 -6 - 6 C1 C 2 臌犏(3. 70� 10 F) (5 00 10 F) =1 . 133椿 10-4 J 1 . 13 10 - 4 J (c) Subtract the two energies to find the change.

-4 - 4 - 4 DPE = PEfinal - PE initial =1 . 133� 10 J . 2 � 664 - 10 J . 1 53 10 J

59. To turn decimal into binary, we need to use the largest power of 2 that is smaller than the decimal number with which we start. That would be 26 = 64, so there is a “1” in the 64’s place. Subtracting 64 from 116 leaves 52. The next power of 2 is 25 = 32, so there is a “1” in the 32’s place. Subtracting 32 from 52 leaves 20. The next power of 2 is 24 = 16, so there is a “1” in the 16’s place. Subtracting 16 from 20 leaves 4, so there is a “0” in the 8’s place, a “1” in the 4’s place, a “0” in the 2’s place, and a “0” in the 1’s place. 64+ 32 + 16 + 0 + 4 + 0 + 0 = 116, so 116 decimal= 1110100 binary.

60. The given binary number has a “1” in the 1’s place, the 4’s place, the 16’s place, and the 64’s place, so we have 1+ 4 + 16 + 64 = 85, and 01010101 binary = 85 decimal . Another way to calculate it is that 01010101= 20 + 2 2 + 2 4 + 2 6 = 1 + 4 + 16 + 64 = 85.

61. Each “1” in the given binary contributes a certain power of 2 to the total value. 1010101010101010= 21 + 2 3 + 2 5 + 2 7 + 2 9 + 2 11 + 2 13 + 2 15 =2 + 8 + 32 + 128 + 512 + 2048 + 8192 + 32768 = 43,690 decimal

62. (a) A 4-bit number can represent 16 different values, from 0 to 15. Thus if 5.0 volts is to be the maximum value, then each binary number would represent 5.0 V/15= 1/3 V. Since 1011 binary bina ry 11(1 V) 3 2 V . =1 + 2 + 8 = 11 decimal, 1011 = 3= 3 (b) 2.0 volts is 6 units of 1/3 volt each, so the binary representation is 0110, which is equal to 6 in the decimal system.

63. (a) A 16-bit sampling would be able to represent 216 = 65,536 different voltages.

(b) A 24-bit sampling would be able to represent 224 = 16,777,216 different voltages. (c) For 3 subpixels each with an 8-bit value, 24 bits are needed. This is the same as part (b), so 224 = 16,777,216 different colors are possible.

64. We see that they have a base 4 counting system, because the decimal number 5 is represented as 11, which would be a “1” in the 1’s place and a “1” in the 4’s place. Making an analogy to humans, we have 5 digits on each hand and use base 10. Since the extraterrestrials use base 4, they must have 2 fingers on each hand, for a total of 4 fingers.

65. (a) The monitor screen is redrawn at regular intervals. Even though the mouse cursor moves in a continuous motion, its location is only drawn at those regular intervals. It is somewhat like a strobe light effect.

(b) Let the interval between images be “T” and construct a table. Position #1 Time= 0 Position #2 Time= T Position #3 Time= 2T Position #4 Time= 3T Position #15 Time= 14T = 0.25 seconds Thus there are 15 images in 0.25 seconds. The refresh time would be T =0.25/14 = 0.018 second. The refresh rate is the reciprocal of this, which is 56 Hz.

66. Consider three parts to the electron’s motion. First, during the horizontal acceleration phase, energy will be conserved. The horizontal speed of the electron u can x  22cm be found from the accelerating potential V. Next, E 11cm during the deflection phase, a vertical force will be applied by the uniform electric field which gives the vx 30o electron an upward velocity, uy . We assume that there Dx is very little upward displacement during this time. screen Finally, after the electron leaves the region of electric Dxfield field, it travels in a straight line to the top of the screen, moving at an angle of approximately 30°. Acceleration: PE KE eV1 mv2 v 2 e V / m initial= final � �2 ux u x Deflection:

Dxfield Time in field: Dxfield = ux t field� t field ux

eE eED xfield Fy= eE = ma y� a y =u y + u y0 = a y t field + 0 m mux Screen:

