In a plane electromagnetic wave, the amplitude of its electric field is 1.0 V/m. The frequency of the wave is 0.5 × 1015 Hz. The average energy density of the wave is
(Given: ε0 = 8.85 × 10−12 C2N−1m−2)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.42 × 10−12 Jm−3
The instantaneous energy density of the electric field is \( \frac{1}{2}\varepsilon_0 E^{2} \). Over one cycle the average of \( E^{2} \) is \( E_0^{2}/2 \), so the electric field alone contributes \( \frac{1}{4}\varepsilon_0 E_0^{2} \) on average. The magnetic field carries an equal share, so the average energy density of the wave is
\( \displaystyle u_{\mathrm{av}} = \frac{1}{2}\varepsilon_0 E_0^{2} \)
\( \displaystyle u_{\mathrm{av}} = \frac{1}{2} \times 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \times (1.0\,\mathrm{V\,m^{-1}})^{2} \)
\( \displaystyle u_{\mathrm{av}} \approx 4.42\times10^{-12}\,\mathrm{J\,m^{-3}} \)
The frequency of the wave is not needed.