Re: visualizing fields near charged objects

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

In this thread we had a discussion on how charges are
distributed on conductors in electrostatic equilibrium.
A misconception that the "a dominant part of the total
charges is at sharp points" was exposed. Then David
Bowman derived the formula for the capacity of a
circular disk, with respect to infinity:

C = 8*[epsilon_o*R.

He also showed that that the radial distribution of
surface density should be:

[sigma](r) = Q/(2*[pi]*R^2*sqrt(1 - (r/R)^2))

> where r ranges over the interior of the disk, i.e.
> 0 <= r <= R .  Note the singularly divergent charge
> density on the outer edge of the disk.

I decided to compare Davidís prediction with my
numerical calculations for a flat cylinder. For the
disk whose diameter is inches (R=0.1524 m) the above
formula gives C=10.79 pF. This is close to the value
quoted by JohnM.

Thus, assuming V=100 volts, we have Q=1079 pC. Using
this Q the values of David's sigma are as in column 2 below:

  r (m)    sigma (pC/m^2)  my program
--------------------------------------
0.00          7394            7673
0.02771       7647            7797
0.05542       7937            8227
0.08313       8822            9106
0.11084      11743           10826
0.13855      17751           14458
                              9716 <-- side

The last column shows my numerically calculated sigmas
for a flat cylinder of R=0.1524 m. My value of C turned
out to be 10.69 pF (instead of 10.79 pF).

Let me explain what I mean by the flat cylinder; it is not
exactly the same thing as Davidís flat disk. The cylinder
was a set of 11 cells situated in the center of my large
enclosure (|x|=|y|=150 cells). It is a projection of a 3
dimensional cylinder. I used the cylindrical Lagrangian
of JohnM (see his message at the end). The central cell
represents a circle in the center of the cylinder, the
next cell represents the first ring (r=0.02771 m), etc.

Note that unlike Davidís disk, my cylinder has a side
surface. The height of the cylinder is one cell=0.02771 m.
The charge density at the side surface is shown at the bottom
of the last column above. It turns out that 31.8% of the total
Q=1109 pC (on my cylinder at 100 volts) was located on
the side surface.

The exact match with Davidís formulas was not expected
for a cylinder whose projection has the aspect ratio of
11/1. But even this crude approximation produced very
satisfactory agreement, except when r-->, where Davidís
formula, for the sigma, is not applicable.

I am sorry that Leigh, who initially cooperated with me
on this Q(V) project, decided to take a sabbatical leave
from phys-L. I hope he will come back soon. We worked
together on this disk and helped each other to find bugs
in our independent codes. He used Excel, I used True
Basic. It turns out that the size of my enclosure could have
been reduced by the factor of two without any significant
effect on the outcome. A three times smaller enclosure,
however, produced C=11.02 pF instead of 10.69 pF,
which is significant (~3%).
Ludwik Kowalski

John Mallinckrodt wrote:

> On Fri, 9 Feb 2001, John S. Denker wrote
>
> > At 08:12 AM 2/9/01 -0500, Ludwik Kowalski wrote:
> >
> > > 1) Is it appropriate to perform simple averaging for usual
> > > cylindrical cells (dr=const)?
> >
> > No.
>
> and went on to give the essential theory.  To put it into practice
> here's what I did (and, I assume, what others are doing as well.)
>
> The key difference is the first derivative term in the cylindrical
> Laplacian which involves r itself.  This means that we need a
> radial coordinate which can be simply generated as an index along
> the side of the "universe."
>
>  27     ^ (+ r direction)
>  26     |
>  25     c
>  24   d a b -> (+z direction
>  23     e
>  22
>  21 <-(radial index)
>
> The derivatives at the position of cell a are then numerically
> implemented as follows:
>
> Laplacian (V) = (c+e-2a) + (c-e)/(2*r_a) + (b+d-2a)
>
> which corresponds respectively to the terms in John D's
>
> > ... the Laplacian is ...
> >  (d/dr)^2 + (1/r) (d/dr) + (d/dz)^2.
>
> where r_a is the radial index for cell a.  Setting the Laplacian =
> 0 and solving for a we get
>
>   a = (b+c+d+e)/4 + (c-e)/(2*r_a)
>
> The only difference from the usual Cartesian implementation is the
> last term which becomes progressively less importan (as one should
> expect) as r gets bigger.
>
> Once the calculation is finished, finding the Laplacian for those
> cells whose value was specified as a part of the boundary
> condition gives the charge density (times some appropriate
> constant) in those cells.