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Method of images: electric dipole and infinite plane

  1. Nov 30, 2015 #1
    Hello, a dubt arose while doing some exercise.
    If I have a charge q at a distance d from the above-mentioned plane, i can find the solution to the laplace equations (thanks to the uniqueness theorems) finding a collection of image charges that satisfies the boundary conditions.
    These conditions are V=0 (or V=constant?) on the plane and V=0 at infinity.
    I find that the image charge is -q at distance d on the other side.

    If I have a dipole the solution is a reversed, "specular" dipole on the other side at the same distance again (thanks to the superposition principle, right? I treat each single charge of the dipole like in the first example mentioned).

    Now I found an exercise in which there was a conducting semispace and the dipole orthogonal to the its surface at distance d.

    I saw in the solution that I don't apply the potential 0 at the surface but the parallel component of E must sum to zero. So the image dipole has the same orientation of the real one.

    Why this? I know the "parallel component of E" condition has to be applyed when studying the electic field in the boundary of two dielectics (and a conductor is a dielectric with infinite permittivity) so it makes sense. But why the potential condition doesn't hold here?
    Is it in fact because this is a region of space conductor and not a surface?
    Why again?
    Can you explain please?

    Thank you and sorry for the length.
  2. jcsd
  3. Nov 30, 2015 #2
    The condition of parallel E-field being zero, in both cases makes sense, because the conductor is a equipotential surface.
    Therefore the condition of setting V=0 inside the conducting surface / volume is just a convention.

    I dont know why the solution claimed that. In fact it should not make a difference whether you use conducting surface or a semispace, because the influenced charge is just on the surface.
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