Question

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answer 1

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer 2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

$P_r=\frac{IA}{c}$

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

$P=140W=140\frac{J}{s}\\0.4*140J=56J$

56J is the electromagnetic energy.

The area of the bulb is

$A_b=4\pi r^2=4\pi (14m)^2=2463m^2$

The radiation pressure is

$P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^{-11}Pa$

hope this helps!!