College Physics (4th Edition)

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

Chapter 17 - Problems - Page 656: 77


(a) $C = 7.1\times 10^{-6}~F$ (b) $\Delta V = 1.1\times 10^4~V$

Work Step by Step

(a) We can find the capacitance: $E = \frac{Q^2}{2C}$ $C = \frac{Q^2}{2E}$ $C = \frac{(8.0\times 10^{-2}~C)^2}{(2)(450~J)}$ $C = 7.1\times 10^{-6}~F$ (b) We can find the potential difference: $E = \frac{Q~\Delta V}{2}$ $\Delta V = \frac{2~E}{Q}$ $\Delta V = \frac{(2)(450~J)}{8.0\times 10^{-2}~C}$ $\Delta V = 1.1\times 10^4~V$
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