Exams › JEE Advanced › Physics
Correct answer: 2.45 V/m
Power spreads over a sphere: I = P/(4 pi r²) = 0.1/(4 pi (1)²) = 7.96*10⁻³ W/m². Average intensity in terms of peak field: I = (1/2) c epsilon0 E0². Solve for E0 = sqrt(2I/(c epsilon0)) = sqrt(2*7.96*10⁻³/(3*10⁸ * 8.85*10⁻¹²)) = sqrt(2*7.96e-3/2.655e-3) = sqrt(6.0) ~ 2.45 V/m.