A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions.If 50% of the power input to such a light bulb is emitted as electromagnetic radiation,what is the maximum value of the electric field at a distance of 2.0 m from the light bulb?
A)43 V/m
B)0.11 mV/m
C)21 V/m
D)60 V/m
E)30 V/m

Question

Quizplus · Accepted Answer

The power output of the light bulb is 50% of the power input, which is 30 W. We can use this power and the formula for power density of a spherical wave to find the maximum electric field at a distance of 2.0 m:

$P = \frac{4 \pi r^2}{c} S_{max}^2$

where $P$ is the power output, $r$ is the distance from the source, $c$ is the speed of light, and $S_{max}$ is the maximum amplitude of the electric field.

Plugging in the values, we get:

$30\, W = \frac{4 \pi (2.0\, m)^2}{3	imes 10^8\, m/s} S_{max}^2$

$S_{max} = \sqrt{\frac{3 \cdot 10^7}{4 \pi}}\, V/m \approx 21\, V/m$

Therefore, the maximum value of the electric field at a distance of 2.0 m from the light bulb is approximately 21 V/m. The correct choice is (C).

A 60-W Light Bulb Emits Spherical Electromagnetic Waves Uniformly in All