Potential of a dipole with actual physical extension? I think everybody here knows the equation that gives the potential of a point like dipole, but how does the field look like if you have e.g. a metal sphere with radius $R$ and a certain dipol moment, how does this potential look like?
 A: You can find the answer by considering the multipole expansion of a charge distribution.
The electric potential of any localized charge distribution can be expanded in terms of the distance (or inverse of it) from the origin to the field location in a multipole expansion. In this expansion, what you are actually doing is writing the potential of the charge distribution in terms of ideal multipole potentials. So, for the regions where the expansion is valid, the potential of any charge distribution that all of it's multipole moments except the dipole moment vanish is exactly that of an ideal dipole moment.
Assume in an arbitrary (localized) charge distribution, the most distant charge in the distribution is located at a radius of $r=r'$. For the potential at points with $r>r'$ you can use the equation $(1)$ and for the points with $r<r'$ you can use equation $(2)$:
$$\Phi(\mathbf{r}) = 
\frac{q}{4\pi\varepsilon r} \sum_{l=0}^{\infty}
\left( \frac{r^{\prime}}{r} \right)^{l}
\left( \frac{4\pi}{2l+1} \right)
\sum_{m=-l}^{l} 
Y_{lm}(\theta, \phi)  Y_{lm}^{*}(\theta^{\prime}, \phi^{\prime})\tag{1}$$
$$\Phi(\mathbf{r}) = 
\frac{q}{4\pi\varepsilon r^{\prime}} \sum_{l=0}^{\infty}
\left( \frac{r}{r^{\prime}} \right)^{l}
\left( \frac{4\pi}{2l+1} \right)
\sum_{m=-l}^{l} 
Y_{lm}(\theta, \phi)  Y_{lm}^{*}(\theta^{\prime}, \phi^{\prime})\tag{2}$$
or in a more compact form:
$$\Phi(\mathbf{r}) = 
\frac{q}{4\pi\varepsilon} \sum_{l=0}^{\infty}
\frac{r_<^{l}}{r_>^{l+1}}
\left( \frac{4\pi}{2l+1} \right)
\sum_{m=-l}^{l} 
Y_{lm}(\theta, \phi)  Y_{lm}^{*}(\theta^{\prime}, \phi^{\prime})$$
So, the shape of the distribution doesn't matter, and what matters is it's multipole moments. In your example of a charged sphere with (only) dipole moment, the potential outside will be exactly a pure dipole potential, and inside, it will be constant.
