Openstax College Physics Instructor Solutions Manual Chapter 19

OpenStax College Physics Instructor Solutions Manual Chapter 19

CHAPTER 19: ELECTRIC POTENTIAL AND ELECTRIC FIELD

19.1 ELECTRIC POTENTIAL ENERGY: POTENTIAL DIFFERENCE

1. Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final 1.67 1027 kg. speeds. Take the mass of the hydrogen ion to be

Solution 1 1 m v 2  qV; m v 2  qV, 2 e e 2 H H so that

1/ 2 m v 2 v m 1.671027 kg  e e e  H     42.8 2  1,  31  mHvH vH me  9.1110 kg  or

2. An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce x rays. Non-relativistically, what would be the maximum speed of these electrons?

Solution Use energy conservation: 1 qV = mv2  2 v = 2qV/m = (2  1.6  10-19 C  4.0  104 V / 9.11  10-31 kg) 0.5 = 1.17  108 m/s.

3. 6.641027 kg. A bare helium nucleus has two positive charges and a mass of (a) Calculate its kinetic energy in joules at 2.00% of the speed of light. (b) What is this in electron volts? (c) What voltage would be needed to obtain this energy? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution 1 2 KE  mv2  0.5(6.641027 kg)(0.0200)(3.00108 m/s)  2  1.1951013 J  1.201013 J (a)

 1 eV  (1.19510 13 J)   7.47 105 eV 1.60 10-19 J  (b)

KE 1.1951013 J KE  qV  V    3.73105 V q 2(1.60 10-19 J) (c)

4. Integrated Concepts Singly charged gas ions are accelerated from rest through a voltage of 13.0 V. At what temperature will the average kinetic energy of gas molecules be the same as that given these ions?

Solution 3 2 qV KE  kT  qV  T  , 2 3 k k 1.381023 J/K where is the Boltzmann constant.

2 / 31.601019 C 13.0 V T     1.0048105 K  1.00105 K 1.381023 J/K

5. Integrated Concepts The temperature near the center of the Sun is thought to be 15 (1.5107C) million degrees Celsius . Through what voltage must a singly charged ion be accelerated to have the same energy as the average kinetic energy of ions at this temperature?

Solution 3 3kT 3(1.381023 J/K)(15106 K) kT  qV  V    1.94103 V  1.9103 V 2 2q 2(1.601019 C)

6. Integrated Concepts (a) What is the average power output of a heart defibrillator that dissipates 400 J of energy in 10.0 ms? (b) Considering the high-power output, why doesn’t the defibrillator produce serious burns? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution (a) The power is the work divided by the time, so the average power is: W 400 J P    4.00104 W t 10.010-3 s .

(b) A defibrillator does not cause serious burns because the skin conducts electricity well at high voltages, like those used in defibrillators. The gel used aids in the transfer of energy to the body, and the skin doesn’t absorb the energy, but rather, lets it pass through to the heart.

7. Integrated Concepts A lightning bolt strikes a tree, moving 20.0 C of charge through 1.00102 MV a potential difference of . (a) What energy was dissipated? (b) What

1 5 C mass of water could be raised from to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam. 

Solution  PE  qV  20.0 C100106 V  2.00109 J (a) Energy

Q  mcT  mLv  m(cT  Lv ) Q 2.00109 J m    766 kg cT  L (4186 J/kg  C)(85.0 C)  2256103 J/kg (b) v

(c) The expansion of the steam upon boiling can literally blow the tree apart!

8. Integrated Concepts A 12.0 V battery-operated bottle warmer heats 50.0 g of glass, 2 2 2.5010 g 2.0010 g 20.0C 90.0C of baby formula, and of aluminum from to . (a) How much charge is moved by the battery? (b) How many electrons per second flow if it takes 5.00 min to warm the formula? (Hint: Assume that the specific heat of   baby formula is about the same as the specific heat of water.)

