Answer
$9.6\times10^{-5}F$.
Work Step by Step
Find the capacitance from the stored energy, using equation 17–10.
$$PE=\frac{1}{2}CV^2$$
$$C=\frac{2PE}{V^2}=\frac{2(1200J)}{(5.0\times10^3V)^2}=9.6\times10^{-5}F $$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
Published by
Pearson
ISBN 10:
0-32162-592-7
ISBN 13:
978-0-32162-592-2
Chapter 17 - Electric Potential - Problems - Page 498: 53
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