Question

An electromagnetic wave of frequency $1\times {10}^{14}\text{Hz}$ is propagating along z-axis. The amplitude of electric field is $4\text{V/m}$. If ${ϵ}_0=8.8\times {10}^{-12}{\text{C}}^2{\text{/N-m}}^2$, then the average energy density of electric field will be,

• A. $35.2\times {10}^{-11}{\text{J/m}}^3$
• B. $35.2\times {10}^{-12}{\text{J/m}}^3$
• C. $35.2\times {10}^{-13}{\text{J/m}}^3$
• D. $35.2\times {10}^{-10}{\text{J/m}}^3$

Answer 1

The correct option is B $35.2\times {10}^{-12}{\text{J/m}}^3$

The average energy density of electric field is,
$\frac{\Delta U}{\Delta V}=⟨\frac{1}{2}{ϵ}_0{E}_{rms}^2⟩$
$=\frac{1}{2}{ϵ}_0⟨E_{rms}^2⟩$
$=\frac{1}{2}\times {ϵ}_0\times {\left(\frac{E_0}{\sqrt{2}}\right)}^2(\because E_{rms}=\frac{E_0}{\sqrt{2}})$
$=\frac{1}{4}\times 8.8\times {10}^{-12}\times (4{)}^2$
$=35.2\times {10}^{-12}{\text{J/m}}^3$

Hence, $\left(B\right)$ is the correct answer.