Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

In an electromagnetic wave, the amplitude of electric field is 1V/m. The frequency of wave is 5 x 1014 Hz. The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be:

Options:

1.1 x 10–11

2.2 x 10–12

3.3 x 10–13

4.4 x 10-14

Correct Answer:

2.2 x 10–12

Explanation:

Average energy density of electric field is  $U_E = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 (\frac{E_0}{\sqrt 2})^2= \frac{1}{4} \epsilon_0 E_0^2 = \frac{1}{4} \times 8.85\times 10^{-12} \times 1^2 = 2.2\times 10^{-12} J/m^3$