NEET UG PHYSICS MCQ: ELECTROMAGNETIC WAVES

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An electromagnetic wave of frequency

1\mathrm{ } \times \mathrm{ }{10^{14}}Hz

is propagating along z-axis. The amplitude of the electric field is 4 V/m. If

{\varepsilon _{0\;}} = \mathrm{ }8.8\mathrm{ } \times \mathrm{ }10{\;^{ - 12}}{C^2}\;/\mathrm{ }N - {m^2}

, then the average energy density of electric field will be :

View answer and explanation

Correct option: A

35.2 \times {10^{ - 12}}\,J/{m^3}

Option A (

35.2 \times {10^{ - 12}}\,J/{m^3}

) is correct because it is the average energy density calculated using the electric field amplitude and permittivity of free space.

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\sqrt {\frac{\mathrm{\mu }}{\mathrm{\varepsilon }}}