The electric field in $\mathrm{NC}^{-1}$ of an electromagnetic wave is $\mathrm{E}=36 \sqrt{\pi}…

The electric field in $\mathrm{NC}^{-1}$ of an electromagnetic wave is $\mathrm{E}=36 \sqrt{\pi} \operatorname{Sin}(\omega \mathrm{t}-\mathrm{kx})$. The average energy density of the electromagnetic wave due to the electric field is $\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)$
  1. $36 \times 10^{-9} \mathrm{Jm}^{-3}$
  2. $18 \times 10^{-9} \mathrm{Jm}^{-3}$
  3. $36 \times 10^{-7} \mathrm{Jm}^{-3}$
  4. $18 \times 10^{-7} \mathrm{Jm}^{-3}$

Solution

The average encrgy density of the electromagnetic wave is given by $\begin{aligned} & \mathrm{U}_{\mathrm{av}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2 \\ & =\frac{1}{2} \times 8.86 \times 10^{-12} \times(36)^2 \pi=18 \times 10^{-9} \mathrm{~J} / \mathrm{m}^3 \end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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