College Physics (4th Edition)

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

Chapter 17 - Problems - Page 655: 59


(a) The electric field does not change. (b) The potential difference increases.

Work Step by Step

(a) The electric field between the plates can be expressed as $E = \frac{Q/A}{\epsilon_0}$ The area does not change and obviously the constant $\epsilon_0$ does not change. Since the capacitor is disconnected from the battery, the charge $Q$ on the plates can not change. Therefore, the electric field does not change. (b) $V = E~d$ Since the distance $d$ increases while the electric field does not change, the potential difference increases.
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