Method of images with infinite, earthed, conducting plane

  1. Hello guys,

    I'm having some trouble getting my head around the method of images, partly due to confusing notes and things.

    If I have a charge +q at a fixed point (0,0,a) above a conducting plane that is held at zero potential, it is said that the plane can be replaced with a charge of -q at (0,0,-a) as this solution satisfies the same boundary conditions (as per the uniqueness theorem).

    How is it that the potential on the surface of the plane can be 0 but the electric field be described as q/(2pi x espilon_0 x a^2)?

    I would have thought the potential be non-zero in the case of the mirror charge. Can the two charges not be thought of as a dipole? As in their field lines are both going in the same direction. Only if both charges were the +q or both -q would the potential, in my mind at least, be zero midway between them.

    Any help would be greatly appreciated,

    watty
     
  2. jcsd
  3. Doc Al

    Staff: Mentor

    Note that this is the field immediately above the plane, due to the induced surface charge.

    Yes, the charge and the mirror charge form a dipole, and their field lines do go in the same direction at the plane.
    The potential midway between two like charges is not zero. Recall that potential is a scalar--the potential from each charge is the same, thus the potential midway is non-zero. (Only for unlike charges would the potential contributions cancel at the midway point.)

    Here's a clear discussion of the method of image charges that might help you: Method of Images
     
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