# Dual finite parallel disk capacitor; estimating mathematically that the potential or field or distribution is nearly uniform from first principles

Working through Purcell. In particular, doing my best to put everything on a firm footing, mathematically, so that I can understand it. But, I have struggled all week in establishing from first principles that the potential/field/distribution for a configuration of two capacitive disks of radius 1 and separation s along the same cylindrical axis is "nearly" uniform. In the book, like all the other books at this level, it seems, it is merely evinced by a field line diagram drawn according to a numerically estimated solution to the boundary value problem--none of which is developed in the text or even the exercises.

I don't care to get the exact distribution, or even a sharp estimate; I just want an estimate good enough to establish the claim that this setup, with finite disks at finite separation, does exhibit the sort of behavior of the easy-thought-experiment-examples with infinite plates or infinitesimal separation or where the plates are surfaces of nested spheres so large that the curvature is trivial.

I have attempted to get estimates many ways: using the harmonic relationship between partial derivatives and the knowledge that the field is linear at the origin; using the averaging properties of the laplacian and also the fact that the laplacian of the potential is proportional to charge density; I even successfully got (by hand!) estimates for the charge distribution and electric field on an isolated disk. The only problem in the book that seems related is that of verifying that an infinitesimally thin capacitive rod of finite length will have a uniform distribution (this is trivial when you've done the disk already; I also did it the way suggested in the book: where you divide the rod into N uniform segments and estimate the "correction" needed at location i+1 (of the order of N/2) to account for the field at i due to all other segments... but I also struggled to get this method to work for the case of two disks).

Of course, since I have a crude estimate for the electric field from a single disk, I can see quite evidently that the field lines will have a tenancy to straighten out in the region between two disks. But, I am struggling to put any quantitative measure to this notion. Any ideas?

One thing: from looking at more advanced books, it seems that you can get estimates on the solution by appealing to PDE theory and estimating the special functions involved in the solution you get. I don't care about getting a sharp estimate; is there any first principles way to do this? Feel free to apply extra assumptions--an answer that uses the words "uniform continuity" would not be off putting, if you get my drift.

• I have this possibly related question on MATH.SE. I am also looking for a "first principles" argument that can be made into a rigorous proof. Have you tried using that the field minimises its energy? – Keith McClary Sep 30 '18 at 5:53
• Hi Keith. My tools are limited: the only "energy arguments" that I know how to apply are the usual field-line assertions, which don't admit proper numerical estimates of things. However, if you look below, you will see that symmetry arguments, like the ones used to DERIVE the density of the conducting disk (which as you mention in your post diverges at the edge) from the case of the sphere, coupled with some clever first-principles estimation trickery, might provide an answer to THIS problem. In your problem, I can only say that assuming some kind of regularity might help (deform a disk). – entprise Sep 30 '18 at 21:09

Current best guess:

Consider the case of two insulating disks oppositely and uniformly charged of radius R with separation s along the polar axis. We try to show that they "nearly" satisfy the Laplace equation for any region between the interiors of the plates, for small plate separation s. Take any surface point on the interior of the face of a disk, x. Around this point is a maximal circle of radius r that is still contained in the original disk; there is a corresponding circle above the corresponding point on the other disk.

Neither circle contributes to the radial field, by symmetry. Regardless, the field in the zenith direction at x follows directly from Gauss' Law: $$\sigma/\epsilon = Q/(4\pi\epsilon R^2)$$ (notice that we have used the fact that the charge is spread uniformly on the FACE of the plate, not throughout some finite depth).

The field in the radial direction is to be bounded (rather than estimated directly!) in the following manner. The charge on the larger disks, but outside the smaller disks, is all that may contribute to the radial field at the point in question. Assume x corresponds to the zero angle, in cylindrical coordinates. There is certainly a moderate degree of cancellation in the radial field, even among the contributions from a single plate--I believe this is why the exact solution requires special function theory. But, this need not concern us, because there is ENOUGH cancellation when we compare the corresponding patches from the two oppositely charged plates, evinced as follows.

The charge on one of the uniform disks, but not in the maximal circular neighborhood around our point in question is $$Q(\pi R^2-\pi(R-r)^2)/(\pi R^2)$$. Certainly, if we ignore the internal cancellation of this distribution of charge, and treat it like a point charge directly on the edge of our smaller circle, we will not have decreased the radial field we are trying to get a bound on. We do this for both disk, so that we merely have to look at the radial effects from these two point charges to get a bound on the radial field at x. (While replacing the charge on the other disk with a point charge DECREASES the radial field, it does not do so more than the swap on the first disk INCREASES it [by the usual symmetry/similar triangles/distance squared type argument], so the new system does in fact have a BIGGER radial field).

The first point charge is a distance r from x; the second is a distance $$(r^2+s^2)^{(1/2)}$$ from it, at an inclination $$cos\theta=r/(r^2+s^2)^{1/2}$$. So, the radial component of the field at x, due entirely to the point charges, is:

$$(Q/4\pi\epsilon) * (R2r-r^2)/R^2 * [ 1/r^2 - r/(r^2 + s^2)^{3/2} ]$$

And the ratio of the radial to zenith field components is:

$$(R2r-r^2) * [ 1/r^2 - r/(r^2 + s^2)^{3/2} ] = (2R/r - 1) * (1 - 1/(1+(s/r)^2)^{3/2})$$

It is evident that, for any fixed r less than R, this expression converges to zero as s does; that is to say, for any compact subset of the interior of the disk, we can find a plate separation such that the ratio of the radial and zenith fields at any point in the region is arbitrarily small.

In other words, for any fixed compact subset of the interior of the disks, and for any finite tolerance, we can find a small but finite s so that the uniform distribution satisfies the boundary conditions for the original problem on the subset within the tolerance.

Since solutions to the Laplace equation (with or without forcing) are unique for mixed boundary conditions (I already have proved this using potential theory, not just asserted it as is done in the book), we may interpret this distribution as an approximate solution to the original problem.