Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 22 - Electromagnetic Waves - Problems - Page 641: 27

Answer

$4.50\times10^{-6}J$.

Work Step by Step

Use equation 22–6a to link the instantaneous value of energy density to the electric field strength. $$u=\epsilon_oE^2$$ Now use equation 22–7 to relate this to the intensity of sunlight. $$\overline{u}=\frac{\overline{I}}{c}=\frac{1}{2}\epsilon_oE^2$$ Now find the average energy contained in a volume. Use the intensity value from Example 22-4. $$\overline{U}=\overline{u}V=\frac{\overline{I}}{c}V$$ $$\overline{U} =\frac{1350W/m^2}{3.00\times10^8m/s}(1.00m^3)= 4.50\times10^{-6}J $$
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