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Electric Dipole in the near field region CLOSE to the source

by mrjimbo
Tags: dipole, electric, field, region, source
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Apr29-09, 06:09 AM
P: 2
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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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Apr29-09, 06:31 AM
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PF Gold
Born2bwire's Avatar
P: 1,777
How did you derive it for the far-field?
Apr29-09, 06:56 AM
P: 2
Hey Born2bwire,

Well I was given:

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


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

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)

Those omegas, pis, epsilons etc. aren't superscripts btw just multiply, it looks like they are superscripts in the preview.

Apr29-09, 07:02 AM
P: 39
Electric Dipole in the near field region CLOSE to the source

Pls dont murder me if im 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 im not to sure how it works out...

anybody advance on this for near E and B fields?
Apr29-09, 08:35 AM
Sci Advisor
PF Gold
Born2bwire's Avatar
P: 1,777
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 [tex]kr\gg 1[/tex]. 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 [tex]kr\ll 1[/tex] 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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