2
$\begingroup$

Given two infinitely long perfectly conducting cylinders of radius $r$ with centers separated by $d$, what is the capacitance between them, without assuming $d \gg r$?

I'm familiar with the derivation of the capacitance of parallel cylinders assuming $d \gg r$, which results in:

$$\frac{\epsilon_{r}\epsilon_{0}\pi}{\ln(\frac{d}{r})}$$

I recently had cause to actually apply this formula, and needed the version without the assumption, because the assumption isn't valid for most real cables (which pack the wires quite tightly.) I found this formula from several sources but can't find a derivation:

$$\frac{\epsilon_{r}\epsilon_{0}\pi}{\cosh^{-1}(\frac{d}{2r})}$$

I'd like to understand how to arrive at this result. I can understand intuitively why the formula changes in the absence of the assumption - if the cylinders are close together, charge will not be distributed evenly on the surface, preferring to bunch up on the side nearest the other cylinder. Which means the electric displacement field can't be derived trivially from Gauss's law any longer.

I know that the charge must still be concentrated on the surface of the cylinders, and I know that if I had the equilibrium charge distribution, I could integrate along the hollow shell of one cylinder (once by circular cross section, once along length) to find the $E$ field at each point along a line between the cylinders, then integrate the $E$ field along that line to find the potential difference, and get capacitance from there.

What I'm missing is how to obtain the equilibrium charge distribution. I think it might be an energy minimization problem on the $E$ field, but I'm unsure of both how to set that up or how to solve it.

$\endgroup$

1 Answer 1

1
$\begingroup$

This result follows from a standard problem in the method of image charges. Since it's a fairly common exercise, I will just outline the proof rather than giving it in detail. (This may be why you've had trouble finding a full derivation yourself.) I will work in vacuum, but the extension to a linear dielectric is what you'd expect it to be.

Consider two uniform line charges $\pm \lambda$ running along the $z$-axis at $x = \pm a$ and $y = 0$. We can, from superposition and standard Gauss's Law arguments, show that the potential at a point $(x,y)$ is given by $$ V = \frac{\lambda}{2 \pi \epsilon_0} \ln \frac{r_-}{r_+} $$ where $r_-$ is the distance to the negative charge, $r_+$ is the distance to the positive, and we have set $V = 0$ along the $yz$-plane (where $r_- = r_+$.) Through a large amount of geometry and algebra, one can show that the equipotential surface for any given $V$ in this geometry is a cylinder, centered at $x_0 = a \coth (2 \pi \epsilon_0 V/\lambda)$, with a radius of $r = a \,\mathrm{csch} (2 \pi \epsilon_0 V/\lambda)$. Note that this implies that $x_0/r = \cosh (2 \pi \epsilon_0 V/\lambda)$ for all equipotential surfaces.

Now consider a pair of long cylinders, whose centers are separated by a distance $d$ and with radius $r$, held at a potentials of $\pm V_0/2$ (i.e., their potential difference is $V_0$.) By the uniqueness properties of Laplace's equation, the potential outside the cylinders will be the same as it was in the previous problem, so long as we choose our values of $\lambda$ and $a$ correctly. Specifically, we will want to have our equipotential surfaces at $x_0 = d/2$ above, implying that $d/2r = \cosh(\pi \epsilon_0 V_0 / \lambda)$. Solving the above yields $$ \frac{\lambda}{V_0} = \frac{\pi \epsilon_0}{\cosh^{-1}(d/2r)}. $$ Moreover, the line charge density in the original setup must be the same as the charge per unit length on the cylinders (as can be seen by considering a Gaussian surface just outside the cylinder.) Thus, $\lambda/V_0$ is the capacitance per unit length of the cylinders, and we are done.

Note that it isn't necessary to calculate the surface charge density on the cylinders in this derivation. If you need it, it can be found by calculating $\epsilon_0 \hat{n} \cdot \vec{\nabla} V$ at points along the surface of the cylinders.

$\endgroup$
1
  • $\begingroup$ Thank you! I can try that equipotential surface proof myself now that I know the name for the technique. And this provides a nice abstract visual model too - where the ideal case has an image charge of a line at the center of the cylinder, this case has an image charge of a line slightly offset within the cylinder towards y=0, and by substituting into the original potential formula we're basically asking how far that virtual line charge is offset. $\endgroup$
    – Willa
    Commented Aug 2, 2021 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.