Solution (a) Assume that specific heat for baby formula is approximately the specific heat for

Q  (m1c1  m2c2  m3c3 )T  qV water, so that OpenStax College Physics Instructor Solutions Manual Chapter 19

(m c  m c  m c )T q  1 1 2 2 3 3 V (0.0500 kg)(840 J/kg  C)  (0.200 kg)(900 J/kg  C)(70.00 C)  12.0 V (0.250 kg)(4186 J/kg  C)(70.0 C)  12.0 V  7.40103 C

N t (b) Let equal the number of electrons needed to move the charge, and be the time necessary to move the charge. Then 3 N q / qe q 7.40 10 C 20    19  1.54 10 electron/s t t qet (1.60 10 C)(300 s)

9. Integrated Concepts A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 2.00102 m 25.0 m/s, make it climb a high hill, and then cause it to travel at a 5.00102 N constant 25.0 m/s by exerting a force for an hour.

Solution 1 1 E  mv2  mgh  Fd  mv2  mgh  Fvt tot 2 2 1  (750 kg)(25.0 m/s) 2  (750 kg)(9.80 m/s 2 )(200 m)  (500 N)(25.0 m/s)( 3600 s) 2  4.67 107 J

E 4.67 107 J E  qV  q  tot   3.89 106 C tot V 12.0 V OpenStax College Physics Instructor Solutions Manual Chapter 19

10. Integrated Concepts Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. (a) Calculate the potential energy of two singly 1.001012 m charged nuclei separated by by finding the voltage of one at that distance and multiplying by the charge of the other. (b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?

Solution kq kq2 V  ,so that PE  qV  r r 9 2 2 19 2 (9.0010 N m / C )(1.6010 C) 16 16  12  2.30410 J 2.3010 J (a) 1.0010 m

3 2 (2.3041016 J) KE  kT  T   1.11107 K 2 3 1.381023 J/K (b)

11. Unreasonable Results (a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?

Solution kQ (9.00 109 N  m 2 /C 2 )(8.00 C) V    1.44 1012 V r 0.0500 m (a)

(b) This voltage is very high. A 10.0 cm diameter sphere could never maintain this voltage; it would discharge.

(c) An 8.00 C charge is more charge than can reasonably be accumulated on a sphere of that size.

19.2 ELECTRIC POTENTIAL IN A UNIFORM ELECTRIC FIELD

13. Show that units of V/m and N/C for electric field strength are indeed equivalent. OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution J N  m 1 V  1 J/C ; 1 J  1 N  m  1 V/m 1 1  1 N/C C  m C  m

14. What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of

1.50104 V ?

Solution V 1.50104 V V  Ed  E  AB   1.50 106 V/m AB d 0.0100 m

15. The electric field strength between two parallel conducting plates separated by 4.00 7.50104 V/m cm is . (a) What is the potential difference between the plates? (b)

The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

4 3 Solution VAB  Ed  (7.5010 V/m)(0.040 0 m)  3.0010 V  3.00 kV (a)

4 VAB  Ed  (7.5010 V/m)(0.010 0 m)  750 V (b)

16. How far apart are two conducting plates that have an electric field strength of 4.50103 V/m between them, if their potential difference is 15.0 kV?

Solution V 15.0103 V V  Ed  d  AB   3.33 m AB E 4.50103 V/m OpenStax College Physics Instructor Solutions Manual Chapter 19

17. (a) Will the electric field strength between two parallel conducting plates exceed the 3.0106 V/m breakdown strength for air ( ) if the plates are separated by 2.00 mm 5.0103 V and a potential difference of is applied? (b) How close together can the plates be with this applied voltage?

Solution V 5.0103 V E  AB   2.5106 V/m  3106 V/m. d 2.0010-3 m (a)

No, the field strength is smaller than the breakdown strength for air.

V 5.0103 V d  AB  1.67 10-3 m 1.7 mm . E 3.0106 V/m (b)

18. The voltage across a membrane forming a cell wall is 80.0 mV and the membrane is 9.00 nm thick. What is the electric field strength? (The value is surprisingly large, but correct. Membranes are discussed in Capacitors and Dielectrics and Nerve Conduction—Electrodiograms.) You may assume a uniform electric field.

