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: 19

Answer

(a) it is two times greater

Work Step by Step

The relation between electric field strength and the potential difference is $E=Vd$ In the first case V=6.0 V and in the second case V=12.0 V. d is the same in both cases. Therefore, $E_{1}=6.0\,V\times d$ and $E_{2}=12.0\,V\times d=2(6.0\,V\times d)=2E_{1}$. In the second case, the electric field strength is two times greater.
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