P26_083

# P26_083 - V 2 = q 2 /C 2 = 14 . 3 V. (e) The initial energy...

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83. (Second problem of Cluster ) (a) The change (from the previous problem) is that the initial charge (before switch S is closed) is Q + Q 0 where Q is as before but Q 0 = C 2 (10 V) = 600 µ C. We assume the polarities of these capacitor charges are the same. With this modiFcation, we follow steps similar to those in the previous solution: Q + Q 0 = q 1 + q 2 = q 1 + µ 3 4 q 1 which yields q 1 = 571 µ C. (b) The relation 3 4 q 1 = q 2 gives the result q 2 = 429 µ C. (c) We apply Eq. 27-1: V 1 = q 1 /C 1 =14 . 3V .
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Unformatted text preview: V 2 = q 2 /C 2 = 14 . 3 V. (e) The initial energy now includes 1 2 C 2 (20 V) 2 in addition to the 1 2 C 1 V 2 bat computed in the previous case. Thus, the total initial energy is 8 . 00 × 10 − 3 J. And the Fnal stored energy is 1 2 C 1 V 2 1 + 1 2 C 2 V 2 2 = 7 . 14 × 10 − 3 J. The decrease is therefore 8 . 6 × 10 − 4 J, as it was in the previous problem....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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