Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

In an electromagnetic wave, the amplitude of electric field is 1 V/m. The frequency of wave is 5 × 10–14 Hz. The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be:

Options:

1.1 × 10–11

2.2 × 10–12

3.3 × 10–13

4.4 × 10–14

Correct Answer:

2.2 × 10–12

Explanation:

Average energy density is given by

$U_{E}=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{E_0}{\sqrt{2}}\right)^2=\frac{1}{4} \varepsilon_0 E_0^2$

$=\frac{1}{4} \times 0.85 \times 10^{-12} \times(1)^2$

= 2.2 × 10–12 J/m3