University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 24 - Capacitance and Dielectrics - Problems - Discussion Questions - Page 808: Q24.7


See explanation.

Work Step by Step

The capacitance is halved. For a parallel-plate capacitor, $C=\frac{\epsilon_oA}{d}$. The electric field stays the same, because for a parallel-plate capacitor, $E=\frac{Q}{\epsilon_oA }$. The potential difference is doubled. The electric field stays the same, and the potential difference is proportional to the electric field strength (same) and the distance between the parallel plates (doubled). The stored energy is doubled, because $U=\frac{1}{2}CV^2$ and the capacitance is halved while the potential difference increased by a factor of 2.
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