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

Answer:

1097.8 V/m

Explanation:

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

$I=\frac{1}{2}c\epsilon_0 E^2$

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:

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