Question

In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is $5\times {10}^{14}$Hz. The wave is propagating along z-axis. The average energy density of electric field, in $Joule{m}^3$, will be

• A. $1.1\times {10}^{-11}$
• B. $2.2\times {10}^{-12}$
• C. $3.3\times {10}^{-13}$
• D. $4.4\times {10}^{-14}$

Answer 1

The correct option is B

$2.2\times {10}^{-12}$

Average energy density of electric field is given by

$u_e=\frac{1}{2}{ϵ}_0{E}^2=\frac{1}{2}{ϵ}_0{\left(\frac{E_0}{\sqrt{2}}\right)}^2=\frac{1}{4}{ϵ}_0{E}_0^2$

$=\frac{1}{4}\times 8.85\times {10}^{-12}(1{)}^2=2.2\times {10}^{-12}J/m^3.$