Given, electromagnetic wave frequency, \(f_m=1.0 \times 10^{14} \mathrm{~Hz}\)
amplitude of the electric field, \(E_0=4 \mathrm{Vm}^{-1}\)
permittivity of free space, \(\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)
The value of energy density (energy/volume) is given by
\(\therefore \quad u=\frac{1}{2} \varepsilon_0 E_0^2\)
Putting the given values, we get
\(\begin{aligned}
=\frac{1}{2} \times 8.8 \times 10^{-12} \times(4)^2 \mathrm{~J} / \mathrm{m}^3 =\frac{1}{2} \times 8.8 \times 16 \times 10^{-12} \\
\therefore \quad u =70.4 \times 10^{-12} \mathrm{Jm}^{-3}
\end{aligned}\)
Hence, the energy density of the electric field will be \(70.4 \times 10^{-12} \mathrm{Jm}^{-3}\).