The rms value of the electric field of an electromagnetic wave emitted by a source is $660 \mathrm{NC}^{-1}$…

The rms value of the electric field of an electromagnetic wave emitted by a source is $660 \mathrm{NC}^{-1}$. The average energy density of the electromagnetic wave is
  1. $1.75 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}$
  2. $2.75 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}$
  3. $4.85 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}$
  4. $3.85 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}$

Solution

$\begin{aligned} & \mathrm{E}_{\mathrm{rms}}=660 \mathrm{NC}^{-1} \\ & \therefore \mathrm{E}_0=\sqrt{2} \mathrm{E}_{\mathrm{rms}}=\sqrt{2} \times 660 \mathrm{NC}^{-1} \\ & \therefore \text { Average density, } \mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \\ & =\frac{1}{2} \times 8.85 \times 10^{-12} \times(\sqrt{2} \times 660)^2 \\ & =3.85 \times 10^{-6} \mathrm{~J} \mathrm{m}^{-3}\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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