electromagnetism - Dipole in a spherical cavity in an infinite dielectric

In this example the assumption is made that inside a spherical cavity (with an infinite dielectric) with a dipole placed at its centre, the potential takes the following form: $$\phi_{in}= \frac{p cos(\theta)}{r^2}-E_{in}r \cos(\theta)$$ and $$\phi_{out}=\frac{p' cos(\theta)}{r^2}-E_{\infty}r cos(\theta)$$ I cannot see the justification for these forms of the potentials, so please can someone explain it to me?

I am specifically uncertain on:

1. 
   

   As $r\rightarrow 0$ , why $\phi_{in} \rightarrow \frac{p cos(\theta)}{r^2}$ when you should have from polarization charges on the surface of the cavity and also from $E_\infty$.

   
   


2. 
   

   Why can we assume the form of the first term in $\phi_{out}$?

   
   

Edit

I found this solution to Griffiths problem 3.34, which answers this question expect for the point about when $r \rightarrow 0$ the potential is only that of the dipole. I still do not understand why this has to be, you are inducing charges that are going to produce a field in the sphere and thus will not necessarily give you 0 potential at the centre.

Answer

Let me try to offer a different way to see it based on linearity. Note that your two links are reversed in the situation they consider; your second link has the dipole embedded in the dielectric embedded in vacuum, whereas your first has the dipole in a vacuum embedded in a dielectric. So, equation (3) contains an $\epsilon$ where it should have $\epsilon_0.$ The process I'm asking you to visualize looks like this:

1. Freeze the polarization field.


2. Remove the dipole charge and its vacuum field.


3. Examine what the remaining field must be doing at $r = 0$.


4. Re-insert the dipole charge using the principle of superposition.

During step (3) we know that within the cavity we are solving Laplace's equation due to the bound charges on the sphere -- but the solution needs to be continuous at $r=0$ because there is nothing there! We know that the general solution is: $$ V(r, \theta) = \sum_{\ell=0}^\infty \left[A_\ell~ r^\ell + B_\ell ~ r^{-\ell - 1}\right]~P_\ell(\cos\theta).$$ Continuity is obviously broken if any $B_\ell \ne 0$ as a singularity is not continuous.

Therefore during step 4 when we use the principle of superposition to re-insert the dipole, the voltage inside the cavity due to the surface charge must be simply: $$ V(r, \theta) = \frac {p \cos\theta}{r^2} + \sum_{\ell=0}^\infty A_\ell~ r^\ell ~P_\ell(\cos\theta).$$

Furthermore, as your second link discovers, the fact that $V$ and $\vec D \cdot \hat r$ are continuous across the boundary means that essentially the voltage fields induced by the contained moments must have the same $\theta$-dependence as the moments themselves: quadrupole moments will induce surface charges characteristic of quadrupole moment fields just the same as a dipole moment inducing a surface charge characteristic of the dipole moment field.