The rms value of the electric field of an ele | vedclass

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(D) The average energy density $U_{av}$ of an electromagnetic wave is given by the formula $U_{av} = \frac{1}{2} \varepsilon_0 E_0^2$,where $E_0$ is t

Vedclass · Accepted Answer

$3.85 	imes 10^{-6} \ J \ m^{-3}$

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(D) The average energy density $U_{av}$ of an electromagnetic wave is given by the formula $U_{av} = \frac{1}{2} \varepsilon_0 E_0^2$,where $E_0$ is the peak electric field.Given $E_{rms} = 660 \ NC^{-1}$.We know that $E_0 = \sqrt{2} E_{rms} = \sqrt{2} 	imes 660 \ NC^{-1}$.Substituting this into the energy density formula:$U_{av} = \frac{1}{2} \varepsilon_0 (\sqrt{2} E_{rms})^2 = \frac{1}{2} \varepsilon_0 (2 E_{rms}^2) = \varepsilon_0 E_{rms}^2$.Using $\varepsilon_0 = 8.85 	imes 10^{-12} \ C^2 N^{-1} m^{-2}$:$U_{av} = 8.85 	imes 10^{-12} 	imes (660)^2$.$U_{av} = 8.85 	imes 10^{-12} 	imes 435600$.$U_{av} \approx 3.85 	imes 10^{-6} \ J \ m^{-3}$.

The rms value of the electric field of an electromagnetic wave emitted by a source is $660 \ NC^{-1}$. The average energy density of the electromagnetic wave is

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