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

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Question

(A) The average energy density $u$ of an electromagnetic wave is given by the sum of the electric and magnetic energy densities.$u = u_E + u_B = \frac

Vedclass · Accepted Answer

$4.58 	imes 10^{-6}$

Vedclass · Answer

(A) The average energy density $u$ of an electromagnetic wave is given by the sum of the electric and magnetic energy densities.$u = u_E + u_B = \frac{1}{2} \varepsilon_0 E_0^2 = \varepsilon_0 E_{rms}^2$.Given $E_{rms} = 720 \ N \ C^{-1}$ and $\varepsilon_0 = 8.854 	imes 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$.Substituting the values:$u = 8.854 	imes 10^{-12} 	imes (720)^2$$u = 8.854 	imes 10^{-12} 	imes 518400$$u \approx 4.589 	imes 10^{-6} \ J \ m^{-3}$.Thus,the correct option is $A$.

The $rms$ value of the electric field of the light coming from the sun is $720 \ N \ C^{-1}$. The average energy density of the electromagnetic wave is $...... \ J \ m^{-3}$.

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