If the peak value of the electric fields produced by the radiation coming from a 100 W bulb of 3% efficiency is P V/m at a distance of 3 m from the bulb, then find P = ? (Take `sqrt(5) = 2.24`)

Answer

`04.48`

Explanation

Intensity, `I = ("Half")/("area")`
`=(100 xx 3)/(4pi(3)^(2) xx 100) = 1/(12pi) Wm^(-2)`
Half of this intensity is attributed to electric field and half of that to magnetic field.
`therefore 1/2 =1/4 epsilon E_(0)^(2) c`
`rArr E_(0) = sqrt((2l)/(epsilon_(0)c) = sqrt(1/(4pi xx 9 xx 10^(9)) xx 3 xx 10^(8))`
` = sqrt(20)`
`= 2sqrt(5)`
`= 4.48 Vm^(-1)`