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A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance $4.0 m$ from the source in $\frac{V}{m}$ is $\sqrt{1000 K} .$ Find the value of $k$ approximated to nearest integer.

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answer is 3.

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Detailed Solution

Intensity of e.m. wave is $I = \frac{P_{a v}}{4 π r^{2}} = \frac{1}{2} \in_{0} E_{0}^{2} C$

$∴ E_{0} = (\sqrt{3000}) \frac{V}{m}$

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