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Standard XII

Physics

Wave Equation of EM Waves

In an electro...

Question

In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is $5 \times 10^{14}$ Hz. The wave is propagating along z-axis. The average energy density of electric field, in $J o u l e m^{3}$ , will be

1.1 X 10^-11

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2.2 X 10^-12

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3.3 X 10^-13

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4.4 X 10^-14

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Solution

The correct option is B 2.2 X 10^-12
Average energy density of electric field is given by

$u_{e} = \frac{1}{2} ϵ_{0} E^{2} = \frac{1}{2} ϵ_{0} {(\frac{E_{0}}{\sqrt{2}})}^{2} = \frac{1}{4} ϵ_{0} E_{0}^{2}$

$= \frac{1}{4} \times 8.85 \times 10^{- 12} (1)^{2} = 2.2 \times 10^{- 12} J / m^{3} .$

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In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is 5× 1014Hz. The wave is propagating along z-axis. The average energy density of electric field, in Joulem3, will be

The correct option is B 2.2 X 10-12Average energy density of electric field is given by ue=12ϵ0E2=12ϵ0(E0√2)2=14ϵ0E20 =14× 8.85× 10−12(1)2=2.2× 10−12J/m3.

In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is $5 \times 10^{14}$ Hz. The wave is propagating along z-axis. The average energy density of electric field, in $J o u l e m^{3}$ , will be

The correct option is B 2.2 X 10^-12
Average energy density of electric field is given by

$u_{e} = \frac{1}{2} ϵ_{0} E^{2} = \frac{1}{2} ϵ_{0} {(\frac{E_{0}}{\sqrt{2}})}^{2} = \frac{1}{4} ϵ_{0} E_{0}^{2}$

$= \frac{1}{4} \times 8.85 \times 10^{- 12} (1)^{2} = 2.2 \times 10^{- 12} J / m^{3} .$