StreakPeaked· Practice

ExamsJEE AdvancedPhysics

An electromagnetic wave of frequency 1 x 10¹⁴ Hz propagates along the z-axis with an electric field amplitude of 4 V/m. Given epsilon₀ = 8.8 x 10⁻¹² C²/(N*m²), what is the average energy density of the electric field?

  1. 35.2 × 10⁻¹⁰ J/m³
  2. 35.2 × 10⁻¹¹ J/m³
  3. 35.2 × 10⁻¹² J/m³
  4. 35.2 × 10⁻¹³ J/m³

Correct answer: 35.2 × 10⁻¹² J/m³

Solution

For a sinusoidal EM wave, the time-averaged electric energy density is u_E = (1/4)*epsilon₀*E0² = (1/4) * 8.8e-12 * 16 = 35.2e-12 J/m³.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →