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

Average intensity of electromagnetic waves is
`I=(P)/(4pir^(2))=(1)/(2)epsilon_(0)E_(0)^(2)c" or "E_(0)=[(P)/(2pir^(2)epsilon_(0)c)]^(1//2)`
Substituting the given values, we get
`E_(0)=[(1500)/(2pi(3)^(2)xx[(1//4pixx9xx10^(9))]xx3xx10^(8))]^(1//2)=100Vm^(-1)`