# Electric field on surface of the Earth due to solar radiation

1. Oct 29, 2014

### fayled

This problem is actually an example I'm working through...

1. The problem statement, all variables and given/known data

We are trying to estimate the (magnitude of the) electric field in the light waves from the sun hitting the surface of the Earth.

2. Relevant equations
The Poynting vector for a sinusoidal EM wave has magnitude cε0E02/2 (this was derived before - I understand this).
The intensity of sunlight at the surface of the Earth is 1300Wm-2

3. The attempt at a solution
So the idea is that the magnitude of Poynting vector for a single EM wave will represent the power flow per unit area for this single EM wave at the surface of the Earth. The example then just equates this to the intensity and solves for E0.

But I'm missing something. We have equated an expression for the power per unit area (Poynting vector) for a single EM wave to an expression for the power per unit area for all EM waves incident on that unit area, so we are attributing this intensity all to one wave when we shouldn't be and so are overestimating the field. Can anybody explain? Would it make more sense if the example meant we were trying to find the electric field at the surface due to all waves, not just one?

Thankyou.

2. Oct 29, 2014

### RUber

I believe they are one and the same. $c\epsilon_0E_0^2/2$ can represent a single wave or the entire field. By solving for the $E_0$ in this way, based on total intensity, you are indeed finding the total field.

3. Oct 29, 2014

### fayled

Actually wouldn't a single EM wave have the same electric field as the field produced by all the EM waves put together? Because if the waves are emitted in all directions from a source point, at a point on a sphere centred on this source point, you get basically one wave at each point (and if we go into the continuous limit, we get a 'spread' of EM waves), and so the field vectors don't really add they just act next to one another, i.e we get a 'spread' of the field, and the vectors don't add.

Last edited: Oct 29, 2014
4. Oct 29, 2014

### rude man

I'd say there is an equivalent electric field amplitude such as to give an intensity of 1300W.

So you could start with the expression relating instantaneous power of a plane wave Emcos(wt - kx), then integrate to get the average power in the wave per sq. meter and then the effective E amplitude. In other words you are assuming an equivalent plane wave of one frequency and with the amplitude necessary to give the given intensity density (W/sq. m.).

In reality there are an infinite number of oscillators of random phase, each of frequency df, as determined by the Planck radiation law. That math would seem to be beyond introductory physics.