Question

B. In the vicinity of Earth’s orbit around the Sun, the energy intensity of sunlight is about 1600 W/m2 . What is the approximate magnitude of the electric field in the sunlight?

Answer 1

1097.8 V/m

Step-by-step explanation:

The equation that relates the intensity of an electromagnetic wave with the magnitude of the electric field is:

`I=(1)/(2)c\epsilon_0 E^2`

where

c is the speed of light

`\epsilon_0` is the vacuum permittivity

E is the peak magnitude of the electric field

In this problem, we know the intensity:

I = 1600 W/m^2

So we can rearrange the formula to find E:

`E=\sqrt{(2I)/(c\epsilon_0)}=\sqrt{(2(1600 W/m^2))/((3\cdot 10^8 m/s)(8.85\cdot 10^(-12) F/m))}=1097.8 V/m`