1 31. The energy stored by a capacitor is given by U = 2 CV 2 , where V is the potential difference across its plates. We convert the given value of the energy to Joules. Since 1 J = 1 W s, we multiply by (103 W/kW)(3600 s/h) to obtain 10 kW h = 3.6 107 J
33. The energy per unit volume is 1 1 e u = 0E 2 = 0 2 2 4 0 r 2
FG H
IJ K
2
=
e2 . 32 2 0r 4
(a) At r = 1.00 103 m , with e = 1.60 1019 C and 0 = 8.85 1012 C2 /N m 2 , we have u = 9.16 1018 J/m3 . (b) Similarly, at r = 1.00 106 m , u = 9.16 106 J/m3 . (c
34. (a) The potential difference across C1 (the same as across C2) is given by V1 = V2 =
(15.0 F ) (100V ) = 50.0V. C3V = C1 + C2 + C3 10.0 F+5.00 F+15.0 F
Also, V3 = V V1 = V V2 = 100 V 50.0 V = 50.0 V. Thus, q1 = C1V1 = (10.0 F ) ( 50.0V ) = 5.00 104 C
35. (a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by 0 A d i , the charge is q = CV = 0 AVi d i . After the plates are pulled apart, their separation is d f and the potential difference is Vf.