Dxscreen D x screen Dxscreen =ux t screen� t screen D = y screen = u y t screen u y ux u x

eED xfield Dyu mu eE D x screen=y =x = field 2 Dxscreen ux u x mux 2eV Dy m Dy mu 2 screen 2 V D y 2(9.0� 103 V)(0 11 m) E =screenx =m = screen = Dxscreen e D x field D x screen e D x field D x screen D x field [(0 . 22 m)cos 30� ](0 028 m) = 3 . 712椿 105 V/m 3 . 7 10 5 V/m

As a check on our assumptions, we calculate the upward distance that the electron would move while in the electric field.

2 2 2 12 1 骣eE 骣Dxfield eE( D x field ) E ( D x field ) Dy =uy0 t field + a y t field =0 +琪琪 = = 2 2 桫m桫 u 骣2eV 4 V x 2m琪桫m (3. 712� 105 V/m)(2 8 10- 2 m) 2 = = 8 . 1 10-3 m 4(9000 V)

This is about 7% of the total 11-cm vertical deflection. For an estimation, our approximation is acceptable.

67. If there were no deflecting field, then the electrons would hit the center of the screen. If an electric field of a certain direction moves the electrons toward one  extreme of the screen, then the opposite field will E Dyscreen = 15cm move the electrons to the opposite extreme of the screen. So we solve for the field to move the electrons vx to one extreme of the screen. Consider three parts to Dx = 34cm the electron’s motion and see the diagram, which is a screen

top view. First, during the horizontal acceleration Dxfield = 2.6cm phase, energy will be conserved. The horizontal speed of the electron ux can be found from the accelerating potential V. Next, during the deflection phase, a vertical force will be applied by the uniform electric field which gives the electron a leftward velocity, uy . We assume that there is very little leftward displacement during this time. Finally, after the electron leaves the region of electric field, it travels in a straight line to the left edge of the screen. Acceleration: PE KE eV1 mv2 v 2 e V / m initial= final � �2 ux u x Deflection:

Dxfield time in field: Dxfield = ux t field� t field ux

eE eED xfield Fy= eE = ma y� a y = u y + u y0 = a y t field + 0 m mux Screen:

Dxscreen D x screen Dxscreen =ux t screen� t screen D = y screen = u y t screen u y ux u x

eED xfield Dyu mu eE D x screen=y =x = field 2 Dxscreen ux u x mux 2eV Dy m Dy mu 2 screen 2 V D y 2(6. 0� 103 V)(0 15 m) E =screenx =m = screen = Dxscreen e D x field D x screen e D x field D x screen D x field (0 . 34 m)(0 . 026 m) =2 . 04椿 105 V/m 2 . 0 10 5 V/m

As a check on our assumptions, we calculate the upward distance that the electron would move while in the electric field.

2 2 2 12 1 骣eE 骣Dxfield eE( D x field ) E ( D x field ) Dy =u0 t field + ay t field =0 +琪琪 = = 2 2 桫m桫 u 骣2eV 4 V x 2m琪桫m (2. 04� 105 V/m)(2 6 10- 2 m) 2 = = 6 10-3 m 4(6000 V)

This is about 4% of the total 15-cm vertical deflection. Thus for an estimation, our approximation is acceptable. The field must vary from +2 . 0� 105 V/m to . 2 0 10 5 V/m . 68. (a) The energy is related to the charge and the potential difference by Eq. 17–3. DPE 5 . 2 106 J DPE =q D V瓺 V = = = 1 . 3 106 V q 4. 0 C (b) The energy (as heat energy) is used to raise the temperature of the water and boil it. Assume that room temperature is 20°C. Use Eq. 14–2 and Eq. 14–4.

Q= mc D T + mLF Q 5. 2 106 J m = = =2 . 0 kg cD T + LF 骣J 骣 5 J 琪4186 (80 C� ). 琪 22 6 10 桫kg装 C 桫 kg

69. The electric force on the electron must be the same magnitude as the weight of the electron. The magnitude of the electric force is the charge on the electron times the magnitude of the electric field. The electric field is the potential difference per meter: E= V/ d .

