An electromagnetic wave of frequency ( I 10¹⁴ ~Hz ) is propagating along ( z )-axis. The amplitude of electric field is (4 Vm ⁻¹ ), then energy density of the electric field will be (Permittivity of free space (=8.8 10⁻¹² C ^2 ~N ⁻¹ ~m ⁻² ) )

Step-by-step solution

Given, electromagnetic wave frequency, (f_m=1.0 10¹⁴ ~Hz ) amplitude of the electric field, (E₀=4 Vm ⁻¹ ) permittivity of free space, ( ₀=8.8 10⁻¹² C ^2 ~N ⁻¹ ~m ⁻² ) The value of energy density (energy/volume) is given by ( u= 1 2 ₀ E₀^2 ) Putting the given values, we get ( aligned = 1 2 8.8 10⁻¹² (4)^2 ~J / m ^3 = 1 2 8.8 16 10⁻¹² u =70.4 10⁻¹² Jm ⁻³ aligned ) Hence, the energy density of the electric field will be (70.4 10⁻¹² Jm ⁻³ ).