In a plane electromagnetic wave,the electric field oscillates with a frequency $2 \times 10^{10} \,s^{-1}$ and amplitude $40 \,Vm^{-1}$. The energy density due to the electric field is (given $\varepsilon_0 = 8.85 \times 10^{-12} \,Fm^{-1}$):

  • A
    $1.52 \times 10^{-9} \,Jm^{-3}$
  • B
    $2.54 \times 10^{-19} \,Jm^{-3}$
  • C
    $3.54 \times 10^{-9} \,Jm^{-3}$
  • D
    $4.56 \times 10^{-9} \,Jm^{-3}$

Explore More

Similar Questions

The electric field in a plane electromagnetic wave is given by $E_z = 60 \cos(5x + 1.5 \times 10^9 t) \text{ V/m}$. Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

An electromagnetic wave in vacuum has the electric and magnetic field $\vec{E}$ and $\vec{B}$,which are always perpendicular to each other. The direction of polarization is given by $\vec{X}$ and that of wave propagation by $\vec{k}$. Then:

The intensity of light from a source is $\left( \frac{500}{\pi} \right) \, W/m^2$. Find the amplitude of the electric field in this wave.

$A$ point source of electromagnetic radiation has an average power output of $800 \, W$. The maximum value of the electric field at a distance $4.0 \, m$ from the source is .... $V/m$.

Which of the following do not require a medium for transmission?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo