Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 29 - Potential and Field - Exercises and Problems - Page 863: 28

Answer

$E = 1200~N/C$

Work Step by Step

Let $V$ be the volume of the space. We can find the strength of the electric field $E$: $\frac{U}{V} = \frac{1}{2}\epsilon_0~E^2$ $E^2 = \frac{2U}{V\epsilon_0}$ $E = \sqrt{\frac{2U}{V\epsilon_0}}$ $E = \sqrt{\frac{(2)(50\times 10^{-12}~J)}{(0.020~m)^3(8.854\times 10^{-12}~F/m)}}$ $E = 1200~N/C$
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