Answer:
a. 0.332 W/m² b. 11.2 V/m c. 15.8 V/m
Explanation:
(a) the average intensity of the light,
Intensity, I = P/A where P = average power = 150.0 W and A = area through which the power emits = 4πr² where r = distance from bulb = 6 m.
So, I = P/A = P/4πr²
Substituting the values of the variables into the equation, we have
I = P/4πr²
I = 150.0 W/4π(6 m)²
I = 150.0 W/4π(36 m²)
I = 150.0 W/452.39 m²
I = 0.332 W/m²
(b) the rms value of the electric field,
Since Intensity, I = E²/cμ₀ where E = rms value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m.
Making E subject of the formula, we have
E² = Icμ₀
E = √(Icμ₀)
Since I = 0.332 W/m², substituting the other terms into the equation, we have
E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)
E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)
E = √(12.5 × 10)
E = √125 V/m
E = 11.18 V/m
E ≅ 11.2 V/m
(c) the peak value of the electric field.
The peak value of electric field, E' is gotten from E = E'/√2 where E = rms value of electric field.
So, E' = √2E
= √2 × 11.2 V/m
= 15.81 V/m
≅ 15.8 V/m
