While studying the method of images for an infinite grounded plate, I came upon two assumption I can't seem to find the logic for:

  1. The test charge doesn't induce a charge density on the plate which makes the potential anisotropic in space. In other words, the charge doesn't induce a constant charge density (or any other such charge density) on the plate that makes potential at infinity different for different directions.

  2. The grounded plate is a conductor, and hence for each induced negative charge, there should be an equal positive charge residing somewhere on the plate. But the charge distribution derived using this method doesn't have any positive charge density. [Here, I am assuming that the plate doesn't have to be grounded by any special means - since it extends to infinity (where $V=0$), and being a conductor it is equipotential, hence all points on the plate are automatically at zero potential.]


1 Answer 1



Based on the comments below I agree that my original answer is incorrect, I wrote it too quickly. I'll leave the original post beyond the horizontal line. I'll make some corrected comments, but maybe it's better if someone else writes an answer.

Corrected answer:

The basic idea of the method of images is to find a solution to the original BVP by considering a different problem in an enlarged space that gives a solution to the original BVP when restricted to the original space.

There is of course a surface charge induced on the conductor--that is given by the sum of the image charges. This can be seen by Gauss's law--the field lines end on the surface of the conductor so there is a non-zero divergence for E there.

The surface charge induced is usually finite--that basically answers part 1. If the charge is isolated, the fields should die off asymptotically at infinity, so you don't induce an infinite amount of surface charge. Physically speaking, an infinite plane with a uniform surface charge has an infinite amount of energy--a single point charge coupled to a grounded conductor should not be able to do an infinite amount of work on the conductor.

The basic justification for the method of images is the uniqueness theorems for BVPs with Neumann boundary conditions--it doesn't matter how you find a solution to Poisson's equations with given boundary conditions, so long as you find a potential that solves Poisson's equation, and satisfies the boundary conditions (appropriate boundary conditions at the conductor, asymptotic fall off far away) you have found the unique solution. The proposal in the method of images is to enlarge the original space in which you are solving the equation, so you solve it, say, above and below the conductor. Below the conductor is an "unphysical" space in which you can put image charges, if you can find a collection of image charges such that (1) Poisson's equation is still satisfied in the physical region (that's guaranteed so long as the image charges stay in the unphysical region), and (2) the original boundary conditions are satisfied, then you have found a solution.

Original, incorrect answer:

The conductor is neutral. No charges are induced on the surface of the conductor at all.

The method of images instead says that there is an "effective" charge behind the surface of the conductor. The charge does not really exist, and it is crucial that the charge not appear on the surface of the conductor (that would ruin the boundary conditions).

In the end, you are just trying to solve Poisson's equation with a Neumann boundary condition (Neumann because of the conductor), in the region "above" the conductor. The method of images says that is actually easier to solve a different problem: you solve Poisson's equation in an extended space (above and below the conductor), with no boundary put in explicitly. This works so long as you can find a charge distribution in the lower half of the space so that the potential and the derivatives of the potential on the surface where the conductor would be in the extended space match the boundary conditions you would put on the surface of the conductor in the original problem. It comes down to an ingenious way of guessing the solution for a boundary value problem, which is guaranteed to be unique by standard theorems.

That in principle answers both 1 and 2, but just to make an extra comment about 1--putting any surface charge density on the conductor would change the conditions of the problem, because surface charges change the boundary conditions. In the case of an infinite surface charge density, you'd have to add the field of a plane of charge to the field induced by the original point charge--but that's not the problem that was originally asked. If you have a point charge and a conductor, the fields should die off asymptotically at infinity.

  • 1
    $\begingroup$ The first sentence in this answer is simply incorrect. If the conductor is grounded then a net charge is induced on its surface. $\endgroup$ Jun 22, 2015 at 16:06
  • $\begingroup$ @EmilioPisanty: Why would that be? $\endgroup$
    – Sidd
    Jun 22, 2015 at 16:08
  • $\begingroup$ Oh you are right, never mind. I got fooled because the standard example for the method of images has the field lines perpendicular to the surface so it looks like div E = 0 naively, but of course that's wrong since the lines end on the conductor. It's been too long since I took E/M. Can I withdraw the answer? $\endgroup$
    – Andrew
    Jun 22, 2015 at 16:09
  • $\begingroup$ @Andrew You can delete and undelete your own posts at your discretion. You can also edit deleted posts if you wish to do so before undeleting them. $\endgroup$ Jun 22, 2015 at 16:16
  • 1
    $\begingroup$ ... The reason for this boundary condition is ultimately physics, not math. If we started with the point charge plus a neutral infinite conductor, we only want to consider configurations that can be built with a finite amount of work. Alternatively, we really we have in mind that the charge + infinite conductor is an approximation to a charge + large finite conductor. Intuitively, if the conductor is large enough, the answer shouldn't depend on the boundaries/size of the conductor. $\endgroup$
    – Andrew
    Jun 22, 2015 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.