Pre-Quantum Electrodynamics

• PM 3.4
• Gr 3.2

This section could almost be called "guessing the potential" or "pulling the potential out of a hat". We will rely on two things:

• our ability to assemble potentials by combining simple potentials we already know (invoking the principle of superposition)
• our certainty that if the potentials we assemble fulfill the desired boundary conditions, we have found the one and true potential we were originally seeking.

One type of physical system which gives us particularly simple conditions at boundaries are conductors: the fact that conductors are equipotentials means that we can fomally solve loads of electrostatic problems involving conductors of various shapes.

Let's consider one of the simplest electrostatic setups we can think of which goes beyond a point charge, namely a system of two point charges \(\pm q\) separated by distance \(d\). For definiteness, we put a charge \(q\) at coordinate \(\frac{d}{2} ~\hat{z}\), and a charge \(-q\) at \(-\frac{d}{2} ~\hat{z}\).

By superposition, we have that

\begin{align} \phi(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z-d/2)^2]^{1/2}} - \frac{q}{[x^2 + y^2 + (z+d/2)^2]^{1/2}} \right] \tag{p_di_z}\label{p_di_z} \end{align}

Now it's obvious that \(\phi = 0\) when \(z = 0\). Therefore, in the region \(z > 0\), this problem is completely equivalent to a second problem: a point charge \(q\) at \(\frac{d}{2} ~\hat{z}\), and a grounded conductor on the whole plane \(z = 0\). Yet another equivalent problem in the region \(z < 0\) is that of a charge \(-q\) at \(-\frac{d}{2} ~\hat{z}\) with a grounded conductor on the plane \(z = 0\).

We can go one step further. In the two point charges problem (dipole), we could put a conductor on any of the equipotential lines, and still have an explicit solution for the potential in terms of the potential from the original two point charges.

Our next step in such image problems is to determine what the induced surface charge is.