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Method of image for charge and conductor

  1. May 21, 2012 #1
    Let's say we have a charge a distance d above the xy plane, which is a grounded conductor. Find the potential at every point. Now this is not a hard problem if you know the method of images. I do however have some questions:

    1) The idea is, as reader probably knows, to solve poissons equation by considering a completely different distribution, which obeys the same boundary conditions. The uniqueness theorem guarentees then that this is the right solution.
    I just have a little question to this poisson equation thing. As you know the potential must satisfy it because the electric field obeys the maxwell equation
    ∇ [itex]\bullet[/itex] E = ρ/ε , which in turn comes from Gauss' law. But this is only derivable if we assume that the charge we deal with is a continuos distribution. In that case we can integrate over a volume using the charge density and apply the divergence theorem to obtain the above. In our case however, there is also a point charge, apart from the induced charges on the conductor. Wont that wreak havoc?

    2) Solving the problem you find that the total charge induced on the grounded conductor is -q. My book says that this is obvious if you think about it. Unfortunately I fail to see why it is obvious - can someone explain why the induced charge must be -q? :)
     
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  3. May 21, 2012 #2

    rude man

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    Because only that way is the potential identically zero everywhere in front of the conductor.

    Yes, Laplace's equation encounters a singularity at the charge location, which is of ZERO magnitude. Whereas, let dv be an infinitesmally small, BUT NOT ZERO, volume element. Then (del*D)*dv = ρ*dv = q nevertheless holds since dv is infinitesmally small BUT NOT ZERO. So there is a finite charge density involved, which is q/dv. Similarly, Gauss' law holds for a point charge since the surface element dσ surrounding dv is infinitesmally small but not zero.
     
    Last edited: May 21, 2012
  4. May 22, 2012 #3

    gabbagabbahey

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    Actually, this Maxwell equation is Gauss' Law, just in differential form. It is mathematically equivalent to The integral form of Gauss' Law you are more familiar with.

    I'm not sure why you think this. Surely your textbook does not make this assumption?

    There is no trouble using the divergence theorem when the Electric field is due to some combination of point charges and continuous distributions. The divergence theorem only requires that the vector field (electric field in this application) is continuously differentiable. The field of a point charge meets this requirement.

    As for integrating the charge density over a volume, there is also no problem there, as the charge density is defined such that this integral must always equal the total charge enclosed in the volume.

    Iif it were something other than [itex]-q[/itex], then any test charge placed on the other side of the conductor would "see" a net charge that is nonzero. The monopole term of the potential only vanishes at finite distance from a distribution if the net charge of the distribution is zero (take a look at the multipole expansion of the electrostatic potential to convince yourself of this). When nonzero, the monoplole term dominates all higher order terms, at large distances, so they will not magically cancel it out and produce zero everywhere.

    No. The charge density at a point charge is necessarily infinite/undefined and best described mathematically using the dirac delta distribution.
     
  5. May 22, 2012 #4
    Well I wasn't refering to the requirements for E but rather the fact that in order to establish the differential form you put up an argument like this:

    ∫E[itex]\bullet[/itex]dA = ∫ρdV/ε

    and using the divergence theorem:

    ∫∇[itex]\bullet[/itex]E dV = (∫ρdV)/ε

    Now since these integrals are over the same volume they must be identical and then produce the maxwell equation. If there is a point charge I don't really see how you can use this argument, since there will not be an integral over any volume! Unless you replace it with a delta function..
     
  6. May 22, 2012 #5

    gabbagabbahey

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    I assume you meant to say that "since these two integrals are the same over any volume, the integrands must be equal"?

    Which integral do you see a problem with in the case of a point charge? The one on the right?

    If so, do you think that the integral itself doesn't exist for some reason, or do you think that you can't compare the integrands since [itex]\rho[/itex] isn't a nice smooth continous function?
     
  7. May 22, 2012 #6
    Exactly, you can't assign a nice density to it and hence make a volume integral only to cancel that out.

    Then you would get something like ∇[itex]\bullet[/itex]E = infinite density/ε

    But on the other hand I know that the divergence of the field of a point charge is given by the dirac delta so maybe that is part of the explanation. Overall it just bothers me how mathematical rigorousness lacks in my physics courses....


