Answer: To determine the average intensity of the light, the RMS value of the electric field, and the peak value of the electric field, we'll use the following formulas and relationships:
Average intensity (I_avg) is given by:
I_avg = P_avg / A
where P_avg is the average emitted power and A is the area of a sphere with radius r (distance from the bulb).
RMS value of the electric field (E_rms) is related to the intensity by:
E_rms = √(2 * μ₀ * I_avg)
where μ₀ is the permeability of free space (approximately 4π × 10^-7 T·m/A).
Peak value of the electric field (E_peak) is related to the RMS value by:
E_peak = √2 * E_rms
Let's proceed with the calculations:
Given data:
Average emitted power (P_avg) = 150.0 W
Distance from the bulb (r) = 7 m
(a) Average intensity (I_avg):
The area of a sphere with radius r is A = 4πr².
A = 4π * (7 m)² ≈ 4π * 49 m² ≈ 193.62 m²
I_avg = P_avg / A
I_avg = 150.0 W / 193.62 m² ≈ 0.774 W/m²
The average intensity of the light at a distance of 7 m from the bulb is approximately 0.774 W/m².
(b) RMS value of the electric field (E_rms):
μ₀ = 4π × 10^-7 T·m/A
E_rms = √(2 * μ₀ * I_avg)
E_rms = √(2 * 4π × 10^-7 T·m/A * 0.774 W/m²)
E_rms ≈ √(3.094 × 10^-7 T·m²/A·W)
Now, we need to convert the units of E_rms to volts per meter (V/m) since the electric field is typically represented in volts per meter.
1 T·m/A = 1 V·s/m² (by definition)
E_rms ≈ √(3.094 × 10^-7 V·s/m²·W)
E_rms ≈ √(3.094 × 10^-7 V²/m²)
E_rms ≈ √(3.094 × 10^-7) V/m
E_rms ≈ 5.55 × 10^-4 V/m
The RMS value of the electric field is approximately 5.55 × 10^-4 V/m.
(c) Peak value of the electric field (E_peak):
E_peak = √2 * E_rms
E_peak = √2 * 5.55 × 10^-4 V/m
E_peak ≈ 7.84 × 10^-4 V/m
The peak value of the electric field is approximately 7.84 × 10^-4 V/m.