Optics Question 177
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 Joule/m3, will be
Options:
A) $ 1.1\times {{10}^{-11}} $
B) $ 2.2\times {{10}^{-12}} $
C) $ 3.3\times {{10}^{-13}} $
D) $ 4.4\times {{10}^{-14}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Average energy density of electric field is given by $ u _{e}=\frac{1}{2}{\varepsilon _{0}}E^{2}=\frac{1}{2}{\varepsilon _{0}}{{( \frac{E _{0}}{\sqrt{2}} )}^{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}. $
