Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 19 - Electric Potential Energy and the Electric Potential - Problems - Page 539: 43


$1.1\times10^{3}$ V

Work Step by Step

Recall: U= $\frac{1}{2}CV^{2}$. Rearranging the equation, we get: V= $\sqrt {\frac{2U}{C}}$ Plugging in the values, we have: Potential difference: V= $\sqrt {\frac{2\times73J}{120\times10^{-6}F}}= 1.1\times10^{3}V$
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