Question

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

• A. $64.7V{m}^{-1}$
• B. $57.8V{m}^{-1}$
• C. $56.72V{m}^{-1}$
• D. $54.77V{m}^{-1}$

Answer 1

The correct option is D $54.77V{m}^{-1}$

Given,
${\mu }_0=4\pi \times {10}^{-7},P_{av}=800W,r=4m,c=3\times {10}^{8}m{s}^{-1}$
Intensity of electromagnetic wave is
$I=\frac{P_{av}}{4\pi r^2}=\frac{1}{2}\frac{E_0^2}{{\mu }_{0}c}$
or
$E_0=\sqrt{\frac{{\mu }_{0}c{P}_{av}}{2\pi r^2}}$

$E_0=\sqrt{\frac{(4\pi \times {10}^{-7})(3\times {10}^8)\left(800\right)}{2\pi \times 4^2}}$
$E_0=54.77V{m}^{-1}$