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University Physics Volume 2

Chapter 8

University Physics Volume 2Chapter 8

Chapter 8

Check Your Understanding

8.1

1.1 × 10 −3 m 1.1 × 10 −3 m

8.3

3.59 cm, 17.98 cm

8.4

a. 25.0 pF; b. 9.2

8.5

a. C=0.86pF,Q1=10pC,Q2=3.4pC,Q3=6.8pCC=0.86pF,Q1=10pC,Q2=3.4pC,Q3=6.8pC;
 b. C=2.3pF,Q1=12pC,Q2=Q3=16pCC=2.3pF,Q1=12pC,Q2=Q3=16pC;
 c. C=2.3pF,Q1=9.0pC,Q2=18pC,Q3=12pC,Q4=15pCC=2.3pF,Q1=9.0pC,Q2=18pC,Q3=12pC,Q4=15pC

8.6

a.4.0×10−13J4.0×10−13J; b. 9 times

8.7

a. 3.0; b. C=3.0C0C=3.0C0

8.9

a. C0=20pFC0=20pF, C=42pFC=42pF; b. Q0=0.8nCQ0=0.8nC, Q=1.7nCQ=1.7nC; c. V0=V=40VV0=V=40V; d. U0=16nJU0=16nJ, U=34nJU=34nJ

Conceptual Questions

1.

no; no

3.

false

5.

no

7.

3.0 μ F , 0.33 μ F 3.0 μ F , 0.33 μ F

9.

answers may vary

11.

Dielectric strength is a critical value of an electrical field above which an insulator starts to conduct; a dielectric constant is the ratio of the electrical field in vacuum to the net electrical field in a material.

13.

Water is a good solvent.

15.

When energy of thermal motion is large (high temperature), an electrical field must be large too in order to keep electric dipoles aligned with it.

17.

answers may vary

Problems

19.

21.6 mC

21.

1.55 V

23.

25.0 nF

25.

1.1 × 10 −3 m 2 1.1 × 10 −3 m 2

27.

500 µC

29.

1:16

31.

a. 1.07 nC; b. 267 V, 133 V

33.

0.29 μ F 0.29 μ F

34.

500 capacitors; connected in parallel

35.

3.08μF3.08μF (series) and 13.0μF13.0μF (parallel)

37.

11.4 μ F 11.4 μ F

39.

0.89 mC; 1.78 mC; 444 V

41.

7.5 μ J 7.5 μ J

43.

a. 405 J; b. 90.0 mC

45.

1.17 J

47.

a. 4.43×10−9F4.43×10−9F; b. 0.453 V; c. 4.53×10−10J4.53×10−10J; d. no

49.

0.7 mJ

51.

a. 7.1 pF; b. 42 pF

53.

a. before 3.00 V; after 0.600 V; b. before 1500 V/m; after 300 V/m

55.

a. 3.91; b. 22.8 V

57.

a. 37 nC; b. 0.4 MV/m; c. 19 nC

59.

a. 4.4μF4.4μF; b. 4.0×10-5C4.0×10-5C

61.

0.0135 m 2 0.0135 m 2

63.

0.185 μ J 0.185 μ J

Additional Problems

65.

a. 0.277 nF; b. 27.7 nC; c. 50 kV/m

67.

a. 0.065 F; b. 23,000 C; c. 4.0 GJ

69.

a. 75.6μC75.6μC; b. 10.8 V

71.

a. 0.13 J; b. no, because of resistive heating in connecting wires that is always present, but the circuit schematic does not indicate resistors

73.

a. −3.00μF−3.00μF; b. You cannot have a negative C2C2 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.

75.

a. 14.2 kV; b. The voltage is unreasonably large, more than 100 times the breakdown voltage of nylon. c. The assumed charge is unreasonably large and cannot be stored in a capacitor of these dimensions.

Challenge Problems

77.

a. 89.6 pF; b. 6.09 kV/m; c. 4.47 kV/m; d. no

79.

a. 421 J; b. 53.9 mF

81.

C=ε0A/(d1+d2)C=ε0A/(d1+d2)

83.

proof

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