Pre-Quantum Electrodynamics

• Gr 3.2

The fact that conductors are equipotentials means that we can fomally solve loads of electrostatic problems involving conductors of various shapes, by looking at combinations of point charges.

Let's consider the simplest electrostatic problem above a single point charge: 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_dip_z}\label{p_dip_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.