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An electromagnetic wave of frequency $1 \times 10^{14}$ hertz is propagating along $\mathrm{z}$-axis. The amplitude of electric field is $4 \mathrm{~V} / \mathrm{m}$. If $\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}-$ $\mathrm{m}^2$, then average energy density of electric field will be:
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Verified Answer
The correct answer is:
$35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3$
$35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3$
Given: Amplitude of electric field,
$$
E_0=4 \mathrm{v} / \mathrm{m}
$$
Absolute permitivity,
$$
\varepsilon_0=8.8 \times 10^{-12} \mathrm{c}^2 / \mathrm{N}-\mathrm{m}^2
$$
Average energy density $u_E=$ ?
Applying formula,
Average energy density $u_E=\frac{1}{4} \varepsilon_0 E^2$
$$
\begin{aligned}
&\Rightarrow \quad u_E=\frac{1}{4} \times 8.8 \times 10^{-12} \times(4)^2 \\
&=35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3
\end{aligned}
$$
$$
E_0=4 \mathrm{v} / \mathrm{m}
$$
Absolute permitivity,
$$
\varepsilon_0=8.8 \times 10^{-12} \mathrm{c}^2 / \mathrm{N}-\mathrm{m}^2
$$
Average energy density $u_E=$ ?
Applying formula,
Average energy density $u_E=\frac{1}{4} \varepsilon_0 E^2$
$$
\begin{aligned}
&\Rightarrow \quad u_E=\frac{1}{4} \times 8.8 \times 10^{-12} \times(4)^2 \\
&=35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3
\end{aligned}
$$
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