In an electromagnetic wave,the amplitude of t | vedclass

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(B) The average energy density of the electric field in an electromagnetic wave is given by the formula:$\rho_E = \frac{1}{2} \epsilon_0 E_{rms}^2$Sin

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$2.2 	imes 10^{-12}$

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(B) The average energy density of the electric field in an electromagnetic wave is given by the formula:$ho_E = \frac{1}{2} \epsilon_0 E_{rms}^2$Since the root-mean-square electric field $E_{rms} = \frac{E_0}{\sqrt{2}}$,where $E_0$ is the amplitude of the electric field:$ho_E = \frac{1}{2} \epsilon_0 \left( \frac{E_0}{\sqrt{2}} ight)^2 = \frac{1}{4} \epsilon_0 E_0^2$Given $E_0 = 1 \ V/m$ and $\epsilon_0 = 8.85 	imes 10^{-12} \ F/m$:$ho_E = \frac{1}{4} 	imes 8.85 	imes 10^{-12} 	imes (1)^2$$ho_E = 2.2125 	imes 10^{-12} \approx 2.2 	imes 10^{-12} \ J \ m^{-3}$.

In an electromagnetic wave,the amplitude of the electric field is $1 \ V/m$ and the frequency of the wave is $5 \times 10^{14} \ Hz$. The wave propagates in the positive $Z$-direction. The average energy density of the electric field in this wave is ........ $J \ m^{-3}$.

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