JEE PHYSICS MCQ: ELECTROMAGNETIC WAVES

1 Easy

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Application +4 | -1

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times 10^{10} \mathrm{~Hz}

and an amplitude of

50 \mathrm{Vm}^{-1}

. The total average energy density of the electromagnetic field of the wave is : [Use

\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2

]

View answer and explanation

Correct option: C

1.106 \times 10^{-8} \mathrm{Jm}^{-3}

Option C (

1.106 \times 10^{-8} \mathrm{Jm}^{-3}

) is correct because the total average energy density of an electromagnetic wave is given by

u = \frac{1}{2} \epsilon_0 E^2

, where E is the amplitude of the electric field.

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