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Question

Physics Question on Electromagnetic waves

Average energy density of electromagnetic field of a wave having amplitude of 48 Vm−1V m^{-1} is

A

2×107J2 \times 10^7J

B

3×109J3 \times 10 ^9 J

C

1×10−5Jm−31 \times 10 ^{-5} Jm^{-3}

D

4×10−4J4 \times 10 ^{-4} J

Answer

1×10−5Jm−31 \times 10 ^{-5} Jm^{-3}

Explanation

Solution

U=12∈0 E2=12(8.85×10−12)482=1×10−8J m−3U=\frac{1}{2}\in_0\,E^2=\frac{1}{2}(8.85 \times 10^{-12})48^2=1\times 10^{-8}J\,m^{-3}