FE= mg; F E = q E = eV / d� eV / d mg mgd (9. 11� 10-31 kg)(9 . 80 m/s 2 )(0 024 m) V = = =1 . 3 10-12 V e 1. 60 10-19 C

Since it takes such a tiny voltage to balance gravity, the thousands of volts in a television set are more than enough (by many orders of magnitude) to move electrons upward against the force of gravity.

PE 1 CV 2. 70. The energy stored in the capacitor is given by Eq. 17–10, = 2 As the plates are separated, the voltage is held constant by the battery. Since the separation distance has doubled, the capacitance has been multiplied by ½, using Eq. 17–8. Thus the stored energy has been multiplied by ½.

71. The capacitor is charged and isolated, so the amount of charge on the capacitor cannot change. PE 1 QV. (a) The stored energy is given by Eq. 17–10, = 2 The charge does not change, and the potential difference is doubled, so the stored energy is multiplied by 2. Q2 (b) The stored energy can also be given by PE = 1 . The charge does not change. The 2 C capacitance would be reduced by a factor of 2 if the plate separation were doubled, according to Eq. 17–8, so the stored energy is multiplied by 2.

72. The energy in the capacitor, given by Eq. 17–10, is the heat energy absorbed by the water, given by Eq. 14–2. PE Q1 CV2 mc T =heat � 2 D 骣 J 2(2. 8 kg)琪 4186 (95� C 21 C) 2mcD T 桫 kg装 C V = = = 658 V 660 V C 4. 0 F

73. The proton would gain half the kinetic energy as compared to the alpha particle. The alpha particle has twice the charge of the proton, so it will have twice the potential energy for the same voltage. Thus the alpha will have twice the kinetic energy of the proton after acceleration.

74. We assume there is a uniform electric field between the capacitor plates, so that V= Ed . Combine Eq. 17–7 with Eq. 17–8 and Eq. 17–4. e A V Q= CV =0 V =e A = e AE =(8 . 85醋 10-12 C 2 /N m 2 )(65 � 10 - 4 m 2 )(3 0 10 6 V/m) d0 d 0 = 1 . 7 10-7 C

-Q B +Q 75. Use Eq. 17–5 to find the potential due to each charge. Since the triangle is equilateral, the 30-60-90 triangle relationship says that the distance from a corner to the midpoint of the opposite side is 3ℓ /2. A C k(- Q ) k ( - 3 Q ) k ( Q ) 2 kQ骣 1 kQ VA = + + =琪 -4 + = - 6 . 85 -3Q ℓ/2 ℓ /2 3ℓ /2 ℓ桫 3 ℓ k(- Q ) k ( Q ) k ( - 3 Q ) kQ kQ VB = + + = -2 3 = - 3 . 46 ℓ/2 ℓ /2 3ℓ /2 ℓ ℓ

k( Q ) k (- 3 Q ) k ( - Q ) 2 kQ骣 1 kQ VC = + + = -琪2 + = - 5 . 15 ℓ/2 ℓ /2 3ℓ /2 ℓ桫 3 ℓ

76. The energy is given by Eq. 17–10. Calculate the energy difference for the two different amounts of charge and then solve for the difference. Q2( Q+ D Q ) 2 Q 2 1 D Q PE=1瓺 PE = 1 - 1 =[(Q + D Q )2 - Q 2 ] = [2 Q + D Q ] 2C 2 C 2 C2 C 2 C C(DPE ) (17 . 0� 10-6 F)(15 2 J) Q= -1 D Q = - 1 (13 . 0� 10-3 C) . 13 4 10 - 2 C DQ 2(13. 0 10-3 C) 2 77. (a) Because of the inverse square nature of the electric x d field, any location where the field is zero must be closer to the weaker charge (q ). Also, in between 2 q2 < 0 q1 > 0 the two charges, the fields due to the two charges are parallel to each other (both to the left) and cannot cancel. Thus the only places where the field can be zero are closer to the weaker charge, but not between them. In the diagram, they are labeled x.

