Q. 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 $Joule/m^{3}$, will be
A) $1.1 \times 10^{- 11}$
B) $2.2 \times 10^{- 12}$
C) $3.3 \times 10^{- 13}$
D) $4.4 \times 10^{- 14}$
Explanation
(b) Average energy density of electric field is given by
$u_{e} = \frac{1}{2}\varepsilon_{0}E^{2} = \frac{1}{2}\varepsilon_{0}\left( \frac{E_{0}}{\sqrt{2}} \right)^{2} = \frac{1}{4}\varepsilon_{0}E_{0}^{2}$
$= \frac{1}{4} \times 8.85 \times 10^{- 12}(1)^{2} = 2.2 \times 10^{- 12}J/m^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.