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In a plane electromagnetic wave, the electric field oscillates with a frequency $2 \times 10^{10} \mathrm{~s}^{-1}$ and amplitude $40 \mathrm{Vm}^{-1}$, then the energy density due to electric field is $\left(\varepsilon_0=8.85 \times 10^{-12} \mathrm{Fm}^{-1}\right)$
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The correct answer is:
$3.54 \times 10^{-9} \mathrm{Jm}^{-3}$
Energy density due to electric field is
$\begin{aligned}
U & =\frac{1}{2} \varepsilon_0 E^2 \\
\text {Here, } E & =\frac{E_0}{\sqrt{2}}=\frac{40}{\sqrt{2}} \\
\therefore \quad U & =\frac{1}{2} \times 8.85 \times 10^{-12} \times \frac{40 \times 40}{2} \\
& =3.54 \times 10^{-9} \mathrm{~J} / \mathrm{m}^3
\end{aligned}$
$\begin{aligned}
U & =\frac{1}{2} \varepsilon_0 E^2 \\
\text {Here, } E & =\frac{E_0}{\sqrt{2}}=\frac{40}{\sqrt{2}} \\
\therefore \quad U & =\frac{1}{2} \times 8.85 \times 10^{-12} \times \frac{40 \times 40}{2} \\
& =3.54 \times 10^{-9} \mathrm{~J} / \mathrm{m}^3
\end{aligned}$
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