Electric Dipole in the near field region CLOSE to the source

AI Thread Summary
The discussion focuses on deriving the electric (E) and magnetic (B) fields of an electric dipole in the near field region, specifically when the distance (r) is much smaller than the wavelength (λ). Participants share their approaches, noting that while they successfully derived the fields for the far field, they are uncertain about the adaptations needed for the near field. Key points include the use of potential and vector potential to find the fields, with the near field characterized by the assumption that kr is much less than one. The conversation emphasizes that the E field will exhibit a 1/r^2 component from the potential gradient and a 1/r component from the time derivative. Overall, understanding the transition from far field to near field involves recognizing the differences in assumptions about the distance relative to the wavelength.
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Homework Statement



Hey everyone, I need to derive what the E and B fields look like for the electric dipole in the near field region close to the source i.e. lambda>>r.

Homework Equations





The Attempt at a Solution



I managed to derive it for the situation far from the source but I'm not sure what adaptations to make for the close region.
 
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How did you derive it for the far-field?
 
Hey Born2bwire,

Well I was given:

\varphi = [-p(0)\varpi/4\pi\epsilon(0)c] ( cos\theta/r ) sin[\varpi(t- r/c )]

and

A = [-\mu(0)p(0)\varpi/4\pir] sin[ \varpi (t - r/c )] z\widehat{}

From there I used the grad of Psi and curl of A and the unit vector z definition to get the E and B fields where:

E = -(grad Psi) - dA/dt
B = (curl of A)

Note:
Those omegas, pis, epsilons etc. aren't superscripts btw just multiply, it looks like they are superscripts in the preview.
 
Pls don't murder me if I am wrong but...

I know for far away fields if you have the definition of the potential as a function of (r, theta, t) you can use the constraints that as r>>lambda, the lambda is approx c/w therefore 1/lambda = w/c (approx). This means that you potential formula can be changed by replacing every w/c with 1/lambda.

This gives the potential at a distance, Which if you get -Div.(your potential) you get your E.

I think you get E going to zero the further you go! But for r<<lambda I am not to sure how it works out...

anybody advance on this for near E and B fields?
 
You should use the Tex now that it is working again.

If you were given the exact potential and vector potential then all you have to do is use them to find the fields and that is the near field. The only difference between the near and far field is that in the far field you assume kr\gg 1. In this case, I can see that the E field will have a 1/r^2 component from the gradient of the potential and then a 1/r component from the time derivative. So the far-field would keep the leading order term. You could assume the opposite, that kr\ll 1 for the near field and drop the appropriate terms but I can't recall that being done. It really would be a matter of preference, especially for such a simple equation as in this case.
 
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