JEE Main2014PhysicsElectromagnetic WavesActual
An electromagnetic wave of frequency 1 10¹⁴ hertz is propagating along z -axis. The amplitude of electric field is 4 ~V / m . If ₀=8.8 10⁻¹² C ^2 / N - m ^2 , then average energy density of electric field will be:
Options
- A35.2 10⁻¹⁰ ~J / m ^3
- B35.2 10⁻¹¹ ~J / m ^3
- C35.2 10⁻¹² ~J / m ^3
- D35.2 10⁻¹³ ~J / m ^3
Correct answer
C. 35.2 10⁻¹² ~J / m ^3
Step-by-step solution
Given: Amplitude of electric field, E₀=4 v / m Absolute permitivity, ₀=8.8 10⁻¹² c ^2 / N - m ^2 Average energy density u_E= ? Applying formula, Average energy density u_E= 1 4 ₀ E^2 aligned & u_E= 1 4 8.8 10⁻¹² (4)^2 &=35.2 10⁻¹² ~J / m ^3 aligned