Electric field on surface of the Earth due to solar radiation

[fayled]

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

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

Homework 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

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.

 

[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.

 

[fayled]

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.

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.

 

[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 E_mcos(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.

 

[HalfLostAndSearching]

I would like to clarify that the electric field on the surface of the Earth due to solar radiation is a complex phenomenon and cannot be accurately estimated using a simple equation. The Poynting vector and intensity of sunlight are only a few factors that contribute to the overall electric field.

To accurately estimate the electric field, we need to consider various other factors such as the Earth's atmosphere, the Earth's magnetic field, the angle of incidence of the sunlight, and the Earth's surface properties. These factors can greatly affect the electric field and cannot be ignored.

Moreover, the electric field is not just affected by a single EM wave, but rather the combined effect of all the EM waves incident on the Earth's surface. Therefore, it would be more accurate to consider the electric field due to all waves rather than just one wave.

In conclusion, while the Poynting vector and intensity of sunlight can provide a rough estimate of the electric field on the surface of the Earth, it is important to consider all the other factors to accurately determine the magnitude of the electric field.