College Physics (4th Edition)

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

Chapter 22 - Problems - Page 864: 36


The average power of the source is $4.88\times 10^4~W$

Work Step by Step

We can use $E_{rms}$ to find the intensity: $I = c~\epsilon_0~E_{rms}^2$ $I = (3.0\times 10^8~m/s)~(8.85\times 10^{-12}~C^2/N~m^2)~(55\times 10^{-3}~V/m)^2$ $I = 8.0314\times 10^{-6}~W/m^2$ We can find the power of the source: $I = \frac{P}{A}$ $P = I~A$ $P = I~(4\pi~r^2)$ $P = (8.0314\times 10^{-6}~W/m^2)(4\pi)~(22,000~m)^2$ $P = 4.88\times 10^4~W$ The average power of the source is $4.88\times 10^4~W$.
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