Solution V 80.010-3 V E  AB   8.89 106 V/m d 9.0010-9 m

19. Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction —Electrocardiograms.) What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m? You may assume a uniform electric field.

6 9 2 Solution VAB  Ed  (5.5010 V/m)(8.00 10 m)  4.4010 V  44.0 mV

20. Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution V 450 V V  Ed  E  AB   5625 V/m  5.63103 V/m  5.63 kV/ m AB d 0.0800 m (a)

VAB  Ed  (5625 V/m)(0.100 m)  562.5 V  563 V (b)

21. Find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sustainable electric field strength 3.0106 V/m in air to be .

6 -3 4 S olution VAB  Ed  (3.010 V/m)(5.00 10 m) 1.510 V 15 kV

22. A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?

Solution KEf  qVAB  qEd  KE (32.0103 e V)(1.6010-19 J/1 eV) E  f   8.00105 V/m qd 2(1.6010-19 C)(0.0200 m)

23. An electron is to be accelerated in a uniform electric field having a strength of

2.00106 V/m . (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution

KE  qV and VAB  Ed, so that

KE  qVAB  qEd  1 eV  1 keV   (1.601019 C)(2.00 106 V/m)(0.400 m)    1.60 1019 J 1000 eV   800 keV (a)

KE (50.0 109 eV) 1.60 1019 J  d     19 6   qE (1.60 10 C)(2.00 10 v/m)  1 eV   2.50 104 m  25.0 km (b)

19.3 ELECTRIC POTENTIAL DUE TO A POINT CHARGE

24. A 0.500 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 40.0 pC charge on its surface. What is the potential near its surface?

Solution kQ (9.00109 N.m2 / C2 )(40.01012 C) V   144 V r 2.50103 m

25. 0.5301010 m What is the potential from a proton (the average distance between the proton and electron in a hydrogen atom)?

Solution kQ 9.00109 N  m2 / C2 1.601019 C V       27.2 V r 0.5301010 m

26. (a) A sphere has a surface uniformly charged with 1.00 C. At what distance from its center is the potential 5.00 MV? (b) What does your answer imply about the practical aspect of isolating such a large charge? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution kQ kQ 9.00109 N  m 2 / C2 1.00 C V   r      1.80103 m  1.80 km r V 5.00106 V (a)

(b) A 1 C charge is a very large amount of charge; a sphere of 1.80 km is impractical!

27. 1.00μC How far from a point charge will the potential be 100 V? At what distance 2.00102 V will it be ?

Solution kQ kQ 9.00109 N  m 2 / C 2 1.00 106 C V   r       90.0 m r V 100 V kQ 9.00109 N  m 2 / C 2 1.0010 6 C r       45.0 m V 200 V

28. What are the sign and magnitude of a point charge that produces a potential of  2.00 V at a distance of 1.00 mm?

Solution kQ rV 1.00103 m 2.00 V V   Q       2.22 1013 C r k 9.00109 N  m2 / C2

29. 5.00102 V If the potential due to a point charge is at a distance of 15.0 m, what are the sign and magnitude of the charge?

Solution kQ rV 15.0 m500 V V  Q    8.33107 C r k 9.00109 N  m 2 / C2 Since , . The charge is positive because the potential is positive.

30. In nuclear fission, a nucleus splits roughly in half. (a) What is the potential 2.001014 m from a fragment that has 46 protons in it? (b) What is the potential energy in MeV of a similarly charged fragment at this distance? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution (a) 9 2 2 19 kQ 9.0010 N  m /C (46)1.6010 C 6 6 V    3.31210 V  3.3110 V r 2.001014 m

 1 eV  PE  qV  461.60 10 19 C3.312106 J/C   1.52 108 eV 1.601019 J   152 MeV (b)

31. A research Van de Graaff generator has a 2.00-m-diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface? (b) At what distance from its center is the potential 1.00 MV? (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV at this distance?

Solution kQ 9.00109 N  m 2 /C 2 5.00103 C V       4.50107 V  45.0 MV r 1.00 m (a)

kQ kQ 9.00109 N  m 2 /C 2 5.00 103 C V   r    45.0 m r V 1.00106 V (b)

PE  qV  3qe 45.0 MV 1.00 MV 132 MeV (c)

32. An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of 25.0 kV that repels paint droplets onto a grounded object. (a) What charge is on the sphere? (b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?

