Q. 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) $54.77\ V/m$
Explanation
(d) Intensity of EM wave is given by
$I = \frac{P}{4\pi R^{2}} = v_{av}.c = \frac{1}{2}\varepsilon_{0}E_{0}^{2} \times c$
$\Rightarrow E_{0} = \sqrt{\frac{P}{2\pi R^{2}\varepsilon_{0}c}}$
$= \sqrt{\frac{800}{2 \times 3.14 \times (4)^{2} \times 8.85 \times 10^{- 12} \times 3 \times 10^{8}}}$
$= 54.77\frac{V}{m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.