A point source of electromagnetic radiation has an average power output of `1500W`. The maximum value of electric field at a distance `3m` from this source in `Vm^-1` is

Options

A. 500
B. 100
C. `(500)/(3)`
D. `(250)/(3)`

Answer

B

Explanation

Total energy density of electromagnetic wave is
`u =(1) /(2) in _(0) R _(m) ^(2) & i=(I) /(C ) =(P) /( AC)`
or `u= (p) /( 4pi r^(2) C) =(1) /(2) in _(0) E _(m)^(2)`
` E _(m) ^(2) =( 2P ) /( 4 pi in _(0) r^(2) C) = (2xx1500xx9xx10^(5) )/( 3xx3xx3xx10^(8))`
`E_(m) = 100 Vm^(-1)`