Question

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 is

• A. 64.7 V/m
• B. 57.8 V/m
• C. 56.72 V/m
• D. 54.77 V/m

Answer 1

The correct option is D 54.77 V/m

Intensity of EM wave is given by

$I=\frac{P}{4\pi R^2}=v_{av}.C=\frac{1}{2}{ϵ}_0{E}_0^2\times c$
$\Rightarrow E_0=\sqrt{\frac{P}{2\pi R^2{ϵ}_{0}c}}$
$=\sqrt{\frac{800}{2\times 3.14\times (4{)}^2\times 8.85\times {10}^{-12}\times 3\times {10}^8}}$
=$54.77\frac{V}{m}$