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Image charge inside a conducting sphere

  1. Nov 16, 2011 #1
    I think understand the concept of the "method of images" to describe a conductor. The conductor has to be an equipotential or else electrons would be moving around inside it. This equipotential can be described by replacing the conductor with an image charge.

    For example imagine the field lines of a positive charge at a distance +x in front of a grounded vertical conductor in the y-z plane. The conductor can be represented by a negative image charge placed at position -x. The combined system of charge and image charge has a zero volt equipotential in the y-z plane as required.

    Now imagine a positive charge anywhere inside a grounded conducting sphere.

    Can the grounded conducting sphere be represented by a negative image charge placed directly over the positive charge? I guess it can as in such a situation all closed surfaces around such a system of charges will be at zero volts.



    John
     
    Last edited: Nov 16, 2011
  2. jcsd
  3. Nov 16, 2011 #2

    vanhees71

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    Yes, the image-charge method also works for a sphere. You have to make an ansatz with an unknown charge and distance of the image charge from the center. Then you use the boundary conditions at the conducting sphere (equipotential surface, continuity of the tangent em. field across the surface) to determine these unknowns.

    You find this in any good textbook on electromagnetism (e.g., Jackson).
     
  4. Nov 16, 2011 #3
    Does this work for the case I describe here where the charge is *inside* the conducting sphere?

    In Feynman Vol II I have seen a worked example where the charge is *outside* the conducting sphere.
     
  5. Nov 16, 2011 #4

    vanhees71

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    Yes, it also works for a charge inside.
     
  6. Nov 16, 2011 #5
    Thanks
     
  7. Nov 16, 2011 #6
    You have to be careful about the image at R=0. As R goes to zero, the image goes to infinity. In the case where the charge is outside, you need 2 images, one at the origin of the sphere and the other at some ratio that depends on the radius of the sphere. If the charge is at infinity, there's no effect on the sphere, so the image charges cancel out. Not quite sure about the inverse. I think Jackson discuss such a case, but I don't have it here.
     
  8. Nov 16, 2011 #7
    In principle, you can use the image method for any problem (in practice, some problems may require an infinite number of charges that are very hard to find by hand). The Green's function method solution, which is one of the most useful, robust, and general solutions to Maxwell's equations, is in a sense just a summation over the effects of appropriately placed image charges.
     
  9. Nov 16, 2011 #8

    clem

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    The image charge equations hold for a charge outside a conductor or inside a HOLLOW CONDUCTING SHELL. It would not work fora charge inside a solid conductor.
     
  10. Nov 17, 2011 #9
    That's just because there are no free charges inside solid perfect conductors (they are all on the surface), not any inherent weakness in the image method.
     
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