Q 12-08-029JEE MainMedium
A point source emits $100$ W of electromagnetic radiation uniformly in all directions. Find the peak electric field, in V/m, at a distance of $2$ m, to the nearest whole number. ($c = 3 \times 10^8$ m/s, $\epsilon_0 = 8.85 \times 10^{-12}$ F/m)
Numerical value type. Enter your answer.
Answer: 39
$I = \dfrac{100}{4\pi(2)^2} \approx 1.99\ \text{W/m}^2$.
$$E_0 = \sqrt{\frac{2I}{\epsilon_0c}} = \sqrt{\frac{2 \times 1.99}{8.85 \times 10^{-12} \times 3 \times 10^8}} \approx 38.7 \approx 39\ \text{V/m}$$
Solution by Sreeraj P, M.Sc Physics