A point source of electromagnetic radiation h | vedclass

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(B) 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 ter

Vedclass · Accepted Answer

$54.77$

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(B) 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: $P = 800\,W$,$R = 4.0\,m$,$\varepsilon_0 = 8.854 	imes 10^{-12}\,F/m$,and $c = 3 	imes 10^8\,m/s$.$E_0 = \sqrt{\frac{800}{2 	imes 3.1416 	imes (4.0)^2 	imes 8.854 	imes 10^{-12} 	imes 3 	imes 10^8}}$.$E_0 = \sqrt{\frac{800}{133.27 	imes 10^{-4}}} \approx \sqrt{60028.5} \approx 54.77\,V/m$.

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

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