Solution kQ rV 0.100 m 25.0103 V V   Q      2.78107 C r k 9.00109 N  m 2 /C 2 (a)

1 mv2 (0.10010 6 kg)(10.0 m/s) 2 mv2  qV  q    2.001010 C 2 2V 2(25.0103 V) (b) OpenStax College Physics Instructor Solutions Manual Chapter 19

33. In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 5.00 MeV, how close to the gold nucleus (79 protons) could it come before being deflected?

Solution The alpha particle approaches the gold nucleus until its original energy is converted 5.00 MeV  8.0010-13 J to potential energy. , so

qkQ E   0 r -19 9 2 2 -17 qkQ (3.2 10 C)(9.0  10 N.m /C )(1.26  10 C) -14 r =  -13 = 4.54 10 m E0 (8.00 10 J)

7  10-15 m (Size of a gold nucleus is about )

34. 3.0μC (a) What is the potential between two points situated 10 cm and 20 cm from a point charge? (b) To what location should the point at 20 cm be moved to increase this potential difference by a factor of two?

Solution V  kQ / r

(a) Relative to origin, find the potential at each point and then calculate the difference.  1 1    V = V1 - V2 = kQ -   r1 r2 

9 2 2 -6  1 1  3 = (9  10 N  m /C ) (3  10 C)    = 135  10 V.  0.1 m 0.2 m 

(b) To double the potential difference, move the charge from 20 cm to .

35. Unreasonable Results (a) What is the final speed of an electron accelerated from rest through a voltage of 25.0 MV by a negatively charged Van de Graaff terminal? (b) What is unreasonable about this result? (c) Which assumptions are responsible? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution 1 PE  KE  qV  mv2 i f 2 (a)

2qV 2(1.601019 C)(25.0106 V ) v   (8.7821018 m2/s2 )1/ 2 m 9.111031 kg  2.96109 m/s

(b) This velocity is far too great. It is faster than the speed of light.

(c) The assumption that the speed of the electron is far less than that of light and that the problem does not require a relativistic treatment produces an answer greater than the speed of light.

19.4 EQUIPOTENTIAL LINES

36. q (a) Sketch the equipotential lines near a point charge + . Indicate the direction of  3 q increasing potential. (b) Do the same for a point charge .

Solution

increasing V (a)

– 3 q

increasing V (b)

37. Sketch the equipotential lines for the two equal positive charges shown in Figure 19.27. Indicate the direction of increasing potential.

Solution

+q +q increasing V

38. q 1 q 2 Figure 19.28 shows the electric field lines near two charges and , the first having a magnitude four times that of the second. Sketch the equipotential lines for these two charges, and indicate the direction of increasing potential.  

Solution To draw the equipotential lines, remember that they are always perpendicular to electric field lines. The potential is greatest (most positive) near the positive charge,

q2 , q1 . and least (most negative) near the negative charge, In other words, the

q1 , potential increases as you move out from the charge and it increases as you

q2 . move towards the charge

39. Sketch the equipotential lines a long distance from the charges shown in Figure 19.28. Indicate the direction of increasing potential. OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution

– 3 q increasing V

40. Sketch the equipotential lines in the vicinity of two opposite charges, where the negative charge is three times as great in magnitude as the positive. See Figure 19.28 for a similar situation. Indicate the direction of increasing potential.

Solution

– 3 q q

increasing V

41. Sketch the equipotential lines in the vicinity of the negatively charged conductor in Figure 19.29. How will these equipotentials look a long distance from the object?

Solution

– – – – – – – – – – – – – – –

A long distance away from the conductor, the equipotential will be circles (spheres in three dimensions). OpenStax College Physics Instructor Solutions Manual Chapter 19

42. Sketch the equipotential lines surrounding the two conducting plates shown in Figure 19.30, given the top plate is positive and the bottom plate has an equal amount of negative charge. Be certain to indicate the distribution of charge on the plates. Is the field strongest where the plates are closest? Why should it be?

