College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 17 - Problems - Page 657: 89


(a) $Q = 0.135~C$ (b) The average power is $3.0\times 10^5~W$

Work Step by Step

(a) We can find the charge: $Q = C~\Delta V$ $Q = (15\times 10^{-6}~F)(9.0\times 10^3~V)$ $Q = 0.135~C$ (b) We can find the energy: $U = \frac{1}{2}C~(\Delta V)^2$ $U = \frac{1}{2}(15\times 10^{-6}~F)~(9.0\times 10^3~V)^2$ $U = 607.5~J$ We can find the average power: $P = \frac{E}{t}$ $P = \frac{607.5~J}{2.0\times 10^{-3}~s}$ $P = 3.0\times 10^5~W$ The average power is $3.0\times 10^5~W$
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