# Classification of equipotential curves between two equal point charges

The equipotential curves of two parallel equal-density line charges (in a plane perpendicular to the line charges) are known to be Cassini ovals. This is because the potential at a point a distance $r_1$ from one line charge and $r_2$ from the other line charge is

$$V = \frac{\lambda}{2\pi\epsilon_0}\left(\log\left(\frac{r_1}{r_0}\right)+\log\left(\frac{r_2}{r_0}\right)\right)=\frac{\lambda}{2\pi\epsilon_0}\log\left(\frac{r_1r_2}{r_0^2}\right)$$

which means that the equipotential curves require that $r_1r_2$ is constant, the same condition that generates Cassini ovals.

However, there doesn't seem to be any analogous well-known geometric form for the equipotentials generated by two equal point charges. The equipotential curves in a plane coplanar with the charges require that $\frac{1}{r_1}+\frac{1}{r_2}$ be constant. Here are the only things I can find out about these curves:

Can we classify these equipotential curves as instances of a single geometric form in a similar fashion to the line-charge case?

• I think the "infinite" aspect of the line charges changes things. It's similar to charged parallel plates in that when finite, the net potential drop across must be zero but can be finite when the plates are infinite. The fields from point charges are inherently 3D but infinite line charges have a symmetry along the length (since there is no end) reducing the system to 2D. – honeste_vivere Jul 29 '17 at 12:23
• @honeste_vivere I know, but the potential of two point charges also has cylindrical symmetry about the line joining the two charges, so I still feel like there should be an elegant explanation. – probably_someone Jul 30 '17 at 3:32
• Suggested line of attack: transform to ellipsoidal coordinates (foci at charges), look at differential equation for equipotential curve, it will probably be exactly solvable. (The equations of motion are separable in this coordinate system.) – Void Aug 1 '17 at 15:51