In a plane electromagnetic wave, the electric field varies with time with an amplitude of \( 1\,\mathrm{V\,m^{-1}} \). The frequency of the wave is \( 0.5\times10^{15}\,\mathrm{Hz} \) and the wave is propagating along the x-axis. What is the maximum energy density of the electric field of the wave?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 4.4\times10^{-12}\,\mathrm{J\,m^{-3}} \)
The energy density of an electric field \( E \) is \( u=\frac{1}{2}\varepsilon_0E^{2} \). This holds for the field of any charge configuration, including the field of a wave. It is maximum where the field has its amplitude \( E_0 \). The frequency and the direction of travel are not needed. Here \( \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \).
\( \begin{aligned} u_{\max} &= \frac{1}{2}\,\varepsilon_0E_0^{2} \\ &= \frac{1}{2}\times8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}\times(1\,\mathrm{V\,m^{-1}})^{2} \\ &\approx 4.4\times10^{-12}\,\mathrm{J\,m^{-3}} \end{aligned} \)
NCERT (2026-27):
Energy Stored in a Capacitor – “… If we define energy density as energy stored per unit volume of space, …
Energy density of electric field, \( u=(1/2)\,\varepsilon_0E^{2} \)\( \qquad (2.73) \)
Though we derived Eq. (2.73) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact, very general and holds true for electric field due to any configuration of charges.”