A point source heating device of 1000 W with 10% efficiency is 7 m from the observational location. Find the peak value of electric field produced by the radiation at that location.

11.05 V/m

The correct answer is Option (3) → 11.05 V/m

$P = 1000 \times 0.1 = 100 \, W$

$I = \frac{P}{4\pi r^2} = \frac{100}{4\pi (7)^2} = \frac{100}{196\pi}$

$I \approx \frac{100}{615.75} \approx 0.162 \, W/m^2$

$I = \frac{1}{2} c \epsilon_0 E_0^2$

$E_0 = \sqrt{\frac{2I}{c\epsilon_0}}$

$c = 3\times10^8,\ \epsilon_0 = 8.85\times10^{-12}$

$c\epsilon_0 = 2.655\times10^{-3}$

$E_0 = \sqrt{\frac{2 \times 0.162}{2.655\times10^{-3}}}$

$= \sqrt{\frac{0.324}{2.655\times10^{-3}}} \approx \sqrt{122} \approx 11 \, V/m$

The peak electric field is $\approx 11 \, V/m$.