Solution

+ + + + + + +

– – – – – – –

The field is strongest where the plates are closest, because the field is inversely proportional to the separation distance.

43. (a) Sketch the electric field lines in the vicinity of the charged insulator in Figure 19.31. Note its non-uniform charge distribution. (b) Sketch equipotential lines surrounding the insulator. Indicate the direction of increasing potential.

Solution

+ + ++ + + + ++ + +

(a), (b)

44. The naturally occurring charge on the ground on a fine day out in the open country is

1.00 nC/m 2 . (a) What is the electric field relative to ground at a height of 3.00 m? (b) Calculate the electric potential at this height. (c) Sketch electric field and equipotential lines for this scenario. OpenStax College Physics Instructor Solutions Manual Chapter 19

Uniform electric field E from ground: Solution σ 1  10-9 C E = = = 1.13  102 N/C ε 8.8510-12 C2 /N m2 (a) (The negative sign indicates the electric field lines are directed inward.)

V  E d  (1.13102 N/C)(3 m)  3.39102 V (b) Electric potential . The electric field lines are perpendicular to the ground and pointing towards earth.

45. The lesser electric ray (Narcine bancroftii) maintains an incredible charge on its head and a charge equal in magnitude but opposite in sign on its tail (Figure 19.32). (a) Sketch the equipotential lines surrounding the ray. (b) Sketch the equipotentials when the ray is near a ship with a conducting surface. (c) How could this charge distribution be of use to the ray?

Solution (a) The electric field lines and equipotential lines are similar to those of the electric dipole of Figure 19.9 .

(b) Near a metal ship, the net electric field lines inside the conductor must be zero, and be perpendicular to the surface, similar to this figure: OpenStax College Physics Instructor Solutions Manual Chapter 19

+ + ++ + + + ++ + +

(c) The ray uses its charge distribution to produce a large electric potential to defend itself.

19.5 CAPACITORS AND DIELECTRICS

46. 180 μF What charge is stored in a capacitor when 120 V is applied to it?

Solution Q  CV  1.80104 F120 V  2.16102  21.6 mC

47. Find the charge stored when 5.50 V is applied to an 8.00 pF capacitor.

Solution Q  CV  8.001012 F5.50 V  4.401011 C  44.0 pC

48. What charge is stored in the capacitor in Example 19.8?

Solution Q  CV  8.00106 F1.00104 V 8.00102 C  80.0 mC

49. 2.00 μF 3.10 μC Calculate the voltage applied to a capacitor when it holds of charge. OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution Q 3.10106 C Q  CV  V   1.55 V C 2.00106 F

50. What voltage must be applied to an 8.00 nF capacitor to store 0.160 mC of charge?

Solution Q 1.60104 C V    2.00104 V  20.0 kV C 8.00109 F

51. 3.00 μC What capacitance is needed to store of charge at a voltage of 120 V?

Solution Q 3.00106 C Q  CV  C    2.50108 F  25.0 nF V 120 V

52. What is the capacitance of a large Van de Graaff generator’s terminal, given that it stores 8.00 mC of charge at a voltage of 12.0 MV?

Solution Q 8.00103 C Q  CV  C    6.67 1010 F  667 pF V 12.0 106 V

53. 5.00 m2 Find the capacitance of a parallel plate capacitor having plates of area that are separated by 0.100 mm of Teflon.

Solution  A 2.1 8.851012 F/m 5.00 m2 C  0      9.29107 F  0.93 μF d 0.100103 m

54. (a) What is the capacitance of a parallel plate capacitor having plates of area 1.50 m2 that are separated by 0.0200 mm of neoprene rubber? (b) What charge does it hold when 9.00 V is applied to it? OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution  A (6.7)(8.851012 F/m)(1.50 m2 ) C  0   4.447106 F  4.4 μF d 0.0200103 m (a)

Q  CV  4.45106 F9.00 V  4.0105 C (b)

55. 80.0 μF Integrated Concepts A prankster applies 450 V to an capacitor and then tosses it to an unsuspecting victim. The victim’s finger is burned by the discharge of the capacitor through 0.200 g of flesh. What is the temperature increase of the flesh? Is it reasonable to assume no phase change?

