A point source of electromagnetic radiation has an average power output of `800 W`. The maximum value of elecvtric field at a distance `4.0 m` from the source is

Options

A. `64.7 V//m`
B. `57.8 V//m`
C. `56.72 V//m`
D. `54.77 V//m`

Answer

d

Explanation

Intensity of EM wave is given by
`I=(P)/(4pi R_(2))=v_(av) c=(1)/(2)epsilon_(0)E_(0)^(2)xxc`
`implies E_(0)sqrt((P)/(2piR^(2)epsilon_(0)c))`
`=sqrt((800)/(2xx3.14xx(4)^(2)xx8.85xx10^(-12)xx3xx10^(8)))`
`=57.77(V)/(m)`