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 865: 64

Answer

$\approx 0.18\;\rm J/m^3$

Work Step by Step

We know that the energy density in the electric field is given by $$u =\dfrac{\epsilon_0E^2}{2}\tag 1$$ where $E$ is the electric field. We also know that the electric field at the surface of a sphere is given by $$E=\dfrac{k_eQ}{R^2}$$ where the potential at the surface of the sphere is given by $V=\dfrac{k_eQ}{R^2}$, so $$E=\dfrac{k_eQ}{R }\dfrac{1}{R}=\dfrac{V}{R}$$ Plug into (1). $$u =\dfrac{\epsilon_0V^2}{2R^2} $$ Plug the known; $$u =\dfrac{(8.85\times 10^{-12})(1000)^2}{2(0.5\times 10^{-2})^2} $$ $$u=\color{red}{\bf 0.177}\;\rm J/m^3$$
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