Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 19 - Electric Potential Energy and the Electric Potential - Problems - Page 539: 48


5.66 V

Work Step by Step

Energy of empty capacitor= Energy of capacitor filled with the dielectric $\implies \frac{1}{2}C_{0}(V_{0})^{2}=\frac{1}{2}CV^{2}$ But $C=\kappa C_{0}$ which gives $\frac{1}{2}C_{0}(V_{0})^{2}=\frac{1}{2}\times\kappa\times C_{0}\times V^{2}$ Dividing both sides by $\frac{C_{0}}{2}$, we get $V_{0}^{2}=\kappa\times V^{2}$ $\implies V=\sqrt {\frac{V_{0}^{2}}{\kappa}}=\frac{V_{0}}{\sqrt {\kappa}}=\frac{12.0\,V}{\sqrt {4.50}}=5.66\,V$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.