A point source of electromagnetic radiation h | vedclass

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(D) The intensity $I$ of an electromagnetic wave at a distance $R$ from a point source is given by $I = \frac{P}{4\pi R^2}$.Also,the intensity in term

Vedclass · Accepted Answer

$54.77$

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(D) The intensity $I$ of an electromagnetic wave at a distance $R$ from a point source is given by $I = \frac{P}{4\pi R^2}$.Also,the intensity in terms of the maximum electric field $E_0$ is $I = \frac{1}{2} \varepsilon_0 E_0^2 c$.Equating the two expressions: $\frac{P}{4\pi R^2} = \frac{1}{2} \varepsilon_0 E_0^2 c$.Solving for $E_0$: $E_0 = \sqrt{\frac{P}{2\pi R^2 \varepsilon_0 c}}$.Substituting the given values: $P = 800 \, W$,$R = 4.0 \, m$,$\varepsilon_0 = 8.85 	imes 10^{-12} \, F/m$,and $c = 3 	imes 10^8 \, m/s$.$E_0 = \sqrt{\frac{800}{2 	imes 3.14 	imes (4)^2 	imes 8.85 	imes 10^{-12} 	imes 3 	imes 10^8}}$.$E_0 = \sqrt{\frac{800}{1331.328 	imes 10^{-4}}} = \sqrt{6008.9} \approx 54.77 \, V/m$.

$A$ point source of electromagnetic radiation has an average power output of $800 \, W$. The maximum value of the electric field at a distance $4.0 \, m$ from the source is .... $V/m$.

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