College Physics (4th Edition)

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

Chapter 17 - Problems - Page 655: 62


(a) $Q = 1.44\times 10^{-8}~C$ (b) The magnitude of the charge on the plates is halved. The electric field is halved.

Work Step by Step

(a) We can find the magnitude of the charge on each plate: $Q = C~\Delta V$ $Q = (1.20\times 10^{-9}~F)(12~V)$ $Q = 1.44\times 10^{-8}~C$ (b) Since the battery remains connected, the potential difference between the plates will be the same as the battery, which is 12 V. $Q = C~\Delta V$ $Q = (\frac{A~\epsilon_0}{d})~\Delta V$ Since the distance $d$ doubles while the other values remain the same, the magnitude of the charge on the plates is halved. $E = \frac{\Delta V}{d}$ Since the potential difference remains the same while the distance doubles, the electric field $E$ is halved.
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