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: 29

Answer

(a) $U = 1.1\times 10^{-7}~J$ (b) $0.70~J/m^3$

Work Step by Step

(a) We can find the energy: $U = \frac{1}{2}C~(\Delta V)^2$ $U = \frac{(\epsilon_0~A)~(\Delta V)^2}{2~d}$ $U = \frac{(\epsilon_0~\pi~r^2)~(\Delta V)^2}{2~d}$ $U = \frac{(8.854\times 10^{-12}~F/m)~(\pi)~(0.010~m)^2~(200~V)^2}{(2)(0.50\times 10^{-3}~m)}$ $U = 1.1\times 10^{-7}~J$ (b) Let $V$ be the volume of the space between the plates. We can find the energy density: $\frac{U}{V} = \frac{1.1\times 10^{-7}~J}{(\pi)~(0.010~m)^2~(0.50\times 10^{-3}~m)} = 0.70~J/m^3$
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