Field Amplitude From Intensity

The average intensity of a wave is related to the peak electric field by I = \frac{1}{2}c\varepsilon_0 E_0^2, since intensity is the time-averaged Poynting vector. Solving for the field amplitude, E_0 = \sqrt{\frac{2I}{c\varepsilon_0}}. Substituting I = 600\ \text{W/m}^2, c = 3.0 \times 10^{8}, and \varepsilon_0 = 8.85 \times 10^{-12} gives E_0 = \sqrt{\frac{2 \times 600}{(3.0 \times 10^{8})(8.85 \times 10^{-12})}} = \sqrt{\frac{1200}{2.655 \times 10^{-3}}} = \sqrt{4.52 \times 10^{5}} \approx 673\ \text{V/m}. The 476 V/m option drops the factor of two and uses I = c\varepsilon_0 E_0^2. The 338 V/m value comes from taking the root-mean-square field rather than the peak. The 1346 V/m answer doubles the correct result through a misplaced factor. This inversion of the intensity formula to recover field strength is a standard JEE exercise built on the NCERT energy relations, and it highlights how a modest power flux corresponds to surprisingly large peak field values of several hundred volts per metre. The reason these fields seem large yet harmless is that they oscillate hundreds of trillions of times per second, so their average effect on matter is gentle. As a check, the corresponding magnetic amplitude would be E_0/c \approx 2.2 \times 10^{-6}\ \text{T}, a physically reasonable value for sunlight, supporting the answer.