You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.2 X 10^-12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.3 X 10^-13

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.4 X 10^-14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

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

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

Suggest Corrections

Similar questions

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

Q. An electromagnetic wave of frequency

1 \times 10^{14}

Hertz is propagating along z-axis. The amplitude of electric field is

4 V / m

. If

\in_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}

, then average energy density of electric field will be:

Q. An electromagnetic wave of frequency,

1 \times 10^{4} Hz

is propagating along

z -

axis. The maximum amplitude of the electric field is

4 V/m

. Average energy density of the electric field will be -

Q. An electromagnetic wave of frequency

1 \times 10^{14} Hz

is propagating along z-axis. The amplitude of electric field is

4 V/m

. If

ϵ_{0} = 8.8 \times 10^{- 12} C^{2} {/N-m}^{2}

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

Q. An electromagnetic wave of frequency

1 \times 10^{14}

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

ϵ_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}

, then average energy density of electric field will be:

Join BYJU'S Learning Program

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

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

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