Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 25 - Electromagnetic Waves - Problems and Conceptual Exercises - Page 902: 49

Answer

$E_{max}=61.4V/m$

Work Step by Step

We know that $E_{max}^2=\frac{2I_{avg}}{c\epsilon_{\circ}}$ $\implies E_{max}=\sqrt{\frac{2I_{avg}}{c\epsilon_{\circ}}}$ We plug in the known values to obtain: $E_{max}=\sqrt{\frac{2(5.002/m^2)}{(3\times 10^8m/s)(8.85\times 10^{-12}C^2/Nm^2)}}$ $E_{max}=61.4V/m$
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