Question:

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of \( 5 \times 10^{10} \, \text{Hz} \) and an amplitude of \( 50 \, \text{Vm}^{-1} \). The total average energy density of the electromagnetic field of the wave is: \([ \text{Use } \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{Nm}^2 ]\)

Updated On: Nov 20, 2024
  • \( 1.106 \times 10^{-8} \, \text{Jm}^{-3} \)
  • \( 4.425 \times 10^{-8} \, \text{Jm}^{-3} \)
  • \( 2.212 \times 10^{-8} \, \text{Jm}^{-3} \)
  • \( 2.212 \times 10^{-10} \, \text{Jm}^{-3} \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

The average energy density of the electric field is given by:

\[ U_E = \frac{1}{2} \epsilon_0 E^2 \]

Substituting the given values:

\[ U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 \]

Calculating:

\[ U_E = 1.106 \times 10^{-8} \, \text{Jm}^{-3} \]

Was this answer helpful?
0
0

Questions Asked in JEE Main exam

View More Questions