Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 26 - Potential and Field - Exercises and Problems - Page 738: 31


$V = 45~V$

Work Step by Step

We can find the required potential: $U = \frac{1}{2}CV^2$ $V^2 = \frac{2U}{C}$ $V = \sqrt{\frac{2U}{C}}$ $V = \sqrt{\frac{(2)(1.0~J)}{1.0\times 10^{-3}~F}}$ $V = 45~V$
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