College Physics (4th Edition)

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

Chapter 17 - Problems - Page 657: 88


$\Delta V = 8200~V$

Work Step by Step

We can find the potential difference: $U = \frac{1}{2}C~(\Delta V)^2$ $(\Delta V)^2 = \frac{2~U}{C}$ $\Delta V = \sqrt{\frac{2~U}{C}}$ $\Delta V = \sqrt{\frac{(2)(300~J)}{9\times 10^{-6}~F}}$ $\Delta V = 8200~V$
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