A point source of Electromagnetic radiation has an average power output of 1500 W. The maximum value of electric field at a distance of 3 m from this source in Vm-1 is
Answer & explanation
Correct answer: option 2
Intensity $ I = \frac{pressure}{area}=\frac{p}{4\pi r^2}$
=average energy density × velocity
$=\frac{1}{2}ε_0E^2_0c$
$∴ E_0 = \sqrt{\frac{P}{4\pi ε_0r^2c}}=\sqrt{\frac{P}{2\pi ε_0r^2c}}$
here r = 3m , P = 1500 W
$E_0 = 73V/m$