# Linear, homogenous and isotropic dielectric in electrostatic field. Why do I consider two potentials (inside & outside sphere)?

## Presentation of the problem :

We have a uniform homogenous isotropic dielectric sphere in an electrostatic field.

To solve this problem, we remark that we have an azimuthal symmetry. So the potential of the problem is $V(r, \theta)$.

Because we are in homogenous isotropic dielectric medium, we have a Laplace equation for the potential inside and outside of this sphere:

$$\Delta V=0$$

By assuming $V=f(r)g(\theta)$, we end up by a potential of the following form:

$$V(r,\theta)=\sum_{l=0}^{\infty} (A_l r^l+B_l r^{-(l+1)} )P_l (\cos(\theta))$$

This solution is done in Plasmonic fundamentals by Stefan Alexander Maier. The image is taken from it.

## My problem :

The moment I don't totally understand is that they say we decompose the potential in two contributions: $V_{in}$ inside the sphere and $V_{out}$ outside.

Then they say that as the potential can't diverge in $r=0$, we have:

$$V_{in}=\sum_{l=0}^{\infty} (A_l r^l)P_l (\cos(\theta))$$

And

$$V_{out}=\sum_{l=0}^{\infty} (C_l r^l+D_l r^{-(l+1)})P_l (\cos(\theta))$$

What I don't understand is: how do we know we have to write two contributions of the potential.

If I had to solve the problem by myself I would only write one potential $V$ as $$V(r,\theta)=\sum_{l=0}^{\infty} (A_l r^l+B_l r^{-(l+1)} )P_l (\cos(\theta))$$

And then saying that the potential can't diverge in $r=0$, thus I would write $A_l=0$.

But of course it would be wrong (not enough "unknown to solve").

But I don't see from a math point of view why I should write two different potentials. Indeed, the potential $V$ is following the same equation $\Delta V=0$ inside and outside of the sphere so how can I know I have to write it in two different ways $V_{in}$ and $V_{out}$ ?