Solution c Let be the specific heat of the human body (see Table 14.1)

CV 2 E   mcT  cap 2 CV 2 (80.0 10-6 F)(450 V)2 T    11.57 C  11.6 C 2mc 2(0.200 10-3 kg)(3500 J/kg  C)

Yes, it is reasonable to assume no phase change because the temperature is still not close to the boiling point.

56. Unreasonable Results (a) A certain parallel plate capacitor has plates of area 4.00 m 2 , separated by 0.0100 mm of nylon, and stores 0.170 C of charge. What is the applied voltage? (b) What is unreasonable about this result? (c) Which assumptions are responsible or inconsistent?

Solution A 3.4(8.851012 F/m)(4.00 m2 ) C    1.20105 F 0 d 1.00105 m (a) Q  CV Now, using , it follows that

Q 0.170 C V    1.42104 V  14.2 kV C 1.2010-5 F

(b) The voltage is unreasonably large, more than 100 times the breakdown voltage of (14106 V/m 14 kV/ mm). nylon

(c) The assumed charge is unreasonably large and cannot be stored in a capacitor of these dimensions.

19.6 CAPACITORS IN SERIES AND PARALLEL

57. Find the total capacitance of the combination of capacitors in Figure 19.33.

Solution

Cp  C1  C2  12.5 μF 1 1 1   Cs Cp C3

CpC3 (12.5 μF)(0.300 μF) Cs    0.293 μF Cp  C3 12.80 μF

58. Suppose you want a capacitor bank with a total capacitance of 0.750 F and you possess numerous 1.50 mF capacitors. What is the smallest number you could hook together to achieve your goal, and how would you connect them?

Solution You must connect the capacitors in parallel, since the total capacitance is greater than the individual capacitances.

Cp 0.750 F N   3  500 capacitors Cp  C1  C2  ...  NC, C 1.50 10 F so that OpenStax College Physics Instructor Solutions Manual Chapter 19

59. 5.00 μF 8.00 μF What total capacitances can you make by connecting a and an capacitor together?

Solution

There are two ways in which you can connect two capacitors: in parallel and in series, the total capacitance in series is:

1 1 1 C1C2 (5.00 μF)(8.00 μF)    Cs    3.08 μF (series) Cs C1 C2 C1  C2 5.00 μF 8.00 μF

and when connected in parallel, the total capacitance is:

Cp  C1  C2  5.00 μF  8.00 μF  13.0 μF(parallel)

60. Find the total capacitance of the combination of capacitors shown in Figure 19.34.

Solution

1 1 1 C1C2 0.300 μF10.0 μF    Cs    0.291 μF Cs C1 C2 C1  C2 10.0 μF 0.300 μF

Cp  C3  Cs  2.50 μF 0.291 μF  2.791 μF  2.79 μF

61. Find the total capacitance of the combination of capacitors shown in Figure 19.35. OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution

C12  C1  C2  15.0 μF  0.750 μF  15.75 μF

1 1 1 C3C12 1.50 μF15.75 μF    C123    1.370 μF C123 C3 C12 C3  C12 1.50 μF 15.75 μF

1 1 1 C5C6 5.00 μF3.50 μF    C56    2.059 μF C56 C5 C6 C5  C6 8.50 μF

C  C56  C4  C123  2.059 μF  8.00 μF 1.370 μF  11.4 μF

62. 8.00 μF Unreasonable Results (a) An capacitor is connected in parallel to another 5.00 μF capacitor, producing a total capacitance of . What is the capacitance of the

second capacitor? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Solution Cp  C1  C2  C2  Cp - C1  5.00 μF  8.00 μF  - 3.00 μF (a)

C2 (b) You cannot have a negative capacitance.

(c) The assumption that they were hooked up in parallel, rather than in series, is incorrect. A parallel connection always produces a greater capacitance, while here a smaller capacitance was assumed. This could only happen if the capacitors are connected in series.

