Q 12-08-022NEETJEE MainMedium
The peak electric field of sunlight at the Earth is about $1000$ V/m. The average intensity of the radiation is about ($c = 3 \times 10^8$ m/s, $\epsilon_0 = 8.85 \times 10^{-12}$ F/m)
Answer: (C) $1.33\ \text{kW/m}^2$
$I = \dfrac{1}{2}\epsilon_0E_0^2c = 0.5 \times 8.85 \times 10^{-12} \times 10^6 \times 3 \times 10^8 \approx 1.33 \times 10^3\ \text{W/m}^2$.
Solution by Sreeraj P, M.Sc Physics