Solving the potential function for a Dielectric Sphere in a constant Electric Field

Question

I was observing the solution to the potential function for a dielectric sphere (dielectric constant= $\epsilon$) of radius $r=a$ at a constant field $E_0$. The boundary conditions were as follows:

[WHERE, $\phi_1(r,\theta)$ is the potential inside the sphere and $\phi_2(r,\theta)$ outside the sphere]

1.$\phi_1(r=\infty)=E^0r \cos(\theta)$;

2.$\phi_1(r=a)=\phi_2(r=a)$

3.$\phi(r=0)$ is finite

4.$\epsilon\frac{\partial \phi_1}{\partial r}=\frac{\partial \phi_2}{\partial r}$ at r=a

The solution is the following potential functions:

$\phi_1(r,\theta)=-\frac{(3E^0r \cos(\theta))}{(\epsilon+2)}$ and $\phi_2(r,\theta)=-E^0r \cos(\theta)+\left(\frac{\epsilon-1}{\epsilon+2}\right)\frac{E^0(a)^3 \cos(\theta)}{r^2}$

My questions are as follows:

a) On which physical ground is the condition 3 taken .

b) It was written that the condition for equality of Tangential Electric field component at the junctional surface is included within condition 2. How is that happening?

c) It was also stated that only at the surface of the dielectrics, the Laplace's equation does not hold. I understand why the equations holds within the sphere and outside it (free charge density is 0), but then what is wrong at the surface?

Answer 1

1. The potential can only diverge if the charge density diverges, and even then only for line or point charges, not surface charges. Since there is not a point or line charge at the origin, the potential has to be finite there.
2. The equality of the tangent components follows from the requirement that the potential be continuous and $\mathbf{E} = -\nabla \Phi$. In order to get a non-vanishing curl with that definition you would have to have a discontinuous change in $\Phi$. See, for example, $\Phi = y\Theta(x)$ ($\Theta$ the Heaviside step function).
3. What's going on at the surface follows from looking at the macroscopic formulation of Maxwell's equations. Notice that in that formulation the dielectric permittivity is inside of the divergence, so if the permittivity changes as a function of position you get a product rule: $$-\nabla \cdot \left( \epsilon \nabla \Phi\right) = - \epsilon \nabla^2 \Phi - (\nabla \Phi) \cdot (\nabla \epsilon)= 0,$$ which is not Laplace's equation. As @CStarAlgebra noted in another answer, this is because of surface charge densities in this case. If $\epsilon$ continuously varies with position, you would get a continuous volume bound charge density.

Answer 2

A) Condition 3 holds because the potential should never blow up to infinity, it would be undefined at that point. This is why when solving Laplace's equation with azimuthal symmetry we choose our solution such that the potential does not diverge in the region we are solving for.

B) The potential should always be continuous, if it was not the electric field, the gradient of the potential, would be infinite, an unphysical situation.

C) At the surface of the dielectric sphere, a surface charge density is induced by the electric field, thus there is charge on the surface of the dielectric, but Laplace's equation is only valid in a region with no sources, so it does not hold at the surface.