Question

An electromagnetic wave of frequency $1 \times 10^{14} Hz$ is propagating along z-axis. The amplitude of the electric field is $4\, V / m$. If $\varepsilon_{0}=8.8 \times 10^{-12} C ^{2} / N - m ^{2}$, then the average energy density of electric field will be :

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

Answer 1

Answer: (A)

$f =10^{14} Hz$
$E _{0}=4\, V / m$
$E _{0}=8.8 \times 10^{-12} C ^{2} / N - m ^{2}$
Energy density of electric field
$=1 / 2$ (Total energy density)
$=\frac{1}{2} \cdot \frac{1}{2} \varepsilon_{0} E ^{2}$
$=\frac{1}{2} \cdot \frac{1}{2}\left\{8.8 \times 10^{-12} \times 4^{2}\right\}$
$=35.2 \times 10^{-12} J / m ^{3}$