Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

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

Options:

500

73

$\frac{500}{3}$

$\frac{250}{3}$

Correct Answer:

73

Explanation:

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

$E_0 = 73V/m$