Electric field on surface of the Earth due to solar radiation

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Homework Help Overview

The discussion revolves around estimating the electric field magnitude of solar radiation impacting the Earth's surface, specifically through the context of the Poynting vector and the intensity of sunlight.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the Poynting vector and the intensity of sunlight, questioning whether the electric field derived from a single wave can accurately represent the total field due to all incident waves. There is a discussion about the implications of attributing intensity to a single wave versus considering the collective effect of multiple waves.

Discussion Status

Some participants suggest that the electric field derived from the intensity can represent the total field, while others express concerns about overestimating the field by equating it to a single wave. The conversation is exploring different interpretations of how the electric field behaves in the context of multiple waves.

Contextual Notes

There is mention of the complexity introduced by the presence of multiple oscillators with random phases and frequencies, which may not be fully addressed within the scope of introductory physics.

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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.
 
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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.
 
RUber said:
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.
 
Last edited:
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.
 

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