The rms value of the electric field of light | vedclass

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Question

(A) The average total energy density $u_{av}$ of an electromagnetic wave is given by the sum of the average electric energy density and average magnet

Vedclass · Accepted Answer

$4.58	imes10^{-6}\, J/m^3$

Vedclass · Answer

(A) The average total energy density $u_{av}$ of an electromagnetic wave is given by the sum of the average electric energy density and average magnetic energy density.$u_{av} = u_{E} + u_{B} = \frac{1}{2} \epsilon_{0} E_{rms}^{2} + \frac{1}{2} \frac{B_{rms}^{2}}{\mu_{0}}$Since $E_{rms} = c B_{rms}$,we have $u_{E} = u_{B}$,so $u_{av} = \epsilon_{0} E_{rms}^{2}$.Given $E_{rms} = 720\, N/C$ and $\epsilon_{0} = 8.854 	imes 10^{-12}\, C^2/(N \cdot m^2)$.$u_{av} = (8.854 	imes 10^{-12}) 	imes (720)^2$$u_{av} = 8.854 	imes 10^{-12} 	imes 518400$$u_{av} \approx 4.589 	imes 10^{-6}\, J/m^3$.

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

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