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 $8.8\times {10}^{-12}{\text{C}}^2/\text{N}.{\text{m}}^2$ , then 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 $35.2\times {10}^{-12}$ ${\text{J/m}}^3$

Given:
Frequency $\nu =1\times {10}^{14}\text{Hz}$
Electric field is $E=4\text{V/m}$
${\epsilon }_0=8.8\times {10}^{-12}{\text{C}}^2/\text{N}.{\text{m}}^2$
We know that,
Energy density is given as
$e=\frac{{\epsilon }_0{E}_0^2}{2}$

$=\frac{8.8\times {10}^{-12}\times 4^2}{2}{\text{J/m}}^3$

$e=70.4\times {10}^{-12}{\text{J/m}}^3$

Therefore , average energy density of electric field,

$=\frac{e}{2}=35.2\times {10}^{-12}{\text{J/m}}^3$

Hence , option $\left(A\right)$ is correct.