Question

The total average energy density of e.m waves whose electric field variation is given by$e=(50\frac{N}{C})\sin (\omega t-kx)$ will be nearly

• A. ${10}^{-8}J/m^3$
• B. ${10}^{-6}J/m^3$
• C. ${10}^{-10}J/m^3$
• D. ${10}^{-12}J/m^3$

Answer 1

The correct option is A

${10}^{-8}J/m^3$

The correct option is A.

The explanation for the correct option:

Step 1:Given Parameters:

electric field variation $e=(50\frac{n}{c})\sin (\omega t-kx)$

Step 2: What to find:

The total average energy density of electromagnetic waves $U_{av}=?$

Step 3: Solution:

from the equation given, $E=50N/C$

The total average energy density of electromagnetic waves is given as,

$U_{av}=\frac{1}{2}{\epsilon }_0{E}^2$

$U_{av}=\frac{1}{2}\times \left(8.85\times {10}^{-12}\right)\times {\left(50\right)}^2$

$U_{av}=1.1\times {10}^{-8}J/m^3$

Hence, The total average energy density of electromagnetic waves is $U_{av}\approx {10}^{-8}J/m^3$.