Question

A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb, determine (a) the average intensity of the light, (b) the rms value of the electric field, and (c) the peak value of the electric field.

Answer 1

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

$\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2$

∴ Average intensity is 0.477 W/m²

b) Rms value

$\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C$

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

$E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C$

∴ Peak value of the electric field is 18.96 N/C