P26_082 - V 1 = q 1 /C 1 = 5 . 71 V. (d) Similarly, V 2 = q...

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82. (First problem of Cluster ) (a) We do not employ energy conservation since, in reaching equilibrium, some energy is dissipated either as heat or radio waves. Charge is conserved; therefore, if Q = C 1 V bat = 400 µ C, and q 1 and q 2 are the charges on C 1 and C 2 after the switch S is closed (and equilibrium is reached), then Q = q 1 + q 2 . After switch S is closed, the capacitor voltages are equal, so that V 1 = V 2 q 1 C 1 = q 2 C 2 which yields 3 4 q 1 = q 2 . Therefore, Q = q 1 + µ 3 4 q 1 which gives the result q 1 = 229 µ C. (b) The relation 3 4 q 1 = q 2 gives the result q 2 = 171 µ C. (c) We apply Eq. 27-1:
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Unformatted text preview: V 1 = q 1 /C 1 = 5 . 71 V. (d) Similarly, V 2 = q 2 /C 2 = 5 . 71 V (which is equal to V 1 , of course since that fact was used in the solution to part (a)). (e) When C 1 had charge Q and was connected to the battery, the energy stored was 1 2 C 1 V 2 bat = 2 . 00 10 3 J. The energy stored after S is closed is 1 2 C 1 V 2 1 + 1 2 C 2 V 2 2 = 1 . 14 10 3 J. The decrease is therefore 8 . 6 10 4 J....
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