Question

Ina n electromagnetic wave, the amplitude of electric field is 1 V/m. What is average energy density of electric field ?

Answer 1

Solution

We know,

(Energy Density) avg = $\frac{1}{2}{ϵ}_0{E}^2=\frac{1}{2}{ϵ}_0(\frac{E_0}{\sqrt{2}}{)}^2$

${\mu }_{ϵ}=\frac{1}{2}{ϵ}_0\frac{E_0^2}{2}=\frac{{ϵ}_0(E_0{)}^2}{4}$

Here, $E_0^2=(1{)}^2{v}^2{m}^{-2}.$ so,

${\mu }_{ϵ}=\frac{1}{2\times 2}{ϵ}_0{E}^2=\frac{8.85\times {10}^{-12}}{4}\times (1{)}^2$

${\mu }_{ϵ}=2.21\times {10}^{-12}J/m^3$