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. $\frac{54.77V}{m}$
• B. $\frac{44.7V}{m}$
• C. $\frac{55V}{2m}$
• D. none of these

Answer 1

The correct option is A $\frac{54.77V}{m}$

$\begin{array}{c}IntensityofEMwaveisgivenbyI=\frac{P}{4\pi R^2}\\ v_{ac}\cdot c=\frac{1}{2}{\in }_{0}E{}_0^2\times c\\ E_0=\sqrt{\frac{P}{2\pi R^2{\in }_{0}c}}\\ =\sqrt{\frac{800}{2\times 3.14\times {\left(4\right)}^2\times 8.85\times {10}^{-12}\times 3\times {10}^8}}=\frac{54.77V}{m}\end{array}$

Hence,

option $\left(A\right)$ is correct answer.