q2 q1 2 2 E= k - k =0� q2 ( d = x ) q 1 x x2( d+ x ) 2

-6 q2 2. 6 10 C x= d =(2 . 5 cm) = 17 . 42 cm 17 cm left of q2 -6 - 6 q1- q 2 3. 4� 10 C . 2 6 10 C (b) The potential due to the positive charge is positive d everywhere, and the potential due to the negative x2 x1 charge is negative everywhere. Since the negative charge is smaller in magnitude than the positive q2 < 0 q1 > 0 charge, any point where the potential is zero must be closer to the negative charge. Consider locations between the charges (x1 ) and to the left of the negative charge (x2 ) as shown in the diagram. kq kq q d (- 2.6 10-6 C)(2.5 cm) V=1 + 2 =0� x = 2 = 1.1 cm location 1 1 -6 (d- x1 ) x 1 ( q 2 - q 1 ) (- 6.0 10 C) kq kq q d (- 2.6 10-6 C)(2.5 cm) V=1 + 2 =0� x - = 2 = 8.1 cm location 2 2 -6 (d+ x2 ) x 2 ( q 1 - q 2 ) (0.8 10 C) So the two locations where the potential is zero are 1.1 cm from the negative charge toward the positive charge and 8.1 cm from the negative charge away from the positive charge. 78. Since the electric field points downward, the surface of the Earth is at a lower potential than above the surface. Call the potential of the surface of the Earth 0. Then a height of 2.00 m has a potential of 300 V. We also call the surface of the Earth the 0 location for gravitational PE. Write conservation of energy relating the charged spheres at 2.00 m (where their speed is 0) and at ground level (where their electrical and gravitational potential energies are 0).

2 骣 qV E= E� mgh = qV �1 muv + u 2琪 gh initial final 2 桫 m

-4 轾 2 (6. 5 10 C)(300 V) uv+ =2犏 (9 . 80 m/s )(2 . 00 m) + = 6 . 3073 m/s 臌犏 (0. 670 kg) -4 轾 2 (- 6 . 5 10 C)(300 V) u- =2犏 (9 . 80 m/s )(2 . 00 m) + = 6 . 2143 m/s 臌犏 (0. 670 kg)

uv+- u - =6 . 3073 m/s - 6 . 2143 m/s = 0 . 093 m/s� 0 09 m/s

79. (a) Use Eq. 17–10 to calculate the stored energy. PE 1CV 2 1 (5 0 10- 8 F)(3 5 10 4 V) 2 30 625 J 31 J =2 = 2 .� � . (b) The power is the energy converted per unit time. Energy 0. 12(30 . 625 J) P = = =5 . 9 105 W time 6. 2 10-6 s

80. The kinetic energy of the electrons (provided by the UV light) is converted completely to potential energy at the plate since they are stopped. Use energy conservation to find the emitted speed, taking the 0 of PE to be at the surface of the barium. KE PE 1 mv2 qV initial= final � =2 u

2qV 2(- 1 . 60� 10-19 C)( . 3 02 V) u = = =1 . 03 106 m/s m 9. 11 10-31 kg

81. Let d1 represent the distance from the left charge to point B, and let d2 represent the distance from the right charge to point B. Let Q represent the positive charges, and let q represent the negative charge that moves. The change in potential energy is given by Eq. 17–3.

PEB- PE A =qV BA = q( V B - V A )

2 2 2 2 d1=12 + 14 cm = 18.44 cm d 2 = 14 + 24 cm = 27.78 cm 轾骣kQ kQ 骣 kQ kQ PEB- PE A =q( V B - V A ) = q 犏琪 + - 琪 + 臌桫0.1844 m 0.2778 m 桫 0.12 m 0.24 m 轾骣1 1 骣 1 1 =kQq 犏琪 + - 琪 + 臌桫0.1844 m 0.2778 m 桫 0.12 m 0.24 m =(8.99� 109 )( 创 1.5 10- 6 C)(38 - 10 - 6 C)( = 3.477 m - 1 ) 1.782 J 1.8 J

82. (a) The initial capacitance is obtained directly from Eq. 17–9. -12 Ke0 A 3. 7(8 . 85� 10 F/m)(0 . 21 m)(0 14 m) -9 C0 = = =8 . 752椿 10 F 8 . 8 nF d 0. 11 10-3 m (b) Maximum charge will occur when the electric field between the plates is equal to the dielectric strength. The charge will be equal to the capacitance multiplied by the maximum voltage, where the maximum voltage is the electric field times the separation distance of the plates. -9 6 - 3 Qmax= C 0 V = C 0 Ed =(8 . 752创 10 F)(15 10 V/m)(0 . 11 10 m) =1 . 444椿 10-5 C 14m C

