Question

An electromagnetic wave of frequency, $1\times {10}^4\text{Hz}$ is propagating along $z-$axis. The maximum amplitude of the electric field is $4\text{V/m}$. Average energy density of the electric field will be -

Answer 1

We know that, average electric energy density is given by,

$(u_E{)}_{\text{avg}}=\frac{1}{4}{ϵ}_0{E}_0^2$

$\Rightarrow (u_E{)}_{\text{avg}}=\frac{1}{4}\times 8.85\times {10}^{-12}\times 4^2$

$\Rightarrow (u_E{)}_{\text{avg}}=3.54\times {10}^{-11}{\text{J/m}}^3$

$\Rightarrow (u_E{)}_{\text{avg}}=35.4\times {10}^{-12}{\text{J/m}}^3$

Hence, option $\left(C\right)$ is the correct answer.