There are multiple ways to compute the potential, and each is suited to a different set of circumstances:

Compute directly

I would consider this a last resort unless you’ve got a point charge.

\[V(r) = \frac{1}{4\pi\epsilon_0}\int \frac{1}{R}\rho(\vec{r}^\prime) d\tau^\prime.\]

where as always \(R = \vec{r} - \vec{r}^\prime\).

Integrating Poisson’s (or Laplace’s) equation’s twice to get the potential

This works great in one dimension when you have boundary conditions, because you can integrate it twice to get rid of the derivatives. You’ll have two constants of integration, which you can figure out by using the boundary conditions.

\[\nabla^2 V = \frac{d^2V}{dx^2} = -\frac{\rho}{\epsilon_0}\]

Method of Images

Create a set of charges outside the region of interest that produce a potential in the region of interest that matches the boundary conditions. It’s useful for point charges near conducting planes and conducting spheres. Note that if what you really want is the field rather than the potential, you can compute it directly from the image charges (rather than computing \(V\) and taking the gradient).

Separation of variables

You assume your solution is the product of two functions, each of which is a function of only one variable, and then you use boundary conditions to find the potential. This works great if you have cartesian boundaries, in which case you get solutions that are products of exponentials and/or sinusoids.

In spherical coordinates this works, too, as long as you have azimuthal symmetry (i.e. nothing in the problem involves \(\phi\)) You end up with Legendre Polynomials in the function of \(\theta\).

You’ll be doing the cylindrical version of separation of variables for homework.

Multipole Expansion

If you’re interested in the potential outside of a particular charge distribution, the multipole expansion is useful because it’ll tell you what the dominant term in the potential is…the next leading term…etc. The general solution to Coulomb’s law (at the top of this page) actually naturally gets you Legendre Polynomials. Note that you don’t need boundary conditions to use a multipole expansion, because the multipole expansion is actually a formula for the potential given a particular charge distribution.

When would you do this? (Ethan asked this question after class on Friday). Here’s my tentative answer (we may want to adjust it over the course of the semester). If you have a weird charge distribution and you don’t actually want the exact potential, but just an approximation, this would be a good method. You would only keep the first couple of terms.