    Something different: I still don't quite understand why the induced charge on the conductor must be zero? You mentioned something about the potential below the conductor but in this problem, we're only interested in z≥0. Can you perhaps try to explain some different way? Perhaps by setting up some contraction that only vanishes if the charge is -q. Evidently for zero potential we must have that the field of the conductor is equal to that of a point charge separated by the same distance as q from the origin but on the -z axis. From that it mathematically follows that this field is exactly produced by an infinite conductor with a charge density such that the total charge on it is -q. I'm writing this to show you that I understand the situation perfectly, I just need to intuition for why it must be -q (intuition! I can see it from the math)
     
  8. May 22, 2012 #7

    rude man

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    The divergence of the E field at the point charge is not differentiable. It is in fact infinite, as is the field itself.
     
  9. May 22, 2012 #8

    gabbagabbahey

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    The thing that you seem to be missing is that (AFAIK), there is no requirement on the continuity of the integrands when making the argument that if integrated over any volume, the two integrals are equal, then the integrands must be equal. The only requirement is that the integral exists and is bounded. The integral of the charge density over any volume will always exist and be finite and will equal the charge enclosed in the volume - this is essentially the definition of charge density. The charge density can be undefined at certain points, but the integral will always be well defined.

    As far as a lack of mathematical rigour goes (are you also Canadian? Americans don't put in the u in rigour), all I can say is get used to it. Very few physics derivations are done completely rigourously in texts and papers.

    The reason that you are only interested in z≥0 is because the potential below the conduction plane will be zero (unless there are additional charges there). The image charge gives the correct potential only in the z≥0 region because it does not change Laplace's equation there or alter the boundary contions for the boundary that reqion. For z<0, there is no charge and the potential is zero at both z=0 and infinity, so the potential is zero everywhere in that region.

    If there were a nonzero net charge, then the monopole term of the potential would be nonzero in the z<0 region and so the potential would not be zero (at least not everywhere in that reqion).
     
  10. May 22, 2012 #9

    gabbagabbahey

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    Yes, you're quite right of course. To show that the result of applying the divergence theorem anyways is correct for point charges, a simple argument is given starting at page 9 in this link.
     
  11. May 22, 2012 #10
    You mention you want zero potential for z<0? Why is that important when our situation only deals with a physical system defined for z≥0?

    Also my book seems to say that the field below the conductor is zero, which obviously relates to what you say, how can I realize that the field is indeed zero for z<0? Evidently I thought that the field of the conductor was the same as that of a minus charge placed at (0,0,-d) since that was our image charge :(
     
  12. May 22, 2012 #11

    gabbagabbahey

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    The potential and the field are both zero for z<0. This can easily be seen by appealing to the uniqueness theorem for Laplace's equation. There is no charge in that region, so [itex]\nabla^2V=0[/itex] there. The potential is also 0 at the two boundaries (z=0 and infinity) of this region. [itex]V=0[/itex] certainly satisfies these conditions, and so must be the correct potantial there.

    The only reason that the potential in z<0 is importnant to you for this problem is that it should be obvious that the net charge everywhere (and hence the sum of the point charge and the induced charge at z=0) must be zero for the potential to be zero everywhere below the plane. Thus you should expect that the total induced charge at z=0 should be -q, and when you solve the problem, that's exactly what you get.
     
  13. May 22, 2012 #12

    rude man

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    Pardon my butting in again, but you can easily convince yourself that the image magnitude, sign and location will give you zero potential anywhere along the conductor.

    Remove the conductor, put the image in place, then calculate the work done to move a unity test charge from z = +infinity to z = 0, i.e. anywhere along where the conducting plate was located. What's the work that's done? And what's the definition of potential?

    BTW you're absolutely right, you're not interested in the region z < 0. Except that that's where the image has to be located.

    Or, even more obvious: what's the formula for potential due to a charge q a distance r away from q? So what happens when you add the potential of q and its image -q, anywhere z = 0?

    (Pardon if I'm being redundant here).
     
    Last edited: May 22, 2012
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