Question

Consult Multiple-Concept Example 5 to review the concepts on which this problem depends. A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5.00 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

a) 0.48 W/m²

b) 13.45 N/C

c) 19.02 N/C

Step-by-step explanation:

P = average emitted power = 150 W

r = distance from the bulb = 5 m

I = Intensity of light

Intensity of light is given as

`I = (P)/(4\pi r^(2))`

`I = (150)/(4(3.14) (5)^(2))`

I = 0.48 W/m²

b)

E = rms value of electric field

intensity is given as

I = ∈₀ E² c

0.48 = (8.85 x 10⁻¹²) (3 x 10⁸) E²

E = 13.45 N/C

c)

Peak value of electric field is given as

E₀ = √2 E

E₀ = √2 (13.45)

E₀ = 19.02 N/C