Solution to Boundary-Value Problems with Green's Function, and Electrostatic Energy

(The following are my notes on Jacksons Classical Electrodynamics, Chapter 1)
In another post of mine we found the solution for the potential function by using Greens identities and transforming the Poisson's equation into an integral equation.  We will be using that solution here. Just as a reminder, here is the solution.

The integral on the right hand side was an over-specification of the problem and we need to modify it for it to be able to be useful for either Neumann or Dirichlet boundary conditions. Recall from the same post that we found the following formula.

The function under the Laplacian is only one of a class of functions called Greens functions that follow the same general formula.

We now make the modification to the "solution" for the potential function.

Now, we need to choose G such that for the given boundary conditions, the above represents a solution that is not over-specified.  Basically this means, letting one of the terms in the surface integral equal zero.  For Dirichlet boundary conditions we demand that G(x,x')=0 for x' on the surface.  With this demand, the first term in the surface integral vanishes and we are left with the following..

For Neumann Boundary conditions we need to be more careful.  The most obvious choice here is to set the normal derivative of greens function equal to zero.  This doesn't work however because it is incompatible with Gauss' law.

Which is obviously not true. So this cannot be the correct approach. Thus the simplest allowable boundary condition is..

Where S is the surface over the enclosed volume.  We can now plug this into our expression for the potential.

Electrostatic Potential Energy and Energy Density

The product of the scalar potential and the charge of a point object could be interpreted as potential energy. If a charge is brought into a field described by the potential Phi, then the work done on the charge (and hence its potential energy) is given by...

This potential can be viewed as being produced by an array of (n-1) charges. Meaning...

This describes the potential energy of one charge in the array of "n" charges.  Thus the total potential energy is..

This step follows by he following logic. The first summation puts the restriction that j < i.  The second summation has no such restriction other than the fact that j and i are not equal (Thus, no division by zero. Also called infinite self energy term.)  If you plot this summation around the i=j line, you will see that the work in the first summation equation forms a triangle, while the second summation equation plots two such triangles because now j can be greater than i.  Thus If we impose no restriction, then we can just divide the result by 2 to get the work.  This is done so that when we go to the continuous case, we can substitute one of the volume integrals for the potential.

We will do that now..