19.7 ENERGY STORED IN CAPACITORS OpenStax College Physics Instructor Solutions Manual Chapter 19

63. 10.0 μF (a) What is the energy stored in the capacitor of a heart defibrillator charged 9.00103 V to ? (b) Find the amount of stored charge.

2 Solution CV 2 10.0 106 F 9.00 103 V E       405 J cap 2 2 (a)

Q  CV  10.0106 F9000 V  9.00102 C  90.0 mC (b)

64. In open heart surgery, a much smaller amount of energy will defibrillate the heart. (a) 8.00 μF What voltage is applied to the capacitor of a heart defibrillator that stores 40.0 J of energy? (b) Find the amount of stored charge.

Solution 2E 2(40.0 J) V  cap   3.16103 V  3.16 kV C 8.00 106 F (a)

6 2 Q  2CEcap  2(8.0010 F)(40.0 J)  2.5310 C  25.3 mC (b)

65. 165 μF A capacitor is used in conjunction with a motor. How much energy is stored in it when 119 V is applied?

Solution CV 2 (165106 F)(118 V)2 E    1.15 J cap 2 2

66. 2.00 μF 7.40 μF Suppose you have a 9.00 V battery, a capacitor, and a capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series. (b) Do the same for a parallel connection. OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution (a) If the capacitors are connected in series, their total capacitance is:

1 1 1 C1C2 (2.00 μF)(7.40 μF)    Cs   1.575 μF . Cs C1 C2 C1  C2 9.40 μF

6 5 Q  CsV  (1.57410 F)(9.00 V) 1.4210 C Then,

To determine the energy stored in the capacitors: C V 2 (1.574 10-6 F)(9.00 V)2 E  s   6.38105 J cap 2 2

CV 2 E  2 Note: by using the form of the equation involving capacitance and

voltage, we can avoid using one of the parameters that we calculated, minimizing our chance of propagating an error.

(b) If the capacitors are connected in parallel, their total capacitance is given by

Cp  C1  C2  2.00 μF 7.40 μF  9.40 μF .

6 5 Q  CpV  (9.4010 F)(9.00 V)  8.46 10 C Again, .

C V 2 (9.40 10-6 F)(9.00 V)2 E  p   3.81104 J cap 2 2 And finally

67. A nervous physicist worries that the two metal shelves of his wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. (a) What is the capacitance of the empty shelves if they have area 1.00102 m2 and are 0.200 m apart? (b) What is the voltage between them if opposite charges of magnitude 2.00 nC are placed on them? (c) To show that this voltage poses a small hazard, calculate the energy stored. OpenStax College Physics Instructor Solutions Manual Chapter 19

Solution  A (8.851012 F/m)(0.100 m2 ) C  0   4.4251012 F  4.431012 F d 0.200 m (a)

Q 2.00109 C Q  CV  V    452 V C 4.4251012 F (b)

CV 2 (4.4251012 F)(452 V)2 E    4.52107 J cap 2 2 (c)

68. Show that for a given dielectric material the maximum energy a parallel plate Volume = Ad capacitor can store is directly proportional to the volume of dielectric ( ). Note that the applied voltage is limited by the dielectric strength.

2 2 Solution CV max  A/ d DSd E   0 max 2 2 DS where is the dielectric strength, and

 0 (DS) Ad Emax  Vmax  (DS)d 2 Ad . Thus, . Since is the volume of dielectric, it

Emax follows that is proportional to the volume of dielectric.

70. 9.60103 J Unreasonable Results (a) On a particular day, it takes of electric energy to start a truck’s engine. Calculate the capacitance of a capacitor that could store that amount of energy at 12.0 V. (b) What is unreasonable about this result? (c) Which assumptions are responsible?

Solution CV 2 2E 2(9600 J) E   C  cap  133 F cap 2 V 2 (12.0 V)2 (a)

(b) Such a capacitor would be too large to carry with a truck. The size of the capacitor would be enormous.

(c) It is unreasonable to assume a capacitor can store the amount of energy needed. OpenStax College Physics Instructor Solutions Manual Chapter 19