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Physics Question on Electromagnetic waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×1010 Hz5 \times 10^{10} \, \text{Hz} and an amplitude of 50 Vm−150 \, \text{Vm}^{-1}. The total average energy density of the electromagnetic field of the wave is: [Use ϵ0=8.85×10−12 C2/Nm2][ \text{Use } \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{Nm}^2 ]

A

1.106×10−8 Jm−31.106 \times 10^{-8} \, \text{Jm}^{-3}

B

4.425×10−8 Jm−34.425 \times 10^{-8} \, \text{Jm}^{-3}

C

2.212×10−8 Jm−32.212 \times 10^{-8} \, \text{Jm}^{-3}

D

2.212×10−10 Jm−32.212 \times 10^{-10} \, \text{Jm}^{-3}

Answer

1.106×10−8 Jm−31.106 \times 10^{-8} \, \text{Jm}^{-3}

Explanation

Solution

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

UE=12ϵ0E2U_E = \frac{1}{2} \epsilon_0 E^2

Substituting the given values:

UE=12×8.85×10−12×(50)2U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2

Calculating:

UE=1.106×10−8 Jm−3U_E = 1.106 \times 10^{-8} \, \text{Jm}^{-3}