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?

  • 2
    $\begingroup$ 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, $\endgroup$
    – Sanya
    Commented Jun 22, 2016 at 16:27
  • $\begingroup$ 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. $\endgroup$ Commented Jun 22, 2016 at 16:32
  • $\begingroup$ @Sanya If $\Phi$ can diverge at point. Is there some approximation or limit of $\Phi(0)$? $\endgroup$ Commented Jun 22, 2016 at 16:38

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.