Question

A point source of electromagnetic radiation has an average power output of 800W. The maximum value of electric field at a distance 3.5m from the source will be

• A. 62.6 V/m
• B. 32.3 V/m
• C. 16.15 V/m
• D. 8.08 V/m

Answer 1

The correct option is A

62.6 V/m

$I=\frac{P_{av}}{4\pi r^2}=\frac{E_0^2}{2{\mu }_{0}c}$

$\therefore E_0=\sqrt{\left[\frac{{\mu }_{0}c{P}_{av}}{2\pi r^2}\right]}$

$=\sqrt{\left[\frac{(4\pi \times {10}^{-7})(3\times {10}^8)\left(800\right)}{2\times 3.14\times (3.5{)}^2}\right]}$

$=62.6V/m$