83. Use Eq. 17–7 with Eq. 17–9 to find the charge. -2 2 A -12 2 2p (0. 55 10 m) - 10 Q= CV = Ke0 V =(3 . 7)(8 . 85醋 10 C /N m ) (12 V) = 3 . 7 10 C d (0. 10 10-3 m)

84. (a) Use Eq. 17–5 to calculate the potential due to the charges. Let the distance between the charges be d. kQ kQ 2k 2(8. 99醋 109 N m 2 /C 2 ) V=1 + 2 =( Q + Q ) = (3 . 5� 10-6 C . 7 2 10 - 6 C) mid(d /2) ( d /2) d 1 2 (0. 23 m) = - 2 . 9 105 V (b) Use Eq. 16–4a to calculate the electric field. Note that the field due to each of the two charges will point to the left, away from the positive charge and toward the negative charge. Find the magnitude of the field using the absolute value of the charges.

kQ1 k Q2 4k Emid= + =( Q 1 + Q 2 ) (d /2)2 ( d /2) 2 d 2 4(8. 99醋 109 N m 2 /C 2 ) =(3 . 5� 10-6 C . 7 � 2 . 10 - 6 C) 7 3 10 6 V/m, left (0. 23 m)2

85. (a) Use Eq. 17–7 and Eq. 17–8 to calculate the charge. e A (8. 85醋 10-12 C 2 /N m 2 )(3 . 0 10 - 4 m 2 ) Q= CV =0 V =(12 V) = 6 . 372 10-11 C d (5. 0 10-4 m)

�6 4 10-11 C

(b) The charge does not change. Since the capacitor is not connected to a battery, no charge can flow to it or from it. Thus Q =6 . 372椿 10-11 C 6 . 4 10 - 11 C . (c) The separation distance was multiplied by a factor of 1.5. The capacitance is inversely proportional to the separation distance, so the capacitance was divided by a factor of 1.5. Since Q= CV , if the charge does not change and the capacitance is divided by a factor of 1.5, then the voltage must have increased by a factor of 1.5.

Vfinal=1 . 5 V initial = 1 . 5(12 V) = 18 V (d) The work done is the change in energy stored in the capacitor. W=PE - PE =1 QV - 1 QV = 1 Q( V - V ) final initial2 final 2 initial 2 final initial 1 (6 372 10-11 C)(18 V 12 V) 1 912 10 - 10 J 2 10 - 10 J =2 .� = . 椿

86. The energy stored in the capacitor is given by Eq. 17–10. The final energy is half the initial energy. Find the final charge and then subtract the final charge from the initial charge to find the charge lost. 2 2 1 1Qfinal 1 1 Q initial 1 Efinal= E initial� � Q final Q initial 2 2C 2 2 C 2

骣1 骣 1 -6 Qlost= Q initial - Q final = Q initial 琪1 - = CV 琪 1 - = (2 . 1� 10 F)(6 . 0 V)(0 2929) 桫2 桫 2 = 3 . 7 10-6 C

87. (a) We assume that Q2 is held at rest. The energy of the system will then be conserved, with the initial potential energy of the system all being changed to kinetic energy after a very long time. kQ Q E= E�PE � KE 1 2 1 m u2 initial final initial finalr 2 1 1 2kQ Q 2(8. 99醋 109 N m 2 /C 2 )(6 . 5 10- 6 C) 2 u =1 2 = =3 . 6 103 m/s 1 -6 - 2 m1 r (1. 5� 10 kg)(4 0 10 m) (b) In this case, both the energy and the momentum of the system will be conserved. Since the initial momentum is zero, the magnitudes of the momenta of the two charges will be equal.

