1.

If the electric field associated with a radiation of frequency 10 MHz is E=10 sin (kx - omega t) mV/mthen its energy density is …. J m^(-3).[epsilon_(0) = 8.85xx10^(-12)C^(2)N^(-1)m^(-2)]

Answer»

`4.425xx10^(-16)`
`6.26xx10^(-14)`
`8.85xx10^(-16)`
`8.85xx10^(-14)`

SOLUTION :`E_(RMS)^(2)=(E_(m))/(sqrt(L))`
`therefore E_(rms)^(2)=(E_(m)^(2))/(2)`
`rho=epsilon_(0)E_(rms)^(2)`
`therefore rho =(8.85xx10^(-12)xx100xx10^(-6))/(2)`
`=4.425xx10^(-16)`


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