- Home
- Standard 12
- Physics
8.Electromagnetic waves
medium
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
$4.4 \times {10^{ - 14}}$
Solution
(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}.$
Standard 12
Physics
Similar Questions
medium
