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Method of image

  1. May 5, 2005 #1

    quasar987

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    The situation is illustrated in the attachment. I try to find the force on q by the method of images.

    I know that the potential is 0 at x=R and at infinity, and I know that The potential obeys a certain Poisson equation between those points. I will try to put a charge q' somewhere at x<R that makes it so V=0 at x=R and at infinity. That way, according to the uniqueness theorem, the potential between x=R and infinity created by the 'sphere-q' system will be the same as the potential created by the 'q'-q' system.

    My intuition tells me that I can pretty much put q' anywhere I want and I'll be able to create V=0 at x=R just by adjusting the magnitude of the charge. So I chose to put it at x=0. The potential of this charge configuration is

    [tex]V(x) = \frac{1}{4\pi \epsilon_0}\left(\frac{q}{|x-a|}+\frac{q'}{x}\right)[/tex]

    and we have V(R) = 0 iff

    [tex]q' = -\frac{R}{|R-a|}q = -\frac{R}{a-R}q[/tex]

    So the force on q is just that exerted by q' at this particular location, i.e.

    [tex]F_{\rightarrow q}=-\frac{1}{4\pi\epsilon_0}\frac{Rq^2}{(a-R)(a)^2}[/tex]

    Yes? Or do I have the method of images all wrong?
     

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    Last edited: May 5, 2005
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  3. May 5, 2005 #2

    OlderDan

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    Intuition is a great thing, but sometimes it just wont do :frown: The charge at x = 0 could be adjusted to make the potential zero at one point on the axis bewteen charges, but not for all r = R, which is what I suspect you need to show. Is that a conducting sphere? The method of images is called that for a good reason. Where might an image of q be formed?
     
  4. May 5, 2005 #3

    quasar987

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    First, I edited a sign error in the original post.

    Yes, the ball is a grounded conductor.

    So you're basically saying that it's not sufficient to get an image that creates the same boundary conditions along the x axis. I need one that creates the same boundary conditions on the whole volume bounded by the surface of the ball and infinity. This is logical. But I thought it would be sufficient to just have the same boundary condition along the x axis in view of this argument:

    Along the line, V(x,y,z) = V(x,0,0)=V(x). And V(x) obeys a Poisson equation along the x axis. So suppose there exist 2 funtions V1(x) and V2(x) that satisfy this Poisson equation and share the same value at the boundary x=R and x=infinity. Then V3 = V2 - V1 satisfies Laplace equation along the x axis and V3(R)=V3(infinity)=0. Therefor, V3(x)=0 for all x (since V3 is harmonic). This means that V1 = V2.

    In other words, the uniqueness theorem holds.

    I suspect that the error is when I said V(x,y,z) = V(x,0,0)=V(x) but I don't see why.
     
  5. May 6, 2005 #4

    OlderDan

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    I'm not totally following your argument, but I don't think it holds up even on the x axis. Suppose you find the magnitude of an image charge located at x = 0 that will result in zero potential at x = R. This should be pretty easy to do based purely on the 1/r dependence of the potential from a point charge. What will the potential be at x = -R? There is no way the given charge with an image point charge at x = 0 can give you zero potential at both x = R and x = -R. The image charge is going to have to be opposite the sign of the given charge, and located at some x such that 0<x<R. If you find an image charge that works for both x = R and x = -R, I think you have done it.

    http://www.phys.ufl.edu/~dorsey/phy6346-00/lectures/lect04.pdf
     
  6. May 6, 2005 #5

    quasar987

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    Sorry for not having been more explicit with my "argument". It's roughly just a copy/paste of the proof of the uniqueness theorem given by Griffiths.

    I have the solution for the problem we're discussing (though the source of the solution is not entirely thrusthworthy) and this is what they do too. They put the charge where V=0 at both x=R and x=-R. But I don't see why! We just want the potential in the region x>R and the boundary of this region is x=R and x=infinity. Anyway, I'll read that pdf document. It looks great, thx!
     
  7. May 6, 2005 #6

    OlderDan

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    Still, it is a multidimensional problem. It would be a totally different problem and a different solution if V = 0 at x = R because there was a plane conductor passing through x = R, or if some other conducting surface passed through that point The boundary conditions of the problem extend off the x-axis, so you cannot solve the problem by looking only along that axis in the region of interest.
     
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