A very small source of light that radiates uniformly in all directions produces an electric field amplitude of 8.45 V/m at a point 61 m from the source. What is the power output from the source? (c = 3.00 × 108 m/s, μ 0 = 4π × 10-7 T · m/A, ε 0 = 8.85 × 10-12 C2/N · m2)

4.43 kW

Explanation:

Since Intensity I = P/A = E²/2cμ₀ where P = Power, A = Area = 4πr² where r = distance from source = 61 m and E = electric field amplitude = 8.45 V/m.

P = E²A/2cμ₀ = E²4πr²/2cμ₀ = 2πE²r²/cμ₀

  = 2π(8.45 V/m)²(61 m)²/3 × 10⁸ m/s × 4π × 10⁻⁷ Tm/A

  = 4428.1 W

  = 4.4281 kW ≅ 4.43 kW

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