College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 22 - Problems - Page 867: 77


The intensity is $1.36 \times 10^{-6}~W/m^2$ The average energy density is $4.53\times 10^{-15}~J/m^3$

Work Step by Step

We can find the intensity: $I = (\frac{E_m}{\sqrt{2}})^2~c~\epsilon_0$ $I = \frac{E_m^2~c~\epsilon_0}{2}$ $I = \frac{(32.0\times 10^{-3}~V/m)^2~(3.0\times 10^8~m/s)~(8.85\times 10^{-12}~C^2/N~m^2)}{2}$ $I = 1.36 \times 10^{-6}~W/m^2$ The intensity is $1.36 \times 10^{-6}~W/m^2$ We can find the average energy density: $U = \frac{I}{c} = \frac{1.36 \times 10^{-6}~W/m^2}{3.0\times 10^8~m/s} = 4.53\times 10^{-15}~J/m^3$ The average energy density is $4.53\times 10^{-15}~J/m^3$.
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