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Electrodynamics Method of imaging

by sleventh
Tags: electrodynamics, imaging, method
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sleventh
#1
Feb7-10, 10:54 PM
P: 64
Hello all,

I have recently been taught the method of imagging with opposite point charges. The stereotypical example of a point charge above an infinit conducting plate comes fine but i can't grasp the sphere example. It is shown HERE , I cant see how they derive the value of the mirror charge, thank you very much any help is appreciated.

sleventh
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Born2bwire
#2
Feb8-10, 02:40 AM
Sci Advisor
PF Gold
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P: 1,776
Who knows? At least we know that the solution to Laplace's equation will be unique. How they came up with the image charge is probably a long and boring process. It could have just been that someone solved the problem by directly finding a solution to the associated Laplace equation and then noticed that you could decompose it into the summation of the charge and an image charge. Lots of times it seems that complicated relationships that can be distilled into more simplified physical situations are usually found after the fact by playing around with the results and trying to decompose them into more meaningful forms.

But apart from however they came up with the relationship, as long as we can prove that it satisifies Laplace's equation and the resulting boundary equations then we know it is the solution to the potential. This part is easier to show as they have done in the Wikipedia article.
torquil
#3
Feb8-10, 03:18 AM
P: 641
Quote Quote by sleventh View Post
Hello all,

I have recently been taught the method of imagging with opposite point charges. The stereotypical example of a point charge above an infinit conducting plate comes fine but i can't grasp the sphere example. It is shown HERE , I cant see how they derive the value of the mirror charge, thank you very much any help is appreciated.

sleventh
It is chosen so as to cancel exactly the field from q at the position R between the two charges.I.e. the position and value of the mirror charge is chosen to accomplish this. You can either have a large mirror charge far away, or a smaller charge closer by. So there is a free parameter that is chosen so that the system describes a sphere. If I assume that the mirror charge is located at R^2/p on the x-axis, I get for the mirror charge Q:

Cancellation of the potential at x=R:
q/(R-p) + Q/(R^2/p - R) = 0

Gives:
Q = -qR/p

In this I already assumed that the mirror charge would be located at R^2/p. If you want to derive this, you need to leave that as a free parameter, write down the total potential as they do, and then determine for which mirror charge position the V=0 equipotential surface is a sphere of the correct radius.

Torquil


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