m1 pinitial= p final � =0 + m 1u 1 m 2 u 2 � - u 2 u 1 m2

Einitial= E final � PE initial KE final kQ Q轾骣 m m 1 21m2 1 m 2 1 m 2 m 1 1 1 ( m m ) 2 =21u 1 + 2 2 u 2 = 2犏 1 u 1 + 2琪 - u 1 = 2 1 + 2 u 1 r臌犏桫 m2 m 2

2kQ Q m 2(8.99醋 109 N m 2 /C 2 )(6.5 创 10- 6 C) 2 (2.5 10 - 6 kg) u =1 2 2 = 1 -6 - 6 - 2 m1( m 1+ m 2 ) r (1.5创 10 kg)(4.0 10 m)(4.0 10 m) = 2.8 103 m/s

88. Calculate VAB= V A - V B. Represent the 0.10-m distance by the variable d.

骣kq1 kq 2 骣 kq 2 kq 1 k 骣 1 VAB= V A - V B =琪 + - 琪 + =( q 1 - q 2 ) 琪 1 - 桫d2d 桫 d 2 d d 桫 2 9 2 2 (8. 99醋 10 N m /C )-6骣 1 5 = (4 . 5� 10 C)琪 1 = . 1 2 10 V 0. 1 m 桫 2

89. The potential of the Earth will increase because the “neutral” Earth will now be charged by the removing of the electrons. The excess charge will be the elementary charge times the number of electrons removed. We approximate this change in potential by using a spherical Earth with all the excess charge at the surface. 骣1. 602� 10-19 C 骣 10 .e - 骣 6 02 10 23 molecules骣 1000 kg Q =琪琪琪 4 p (0 . 00175 m)3 琪- 琪琪琪 3 3 桫e 桫H2 O molecule 桫 0. 018 kg 桫 m = 1203 C 1Q 1203 C V = =(8 . 99醋 109 N m 2 /C 2 ) = 1 . 7 10 6 V 6 4pe0R Earth 6. 38 10 m

90. As an estimate, the length of the bolt would be the voltage difference of the bolt divided by the breakdown electric field of air. 5 107 V =16 . 7 m 20 m 3 106 V/m

If the lightning strikes from a height of 750 m (roughly a half-mile), then 40 or more steps might be involved.

91. If we assume that the electric field is uniform, then we can use Eq. 17–4a to estimate the magnitude of the electric field. From Eq. 16–10 we have an expression for the electric field due to a pair of oppositely charged planes. We approximate the area of a shoe as 30 cm 8 cm. Answers will vary with the shoe size approximation. Vs Q E = = = de0 e 0 A e AV (8. 85醋 10-12 C 2 /N m 2 )(0 . 024 m 2 )(6 . 0 10 3 V) Q =0 = =1 . 3 10-6 C d 1. 0 10-3 m

92. Use Eq. 17–7.

DQ C D V (1200 F)(6 . 0 V)6 骣 1 d DQ = C D V; D t = = = = 7 . 2� 10 s琪 83 d DQ/ D t D Q / D t 1. 0 10-3 C/s 桫 86,400 s

93. To find the angle, the horizontal and vertical components of the velocity are needed. The horizontal component can be found using conservation of energy for the initial acceleration of the electron. That component is not changed as the electron passes through the plates. The vertical component can be found using the vertical acceleration due to the potential difference of the plates and the time the electron spends between the plates. Horizontal: Dx PE KE qV1 mv2 t inital= final � = 2 u x ux

Vertical:

(uy- u0 y ) qE y t qE y D x FE = qEy = ma = m � v y = t m mux Combined:

qEy D x 骣250 V 琪 (0. 065m) uymu qE yD x qE y D x E y D x 桫0. 013 m tanq = =x = = = = = 0 . 284 2 ux u x mux 2qV 2 V 2(2200 V) q =tan2 1 0 . 284 = 16

94. (a) The absolute value of the charge on each plate is given by Eq. 17–7. The plate with electrons has a net negative charge. Q= CV� N( = e ) - CV

CV (35� 10-15 F)(1 5 V) N = = =3 . 281椿 105 3 . 3 10 5 electrons e 1. 60 10-19 C (b) Since the charge is directly proportional to the potential difference, a 2.0% decrease in potential difference corresponds to a 2.0% decrease in charge. DQ =0 . 020 Q DQ0 . 020 Q 0 . 020 CV 0 . 020(35� 10-15 F)(1 5 V) Dt = = = = =3 . 5 s DQ/ D t D Q / D t D Q / D t 0. 30 10-15 C/s

95. (a) From Problem 94, we have C = 35 10-15 F. Use Eq. 17–9 to calculate the area.

