Consider first another problem: An conductor with internal homogeneous isotropic resistance of 1, infinite in all directions. At the origin is a current source with magnitude 1. Thru each concentric sphere about the origin is a current of 1. The sphere of radius r has area 4πr². The current density at a point at radius r is thus away from the origin with magnitude 1/(4πr²). The electric potential there is −1/(4πr). The differential equations that govern these solutions are linear and the sum of two such solution is a solution.

Consider for each integer n a current source at the point with coordinates (0, 0, n) but with current there (−1)ⁿ. For each such n there is solution as above but with the symmetry center at (0, 0, n). The sum of all of these solutions converges to a solution. The potential function at point (x, y, z) for this aggregate solution is Σ((−1)ⁿ/√(x² + y² + (z − n)²)). This series converges due to the alternating signs.

Now twice the solution to the 1st problem may be found within the 2nd problem by considering that portion {(x, y, z) | 0 < z < 1} in the solution to the 2nd problem. The boundary conditions for the first problem is that the z component of the current flow be 0 for z = 0 and z = 1. In the 2nd problem this is ensured by symmetry. In the 1st problem the current is only 1/2 at each side of the plate.

Perhaps the techniques described here could compute this potential efficiently.