The power crossing a surface perpendicular is equal to the intensity at the surf

Question

The power crossing a surface perpendicular is equal to the intensity at the surface times the area of the surface. Since the source emits uniformly in all directions, the light intensity is the same at all points on the imaginary spherical surface. Moreover, the light crosses this surface perpendicularly.
S = ce0E2
relates the average light intensity at the surface to its rms electric field strength (which is known), and the area of the surface can be found from a knowledge of its radius.

A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5.20 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.
(a) S = ?
(b) Erms =?
(c) E0 =?

Explanation / Answer

(a) avg intensity = power / area = 150 / 4 pi * 4.85^2 =

= 0.50745 W/m^2

(b) rms elec field = sqrt (intensity * u c ) = sqrt (0.50745 * 4pi x 10^-7 * 3 x 10^8) =

= 13.83 V/m

(c) peak value = rms value * sqrt2 = 13.83 * sqrt2 =

= 19.56 V/m

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