For the plane electromagnetic wave given by m | vedclass

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Question

$\frac{	ext { Electric energy density }}{	ext { Magnetic energy density }}=\frac{\frac{1}{2} \in_0 \mathrm{E}_{\mathrm{rms}}^2}{\left(\frac{\mathrm{

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$1$

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$\frac{	ext { Electric energy density }}{	ext { Magnetic energy density }}=\frac{\frac{1}{2} \in_0 \mathrm{E}_{\mathrm{rms}}^2}{\left(\frac{\mathrm{B}_{\mathrm{rms}}^2}{2 \mu_0}ight)}$

$=\left(\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{B}_{\mathrm{rms}}}ight)^2 \cdot \mu_0 \in_0 \quad\left[\mathrm{C}=\frac{1}{\mu_0 \epsilon_0}ight]$

$=\frac{\mathrm{C}^2}{\mathrm{C}^2}=1$

For the plane electromagnetic wave given by $\mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$ and $\mathrm{B}=\mathrm{B}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$, the ratio of average electric energy density to average magnetic energy density is

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