Electric Dipole in the near field region CLOSE to the source

In summary, Born2bwire was able to derive the E and B fields for the electric dipole in the near field region close to the source. However, he is not sure how to derive the fields for the close region.
  • #1
mrjimbo
2
0

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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  • #2
How did you derive it for the far-field?
 
  • #3
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 )]

and

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)

Note:
Those omegas, pis, epsilons etc. aren't superscripts btw just multiply, it looks like they are superscripts in the preview.
 
  • #4
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?
 
  • #5
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.
 

FAQ: Electric Dipole in the near field region CLOSE to the source

What is an electric dipole?

An electric dipole is a pair of equal and opposite charges that are separated by a small distance. It is used to describe the distribution of electric charges in a system.

What is the near field region of an electric dipole?

The near field region of an electric dipole is the area close to the source where the electric field is dominant. It is typically considered to be within one wavelength of the source.

How does the electric field vary in the near field region of an electric dipole?

In the near field region, the electric field varies significantly with distance from the source. It is strongest near the dipole and decreases rapidly as the distance increases.

What are some applications of electric dipoles in the near field region?

Electric dipoles in the near field region are used in various applications such as wireless power transfer, near field communication, and medical imaging.

What factors affect the strength of the electric field in the near field region of an electric dipole?

The strength of the electric field in the near field region of an electric dipole is affected by the distance from the source, the orientation of the dipole, and the frequency of the electric field.

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