Question

Find the average energy density(in SI unit) of The electric field of a plane electromagnetic wave varies with time of amplitude $2V{m}^{-1}$ propagating along $z$ -axis.

• A. $13.29\times 10^{-12}$
• B. $8.86\times 10^{-12}$
• C. $17.72\times 10^{-12}$
• D. $4.43\times 10^{-12}$

Answer 1

Answer: (B)

Amplitude of electric field and magnetic field are related by the relation
$\frac{E_0}{B_0}=c$
Average energy density of the magnetic field is
$u_B=\frac{1}{4}\frac{B_0^2}{{\left(\mu \right)}_0}=\frac{1}{4}\frac{E_0^2}{{\left(\mu \right)}_0{c}^2}\left(\because B_0=\frac{E_0}{c}\right)$
$=\frac{1}{4}{\left(ϵ\right)}_0{E}_0^2\left(\because c=\frac{1}{\sqrt{{\left(\mu \right)}_0{\left(ϵ\right)}_0}}\right)$
$=\frac{1}{4}\times 8.854\times {\left(10\right)}^{-12}\times {\left(2\right)}^2=8.854\times {\left(10\right)}^{-12}J{m}^{-3}$
$\approx 8.86\times 10^{-12}J{m}^{-3}$