Question

An electric field in an e.m.wave is given by
`E=200 sin .(2pi)/lambda(ct-x)NC^-1`.
Find the energy contained in a cylinder of crossection `20 cm^2` and length 40 cm along the x-axis.

Answer 1

Correct Answer - `1.42xx10^(-10)J`
Here, `E_0=200NC^-1`,
`A=20cm^2=20xx10^-4m^2 , l=0.40m`
Volume of cylinder, `V=Al=(20xx10^-4)xx0.40`
`=8xx10^-4m^2`
Energy contained in cylunder is
U=volume x energy density
`=Vxx1/2in_0E_0^2`
`=(8xx10^-4)xx1/2xx(8.85xx10^-12)xx(200)^2`
`=1.42xx10^(-10)J`