99 Percentile Qs Bank for JEE MainPhysicsElectromagnetic Waves
An electromagnetic wave of frequency 1 × 10 14 Hz is propagating along z -axis. The amplitude of the electric field is 4 V/m . If ε 0 = 8 . 8 × 1 0 - 1 2 C 2 / N-m 2 , then the average energy density of electric field will be
Options
- A35.2 × 10 − 12 J/m 3
- B35.2 × 10 − 10 J/m 3
- C35.2 × 10 − 11 J/m 3
- D35.2 × 10 − 13 J/m 3
Correct answer
A. 35.2 × 10 − 12 J/m 3
Step-by-step solution
f = 1 0 1 4 Hz E 0 = 4 V/m ∈ 0 = 8.8 × 1 0 - 1 2 C 2 N - m 2 Energy density of electric field = 1 2 (Total energy density) = 1 2 · 1 2 ∈ 0 E 2 = 1 2 . 1 2 8.8 × 1 0 - 1 2 × 4 2 = 1 2 · 1 2 × 16 × 8.8 × 10 − 12 J/m 3 = 35.2 × 10 − 12 J/m 3