Potential of arbitrary charge distribution Imagine this: 
You have a sphere of air where you have no charge and around this sphere you have a charge distribution $\rho(r,\theta,\phi)$. (For instance, this could be $\rho(r,\theta,\phi)=e^{-r}$)
Now my question is: What is the most general equation that will give me the potential inside the sphere?-You can use that we have azimuthal symmetry. I am just interested in the equation. 
Probably this will contain a series with Legendre Polynomials and so on. 
 A: The electric potential $\Phi$ is defined through the following relation:
$$\mathbf{E}=-\nabla \Phi\tag{1}$$
Now consider a vector field $\mathbf{F}$ such that:
$$\nabla.\mathbf{F}=D$$
$$\nabla \times \mathbf{F}=\mathbf{C}$$
According to Helmholtz theorem, if the divergence $D(\mathbf{r})$ and the curl $\mathbf{C(r)}$ are specified and if they both go to zero faster than $\dfrac{1}{ r^2}$ as $r\to \infty$, and if $\mathbf{F(r)}$ goes to zero as  $r\to \infty$ then $\mathbf{F}$ is uniquely given by
$$\mathbf{F}=-\nabla U+ \nabla \times \mathbf{W}$$ 
where 
$$U(\mathbf{r})=\frac{1}{4\pi}\int \frac{D(\mathbf{r'})}{|\mathbf{r-r'}|}d^3r'\tag{2}$$
$$\mathbf{W}(\mathbf{r})=\frac{1}{4\pi}\int \frac{\mathbf{C}(\mathbf{r'})}
{|\mathbf{r-r'}|}d^3r'\tag{3}$$
For a static electric field, $D=\dfrac{\rho}{\epsilon_0}$ and $\mathbf{C}=0$. So, according to $(1)$ and $(2)$ the electric potential of a charge distribution that goes to zero faster than $\dfrac{1}{r^2}$ as $r\to \infty$ can be calculated as
$$\Phi(\mathbf{r})=\frac{1}{4\pi \epsilon_0}\int \frac{\rho(\mathbf{r'})}{|\mathbf{r-r'}|}d^3r'$$
where the integral is over all of space.
$\dfrac{1}{|\mathbf{r-r'}|}$ can be expanded using spherical harmonics to obtain a multipole expansion. So the multipole expansion is valid only under the above conditions too.
If the above condition doesn't hold, you have to use the $(1)$ equation, i.e. you have to find $\mathbf{E}$ first and then perform the integration to find $\Phi$ (as in the case of an infinite uniformly charged wire).
A: The most general expression for the potential (assuming a static charge distribution, as you use) is:
$$V({\vec r}) = C + \int d^{3}r'\frac{\rho({\vec r'})}{4\pi \epsilon_{0}\left|{\vec r}-{\vec r}'\right|}$$
Where C satisfies ${\vec \nabla}C = 0$, and the integral covers the region where $\rho \neq 0$.
If you doubt this, you can work out that $\nabla^{2} \frac{1}{\left|{\vec r}\right|} = \delta^{3}({\vec r})$, and then it should be pretty obvious that this equation satisfies the differential form of Gauss's law.
EDIT: 
I see that you're asking what happens inside a gap inside of a spherically symmetric distribution.  In this case, you can use the high school physics version of Gauss's law to show that:
$$\left|{\vec E}\right| = k\frac{Q_{inc}}{r^{2}}$$
Since, inside of your inner gap, the charge enclosed by any gaussian surface is zero, you have $E = 0$ and $V=$ Constant
