Finite dual disk capacitor: estimating charge distribution

In summary: Laplace equation.In summary, the setup has a nearly uniform charge distribution field in the zenith direction.
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anjogr
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Homework Statement



Working through Purcell (among others) as fun applied math/math modeling refresher. 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-polar axis is "nearly" uniform. In the book, like all the other books at this level, it seems, this is merely evinced by a field line diagram drawn according to a numerically estimated solution to the boundary value problem--despite the fact that the theories of boundary value problems and numerical analysis are not discussed in the text.

I don't care to get the exact distribution, or even a sharp estimate; I just want an estimate of the charge distribution/field/potential good enough to establish the claim that this setup, with finite disks at finite separation, does indeed 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 of infinite radius so that the curvature is infinitesimal.

Homework Equations



To this point, I have proved or derived every result mentioned in Purcell, chapters 1, 2, and 3. I am comfortable with euclidean differential topology, including stokes theorem for forms on manifolds. I am comfortable with the general procedures of mathematical analysis: assuming uniform continuity, or deploying perturbation theory, for instance, would be fair game. However, I expressly want to avoid using PDE theory and/or special functions--can we get a rough estimate without these?

The Attempt at a Solution



I think I might have got it since the original post. I edited things to reflect my current thinking.

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: σ/ϵ=Q/(4πϵR^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(πR^2−π(R−r)^2)/(πR^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 (r2+s2)(1/2) from it, at an inclination cosθ=r/(r^2+s^2)^(1/2). So, the radial component of the field at x, due entirely to the point charges, is:

(Q/4πϵ)∗(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.
 
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At the very least, this shows that it is not implausible that the solution to the problem is nearly uniform in the interior of the disks.
 

Related to Finite dual disk capacitor: estimating charge distribution

1. What is a finite dual disk capacitor?

A finite dual disk capacitor is a type of capacitor that consists of two parallel circular disks separated by a finite distance. It is used to store and release electrical charge.

2. How is the charge distribution estimated in a finite dual disk capacitor?

The charge distribution in a finite dual disk capacitor can be estimated by using the capacitance formula, which takes into account the geometry and distance between the two disks. It can also be estimated by using the Gauss's law and solving for the electric field between the disks.

3. What factors affect the charge distribution in a finite dual disk capacitor?

The charge distribution in a finite dual disk capacitor is affected by the distance between the two disks, the size and shape of the disks, and the material used for the disks. It is also affected by the voltage applied to the capacitor and any dielectric material between the disks.

4. How does the charge distribution in a finite dual disk capacitor differ from other types of capacitors?

The charge distribution in a finite dual disk capacitor is unique because it is limited by the finite distance between the two disks. This means that the electric field and charge distribution are not uniform like in a parallel plate capacitor, but rather vary based on the distance from the disks.

5. What are some practical applications of finite dual disk capacitors?

Finite dual disk capacitors are used in electronic circuits and devices to store and release electrical charge. They are also used in sensors, actuators, and energy storage systems. Their precise charge distribution is important in applications such as touch screens, accelerometers, and microelectromechanical systems (MEMS).

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