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. In an electromagnetic wave, the amplitude of electric field is

10 {Vm}^{- 1}

. The frequency of the wave is

5 \times 10^{14} Hz

and the wave is propagating along

z -

axis, then total average energy density of E.M. wave is

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An electromagnetic wave of frequency 1×1014 Hertz is propagating along z-axis. The amplitude of electric field is 4 V/m. If ∈0=8.8×10−12 C2/N−m2, then average energy density of electric field will be:

The correct option is A 35.2×10−12 J/m3Energy density is given as:e=ϵoE22=8.8×10−12×8=70.4×10−12Jm3Therefore, average energy density = e/2 = 35.2×10−12 J/m3.

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:

The correct option is A $35.2 \times 10^{- 12} J / m^{3}$
Energy density is given as:
$e = \frac{ϵ_{o} E^{2}}{2} = 8.8 \times 10^{- 12} \times 8 = 70.4 \times 10^{- 12} \frac{J}{m^{3}}$

Therefore, average energy density = $e / 2$ = $35.2 \times 10^{- 12} J / m^{3}$ .