Question

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 $\epsilon_{0} = 8.8 \times 10^{-12} \, C^{2}/N-m^{2}$ then average energy density of electric field will be :

• A. $35.2 \times 10^{-10} J/m^3$
• B. $35.2 \times 10^{-11} J/m^3$
• C. $35.2 \times 10^{-12} J/m^3$
• D. $35.2 \times 10^{-13} J/m^3$

Answer 1

Answer: (C)

Energy density is given as:
$e=\frac{\epsilon_{0} E^{2}}{2}=8.8 \times 10^{-12} \times 8$
$=70.4 \times 10^{-12} \frac{J}{m^{3}}$
Therefore, average energy density
$= e / 2=35.2 \times 10^{-12} \,J / m ^{3}$.