College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Multiple Choice Questions - Page 590: 18


(a) it increases.

Work Step by Step

General equation connecting charge and potential difference is $q=CV$ C is constant for the capacitor (as $\kappa$,d and A are kept constant). Given that $V_{1}=6\,V$ and $V_{2}=12\,V$. $q_{1}=C\times6\,V$ $q_{2}=C\times12\,V$ Comparing $q_{1}$ and $q_{2}$, we find that $q_{2}\gt q_{1}$. That is, charge on the plate increases.
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