College Physics (4th Edition)

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

Chapter 17 - Problems - Page 655: 58


(a) $\Delta V = 3000~V$ (b) $Q = 6.0\times 10^{-3}~C$

Work Step by Step

(a) We can find the maximum potential difference before breakdown: $\Delta V = E~d$ $\Delta V = (3\times 10^6~V/m)(0.0010~m)$ $\Delta V = 3000~V$ (b) We can find the magnitude of the greatest charge the capacitor can store: $Q = C~\Delta V$ $Q = (2.0\times 10^{-6}~F)(3000~V)$ $Q = 6.0\times 10^{-3}~C$
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