Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 31 - Electromagnetic Fields and Waves - Exercises and Problems - Page 901: 20


(a) $I = 2.2\times 10^{-6}~W/m^2$ (b) $E_0 = 0.041~V/m$

Work Step by Step

(a) We can find the intensity of the light wave: $I = \frac{P}{A}$ $I = \frac{P}{4\pi r^2}$ $I = \frac{2.5\times 10^4~W}{(\pi) (3.0\times 10^4~m)^2}$ $I = 2.21\times 10^{-6}~W/m^2$ $I = 2.2\times 10^{-6}~W/m^2$ (b) We can find the electric field amplitude: $I = \frac{1}{2}~E_0^2~\epsilon_0~c$ $E_0^2 = \frac{2~I}{\epsilon_0~c}$ $E_0 = \sqrt{\frac{2~I}{\epsilon_0~c}}$ $E_0 = \sqrt{\frac{(2)(2.21\times 10^{-6}~W/m^2)}{(8.854\times 10^{-12}~F/m)~(3.0\times 10^8~m/s)}}$ $E_0 = 0.041~V/m$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.