# Potential energy of a charged ring [closed]

Consider a ring of radius $R$, and charge density $\rho$. What will be the potential energy of the ring in its self field?

The best I can do: $$dq = \rho R \cdot \, d \alpha$$

$$E_p = 2 \pi R \cdot 2 \int_{0}^{\pi - \delta \alpha} \frac{\rho^2 R^2}{r(\alpha)} \, d\alpha$$

where

$$r(\alpha) = \sqrt{2} R \sqrt {1+ \cos (\alpha)}$$

Edit:

This is wrong - $E_p$ doesn't even have the correct dimensions.

You've got basically the right idea. Just for clarity, let me recap the setup: suppose that your ring is centered at the origin and oriented in the xy plane. Consider two differential elements of charge, $\mathrm{d}q$ located at $(R,0)$, and $\mathrm{d}q'$, located at $(R\cos\phi,R\sin\phi)$. The potential energy of these two charge elements is

$$\mathrm{d}^2U = k\frac{\mathrm{d}q\mathrm{d}q'}{r}$$

The distance between the two differential charges is

$$r = \sqrt{(R - R\cos\phi)^2 + (R\sin\phi)^2} = R\sqrt{2 - 2\cos\phi}$$

You can easily generalize this to apply to any two differential elements of charge located at angles $\theta_1$ and $\theta_2$, by just replacing $\phi$ with the angular difference between them, $\theta_1 - \theta_2$.

Now in theory, you should be able to determine the potential energy of the ring by integrating over all possible pairs of charge elements:

\begin{align}U &= \iint\mathrm{d}^2U\\ &= k\int_0^{2\pi}\int_0^{2\pi}\frac{\rho R\mathrm{d}\theta_1\rho R\mathrm{d}\theta_2}{R\sqrt{2 - 2\cos(\theta_1 - \theta_2)}}\\ &= k\rho^2 R\int_0^{2\pi}\int_0^{2\pi}\frac{\mathrm{d}\theta_1\mathrm{d}\theta_2}{\sqrt{2 - 2\cos(\theta_1 - \theta_2)}} \end{align}

But oops, guess what, the integral doesn't converge! So it's clearly not that easy.

In fact, it actually makes sense that this integral shouldn't converge. Think about the potential energy contributed by a pair of charge elements $\mathrm{d}q_1$ and $\mathrm{d}q_2$ which are very close to each other. The denominator of $\mathrm{d}^2U$ becomes very small, and as the separation goes to zero, the contribution to the potential energy becomes infinite. It turns out that if you're doing the equivalent calculation for a surface or volume charge distribution, the "spread" of the charge in 2 or 3 dimensions is enough to keep the integral from diverging, but not so with a line charge. So the bottom line is that the potential energy is infinite.

In practice, this isn't really an issue because any realistic charge distribution is constructed by pushing together existing pieces. You can never actually get the pieces to be right next to each other, so you don't have the problem with $r = 0$ in the denominator. It does interest some theorists, though, to figure out what's going on with this sort of situation and whether it makes sense on some fundamental level to have a theory in which a simple, sensible calculation like this turns out to be infinite.

• Yep, this is the stuff I remember. Would it also be fair to say that an infinitely thin wire loop has infinitely small capacitance? That sounds much less exotic IMO. I think the same can be said for an infinite line charge. We often look for the field around a line charge, but rarely question the potential (per length) of the line itself. – Alan Rominger Nov 7 '11 at 16:50

Note that the Coulomb self-energy of a uniformly charged two-dimensional (2D) disk is FINITE. The following paper treats the same problem for a 3D cylinder. It also gives the result for the 2D disk as well as a general calculation method that can be easily adapted to the 2D disk case: O. Ciftja, Physica B 407, 2803-2807 (2012).

Thanks David. I tried to re-write Your solution for the 2D case. So assume that the inner area of circle is also charged. Then:

$dq = r \rho \, d\alpha \, dr$

and the final integral is

$U = \frac{1}{2} \cdot \rho^2 \int_{0}^{2 \pi} \int_{0}^{2 \pi} \int_{0}^{R} \int_{0}^{R} \frac{r_1 r_2 \, dr_1 \, dr_2 \, d\theta_1 \, d\theta_2 }{\sqrt{(r_1-r_2 \cos(\theta_1 -\theta_2))^2 + r_2^2 \sin^2 (\theta_1 -\theta_2)}}$

here $1/2$ takes into account double counting of interactions.

I cannot take this integral - nor can I take Yours - but I can see that there are cases when denominator is equal to zero.

It seems to me that such an approach will always give divergence. The only thing which works - is a Poisson's equation: it gives finite potential in 3D for a charged ball (internals also charged).

The reason for that could be abstractness of $\rho$ - in reality we always have "discete" electrons - and if we wish to take that into account - maybe we have to work within quantum theory of electricity.

• If you use a log potential in 2d, you get a convergent answer, the same as the self-potential of a cylinder with uniform charge density on the surface. – Ron Maimon Aug 1 '12 at 6:04

$$E=\frac{N}{2}\times\sum_{n=1}^{N-1}\frac{1}{4\pi\epsilon_0}\frac{e^2}{R\sin(n\frac{\pi}{N})}$$

$$=\frac{Ne^2}{8\pi\epsilon_0 R}\sum_{n=1}^{N-1}\frac{1}{\sin(n\pi/N)}$$

$$\approx\frac{Ne^2}{8\pi\epsilon_0R}\frac{N}{\pi}\int_{\pi/(2N)}^{(2N-1)\pi/2N}\frac{dx}{\sin(x)}$$

$$\approx\frac{Q^2}{8\pi^2\epsilon_0R}\ln\frac{\tan((2N-1)\pi/4N)}{\tan(\pi/4N)}$$

$$=\frac{Q^2}{8\pi^2\epsilon_0R}\ln(\frac{1}{\tan^2(\pi/4N)})$$

$$\approx\frac{Q^2}{4\pi^2\epsilon_0R}\ln(\frac{4N}{\pi})$$

$$=\frac{Q^2}{4\pi^2\epsilon_0R}\ln(\frac{4Q}{\pi e})$$

In fact the approximation by using integral instead of the sum may have significantly altered the non-dimensional constant of the final result...