2 A Cd (35� 10-15 F)(2 0 10 - 9 m)骣 10 6 m m C= Ke � A = = . 3 164 10-13 m 2 琪 0 -12 2 2 琪 d Ke0 (25)(8. 85醋 10 C /N m ) 桫 1 m = 0 . 3164m m2� 0 32 m m 2

(b) Half of the area of the cell is used for capacitance, so 1. 5 cm2 is available for capacitance. Each capacitor is one bit. 2 骣106 m m骣 1 bit骣 1 byte 1. 5 cm2琪琪 = 5 . 86椿 10 7 bytes 59 Mbytes 琪 2琪 2 琪桫10 cm桫 0. 32m m 桫8 bits

96. (a) Use Eq. 17–8 to calculate the capacitance. -12 2 2 2 e0 A (8. 85醋 10 C /N m )(2 . 0 m ) -9 C0 = = =5 . 9 10 F d (3. 0 10-3 m) Use Eq. 17–7 to calculate the charge.

-9 - 7 - 7 Q0= C 0 V 0 =(5 . 9� 10 F)(35 V) . 2 椿 065 10 C . 2 1 10 C The electric field is the potential difference divided by the plate separation.

V0 35 V 4 E0 = = = 11667 V/m� 1 2 10 V/m d 3. 0 10-3 m

Use Eq. 17–10 to calculate the energy stored.

PE 1C V 2 1 (5 9 10- 9 F)(35 V) 2 3 614 10 - 6 J 3 6 10 - 6 J 0=2 0 0 = 2 .� . 椿 . (b) Now include the dielectric. The capacitance is multiplied by the dielectric constant.

-9 - 8 - 8 C= KC0 =3 . 2(5 . 9� 10 F) . 1 椿 888 10 F . 1 9 10 F The voltage doesn’t change. Use Eq. 17–7 to calculate the charge.

-9 - 7 - 7 Q= CV = KC0 V =3 . 2(5 . 9� 10 F)(35 V) . 6 椿 608 10 C . 6 6 10 C Since the battery is still connected, the voltage is the same as before. Thus the electric field doesn’t change.

4 E= E0 =1 . 2 10 V/m Use Eq. 17–10 to calculate the energy stored.

PE 1CV2 1 KC V 2 1 (3 2)(5 9 10- 9 F)(35 V) 2 1 2 10 - 5 J =2 = 20 = 2 . .� .

Solutions to Search and Learn Problems

1. (a) An equipotential surface is one in which all points are at the same potential. (b) An equipotential surface must be perpendicular to the electric field at any point. (c) Equipotential lines and surfaces are always continuous. (d) Equipotential lines and surfaces never end. (e) The entire volume (including the surface) of a conductor is an equipotential. (f) No work is required to move a charged particle from one point on an equipotential surface to another point on that surface. (g) Equipotential lines and surfaces cannot cross. (h) Electric field lines point from higher equipotential surfaces to lower equipotential surfaces. (i) Closer spacing of equipotentials indicates a larger electric field.

2. (a) The contour values on the map are in feet (the mountains are about 2 miles above sea level). There are six contour lines between Iceberg Lake and Cecile Lake. Each line is to be interpreted as an 80-V difference, so the voltage difference is 480 V. The two lakes appear to be about 0.5 km apart. The magnitude of the electric field is equal to the change in voltage divided by the separation distance, Eq. 17–4a. V 480 V E =ba = =0 . 96 V/m 1 V/m d 500 m Iceberg Lake is lower than Cecile Lake, so the direction of the electric field is basically in the +y direction. Answers may vary due to interpreting the figure. (b) At the Minaret Mine, the higher elevations are upward and leftward. The elevation rises about 400 ft (five contour lines) in about 500 m as you move in that direction. Thus the electric field points downward and to the right. Calculate the magnitude by Eq. 17–4a. V 400 V E=ab = =0 . 8 V/m, 45� below the x axis d 500 m Again, answers may vary due to estimations in interpreting the figure.

