# Poisson's equation with point source

PDE's are not my thing, but I'm trying to calculate the voltage in a finite conducting sphere, based on a point current source.

With Poisson's equation of $\nabla^2\Phi = -\frac{J}{\sigma}$, where $J$ is a point charge source at (0,0,0) (Dirac delta function), in a finite sphere with boundary conditions $\Phi(R) = 0$ (R is the radius of the sphere).

With some Green function $\frac{1}{x} - \frac{1}{R}$ (that I don't fully understand), I get that $\Phi(x) = \frac{J}{4\pi\sigma}(\frac{1}{x}-\frac{1}{R})$ which matches a numeric solution. However as $x$ tends to 0, $\Phi$ tends to infinity (as $\frac 1 x$ does). It doesn't make sense to me that the voltage in a finite domain should be infinite.

Have I missed something, or butchered to solution somehow?

• I am a bit confused by your question. Normally, $\Phi$ is used for the electrical potential and that is connected via the Poisson equation to the charge, not the current. In general, it is no problem for $\Phi$ to diverge at single points as putting two point charges at exactly the same spatial coordinate is forbidden - thus an infinite energy barrier, Commented Jun 22, 2016 at 16:27
• This might just be where I don't really know what I'm talking about. But the system I have is $\nabla(-\nabla \sigma \Phi) = J$, so I moved the $-\sigma$ into the source term. Commented Jun 22, 2016 at 16:32
• @Sanya If $\Phi$ can diverge at point. Is there some approximation or limit of $\Phi(0)$? Commented Jun 22, 2016 at 16:38

I am still not sure of your problem physically. Mathematically, it seems to be about finding the solution to: $$\nabla^2 \Phi = \frac{J}{\sigma} \delta(\vec{r})$$ with $\Phi(\vec{r}=R\vec{e_r})=0$ for $\vec{e_r}$ being the boundary condition. And you are right, this can be solved using the method of a Green's function, the one obeying the auxiliary problem: $$\nabla^2 \Phi' = \delta(\vec{r})$$ with the same boundary condition $\Phi'(\vec{r}=R\vec{e_r})=0$.
Mathematically, what you have done seems to be fine to me, assuming that your $x$ is the radial coordinate of canonical spherical coordinates.
Now about the physical side of the problem. Take the simple gravitational potential of a point mass or the electrical potential of a point charge - both diverge with $1/r$ (let's neglect the fact that spherical coordinates are not well defined in the origin). Outside $r=0$ we do not need an approximation because we can just calculate the value. For $r=0$ the potential is just $\infty$. This reflects the physical reality that it is not possible to get two different charges arbitrarily close together - they repel each other the more the closer they are. In the case of the attractive potential it should probably be seen as a limitation of our physical model which does not take into account that any particle does have a finite extension and thus no two particles can occupy exactly the same spatial coordinates. But the positive or negative divergence of the potential is no problem per se.