A point source emits electromagnetic radiation uniformly in all directions with an average power of 60 W . The amplitude of the electric field at a distance of 3 m from the source is: (Given: 1 4 ₀ = 9 10^9 N m ^2 C ⁻² , speed of light c = 3 10^8 m s ⁻¹ )

Step-by-step solution

The intensity I of the electromagnetic radiation at a distance r from a point source emitting power P uniformly in all directions is given by the inverse-square law: I = P 4 r^2 Substituting the given values ( P = 60 W , r = 3 m ): I = 60 4 (3)^2 = 60 36 W m ⁻² The intensity of an electromagnetic wave is also related to the peak electric field E₀ by the formula: I = 1 2 ₀ E₀^2 c Equating the two expressions for intensity: 60 36 = 1 2 ₀ E₀^2 c Rearranging to solve for E₀^2 : E₀^2 = 120 36 ₀ c We are given 1 4 ₀ = 9