An electric bulb of power 30 π W has an efficiency of 10 % and it can be assumed to behave like a point source of light. At a distance of 3 m from the bulb, the peak value of the electric field in the light produced by the bulb is [Take ε 0 ≈ 9 × 10 - 12 C 2 N - 1 m - 2 ]

Step-by-step solution

Intensity at distance r from a point source, I = P 4 π r 2 Efficiency, η = O u t p u t I n p u t = P P ′ P ' = η P P ' = 10 100 30 π = 3 π   W I = 3 π 4 π 3 2 = 1 12   W   m - 2 I = 1 2 c ε 0 E 0 2 ⇒ 1 12 = 1 2 c ε 0 E 0 2 E 0 = 1 6 × 3 × 10 8 × 9 × 10 - 12 = 100 9 2   V   m - 1