*This is a bit of an old question but I stumbled upon it and I believe it a great learning experience and quite a beautiful question.*

# What you got right (most of it)

I'll begin by pointing out that you are entirely right! your configuration creates a perfectly uniform field in the ball!

*(I'll use ball to refer to the volume, while sphere would be used for the thin shell. I highly recommend using those definitions to avoid confusion)*

As you pointed out the uniqueness theorem doesn't mean that your solution is unique. In fact there are infinite configurations of charges that would create uniform field. Unfortunately your solution is slightly problematic, because you described your charges as an nonuniform distribution of charge inside the volume of a perfect conductor.

Why is that a problem? Because inside of a perfectly conducting body there are no electric fields! (Any field the body could be subjected to would create surface currents that would immediately cancel the fields for the volume of the body.) Without electric fields, why would the charge ever arrange itself so nicely?
As there are no way to create charge distribution inside perfect conductors we generally treat those solutions as "Non-physical", meaning mathematically correct but not real. Surface charges are on the other hand completely possible!

I will note, that your solution is on the right track. By taking the right limit and changing your notation a bit you can get the charge distribution as a surface charge.

So, now that we understand the problem, let's answer each of your questions

## Is there a unique solution?

Or as you placed it *"Is there some type of uniqueness theorem to determine this situation?"*
I'm going to take some liberties and assume the conductive sphere is neutrally charged, otherwise there are infinite solutions. (differing only by the charge of the sphere) I'll also assume all the charge is on the sphere and not on the ball.
With those assumptions the solution is unique! Although I'm not aware of any named theorem to the problem, there is an easy way to see that there is only a single solution that has only surface charges.

We'll prove it thusly:

lets assume we have two distributions that create the uniform field $\vec E=A\cdot \hat z $ the solutions: $\sigma_1 (u,v)$ and $\sigma_2 (u,v)$. Both of them defined over the same closed, orientable surface.

($u$ and $v$ are parameters that describe our surface. With a sphere we could choose for example $\theta$ and $\phi$. $A$ is an arbitrary constant and $\hat z$ is an arbitrary direction)

Lets examine $\sigma_1 - \sigma_2 (u,v)$. From superposition we can immediately tell that the field inside the volume is zero. What is the field outside the body? Zero. Let's prove it.

You can get the solution immediately here by making an argument about the energy of the new electric configuration, which must be zero. We'll make the same argument in the proof, albeit a bit more formally.

Let's take a small ball around each part of surface, so it is small enough to not touch another part of the surface (there is only one surface connectivity group inside the ball). We'll use Gauss's law to find the charge inside the sphere. If it is zero everywhere then we have no charge and therefore no field and we are done. If there is a some area with positive charge then there must at least a single direction in which the electric field is negative. Let's imagine a "test charge" and move it with the field direction. There are three options:

- We will continue infinitely away from the body. But then it means the body has positive potential. In contradiction to the assumption that the body is neutral.
- We will reach a minimum of electric field somewhere in space. But a minimum of electric field is simply a point of negative charge! In our scenario all charges are on the surface. so this is also a contradiction.
- We will reach a minimum of electric field on the surface. But that means that we increased in potential! since $-\int \vec E \cdot \delta L=\Delta V$. But all points on the conductor have the same potential! Contradiction.

If the field is zero everywhere then there is no electric field anywhere and so no electric charge anywhere. Formally, due to Gauss's Law every possible volume would have zero divergence of electric field and thus zero average charge.
$\sigma_1 - \sigma_2 (u,v)=0 \Rightarrow \sigma_1 = \sigma_2$

and so we proved that $\sigma$ is unique.

## What is the right solution?

You are already on the right track! If we take the limit of $lim{a\rightarrow 0^+}$ and increase $rho$ accordingly so to keep $\rho \cdot a$ constant (meaning our polarity is the same as before) we can immediately get the surface charges.

Let's look at an arbitrary slice of the ball, at height H from the top, we'll consider the top half for now. (Up is $\hat a$ for simplicity) The radius of this slice is $\sqrt{R^2 - H^2}$ the area of this slice is therefore the square of that times $\pi$ meaning $A(H)= \pi \cdot (R^2 - H^2) $.

What is the total charge at a single slice? $Q (H) = \rho (A(H)-A(H+a))$. All the charge is distributed over the thin shell, since it's simple to prove that the charge is zero anywhere else. We immediately get that $\sigma (H) = \rho (A(H)-A(H+a))/{2 \cdot \pi \cdot \sqrt{R^2 - H^2}}$

$\sigma (H) = \rho (2H \cdot a)/{2 \cdot \pi \cdot \sqrt{R^2 - H^2}}$
We can immediately find $(\rho \cdot a)$ since we know the polarity of the final charge distribution.
good luck

## Best Answer

Dipole moment is a vector and can be calculated using formula

$$\vec{p} = \sum_i q_i \vec{r}_i.$$

It can be shown easily using the formula above that in case of two charges separated by distance $d$

$$\vec{p} = q \vec{d},$$

where vector $\vec{d}$ goes starts at negative ends at positive charge.

http://en.wikipedia.org/wiki/Electric_dipole_moment#Dipole_moment_density_and_polarization_density