The rms value of the electric field of the li | vedclass

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Question

(A) The average total energy density $u$ of an electromagnetic wave is the sum of the electric energy density and the magnetic energy density.$u = u_E

Vedclass · Accepted Answer

$4.58 	imes 10^{-6} \; J/m^3$

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(A) The average total energy density $u$ of an electromagnetic wave is the sum of the electric energy density and the magnetic energy density.$u = u_E + u_B = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2$Since $B_{rms} = \frac{E_{rms}}{c}$,we have $u_B = \frac{1}{2\mu_0} \left(\frac{E_{rms}}{c}ight)^2 = \frac{1}{2\mu_0} (E_{rms}^2 \varepsilon_0 \mu_0) = \frac{1}{2} \varepsilon_0 E_{rms}^2$.Thus,the total energy density is $u = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2} \varepsilon_0 E_{rms}^2 = \varepsilon_0 E_{rms}^2$.Given $\varepsilon_0 = 8.85 	imes 10^{-12} \; F/m$ and $E_{rms} = 720 \; N/C$:$u = (8.85 	imes 10^{-12}) 	imes (720)^2$$u = 8.85 	imes 10^{-12} 	imes 518400$$u = 4.58784 	imes 10^{-6} \; J/m^3 \approx 4.58 	imes 10^{-6} \; J/m^3$.

The $rms$ value of the electric field of the light coming from the Sun is $720 \; N/C$. The average total energy density of the electromagnetic wave is

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