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Physics Question on Work-energy theorem

The energy associated with electric field is E and with magnetic field is B for an electromagnetic wave in free space is given by( ∈0∈_0-permittivity of free space μ0μ_0-permeability of free space)

A

UE=E22∈0,UB=μ0B22U_E=\frac{E^2}{2∈_0},U_B=\frac{μ_0B^2}{2}

B

UE=∈E22,UB=μ0B22U_E=\frac{∈E^2}{2},U_B=\frac{μ_0B^2}{2}

C

UE=E22∈0,UB=B22μ0U_E=\frac{E^2}{2∈_0},U_B=\frac{B^2}{2μ_0}

D

UE=∈0E22,UB=B22μ0U_E=\frac{∈_0E^2}{2},U_B=\frac{B^2}{2μ_0}

Answer

UE=∈0E22,UB=B22μ0U_E=\frac{∈_0E^2}{2},U_B=\frac{B^2}{2μ_0}

Explanation

Solution

The correct option is(D): UE=∈0E22,UB=B22μ0U_E=\frac{∈_0E^2}{2},U_B=\frac{B^2}{2μ_0}