3. (a) Use Eq. 17–8. e A (8. 85醋 10-12 C 2 /N m 2 )(110 10 6 m 2 ) C =0 = =6 . 49椿 10-7 F 6 . 5 � 10 - 7 F 65m F d (1500 m) (b) Use Eq. 17–7. Q= CV =(6 . 49� 10-7 F)(3 � 5 . 10 7 V) 22 715 C 23 C (c) Use Eq. 17–10.

PE 1QV 1 (22 715 C)(3 5 107 V) 4 0 10 8 J =2 = 2 . .� .

4. (a) Both of these equations for the potential energy can be obtained from Eq. 17–10 by using Eq. 17–7 first to eliminate the charge from the equation, and second to eliminate the voltage. PE 1QV 1( CV ) V 1 CV 2 =2 = 2 = 2 2 骣Q Q =1QV = 1 Q琪 = 1 2 2桫C 2 C

(PE 1 CV 2 ) (b) The first equation = 2 is used when the capacitance and voltage difference across the capacitor is known, as when the capacitor is connected to a battery. The second equation 骣 Q2 琪PE = 1 琪 2 is used when the capacitance and charge are known, as when a specific charge is 桫 C placed on an isolated capacitor. (c) Since the voltage is constant, we use the first equation for the potential energy, with the capacitance equal to the product of the initial capacitance and the dielectric constant of the paper. PE1CV2 1( KC ) V 2 K ( 1 C V 2 ) K ( PE ) =2 = 20 = 2 0 = 0 The potential energy in the capacitor will increase by a factor equal to the dielectric constant. For a paper dielectric, the potential increases by a factor of 3.7. (d) In this case the charge is constant, so we use the second equation to determine the final energy. Q2 Q 21骣 Q 2 1 PE= = =琪 = ( PE0 ) 2C 2 KC0 K桫 2 C 0 K The potential energy in the capacitor decreases by a factor equal to the dielectric constant. For quartz, the potential energy decreases to 1/4.3= 0.23, or 23% of the initial value.

5. (a) A horsepower is equal to 746 watts.

3骣1 hp 2 P =(75� 10 W)琪 . 100 � 5 hp 1 0 10 hp 桫746 W (b) The energy is the product of the power and time elapsed.

E= Pt =(75� 103 W)(5 = 0 h)(3600 s/h) . 1 椿 35 10 9 J . 1 4 10 9 J (c) Use Eq. 17–10, in terms of the capacitance and voltage, to solve for the capacitance. 2PE 2(1. 35 109 J) PE =1 CV2 � = C = = .3 椿 737 10 3 F . 3 7 10 3 F 2 V 2(850 V) 2

(d) According to the text, the capacitance of 0.1 g of activated carbon is about 1 F. We again use Eq. 17–10 to find the capacitance and then convert to a mass of carbon. 9 2PE 2(1. 35� 10 J)7 7骣 0 1 g 6 3 C = = =2 . 7� 10 F ; 2 � 7 . 10 F琪 � 2 . 7 10 g 2 7 10 kg V 2(10 V) 2 桫1 F (e) Cars have a mass on the order of 1000–2000 kg, so this capacitor would more than double the mass of the car. This is not practical.

6. (a) If N electrons flow onto the plate, then the charge on the top plate is -Ne, and the positive charge associated with the capacitor is Q= Ne . Since Q= CV , we have Ne= CV� V Ne/ C , showing that V is proportional to N. (b) Given DV = 1 mV and we want DN = 1, solve for the capacitance. Ne eD N V=瓺 V = C C DN 1 C= e =(1 . 60� 10-19 C) . 1 椿 60 10 - 16 F 2 10 - 16 F DV 1 10-3 V (c) Use Eq. 17–9 to solve for the side length of the capacitor. A ℓ2 C=e K = e K 0d 0 d Cd (1. 60创 10-16 F)(100 10 - 9 m)骣 10 6 m m ℓ = = =7 . 76� 10-7 m琪 . 0 8m m -12 2 2 琪 e0K (8. 85醋 10 C /N m )(3) 桫 1 m

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