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*From*: Leigh Palmer <palmer@SFU.CA>*Date*: Thu, 8 Feb 2001 10:03:29 -0800

At 8:52 AM -0500 2/8/01, John Denker wrote:

At 04:33 PM 2/7/01 -0800, Leigh Palmer wrote in part:

I have set up a relaxation calculation in Excel based on the Laplace....

equation in cylindrical coordinates.

The charge density on the surface of the disc must now be inferred by

looking at the difference between numbers at the surface. The electric

field intensity is proportional to the surface charge density, and the

gradient of the potential field is the relevant parameter here.

One could argue that the Laplacian is even more relevant than the

gradient. It's also a whole lot easier to calculate on a spreadsheet,

since it's a scalar.

I calculated the potential itself, as you did. The Laplacian is

identically zero everywhere in free space. The surface charge

density itself is proportional to the magnitude of the gradient

of the potential just outside the surface

>When I have a converged result I can make another spreadsheet (not

iterative)

which will show the charge density on the surface. It won't be great

right on the sharp edges

The charge calculation should be just as accurate as the potential

calculation for any shape you can represent.

That is correct for smooth surfaces, but when one gets to sharp

edges (corners in two dimensions) it breaks down (unconscious

irony).

The formula for the Laplacian

that gives the charge density in this second step is essentially the same

formula used in the relaxation algorithm in the previous step, so at the

very least there should be excellent consistency. Checking to see that

overall charge is conserved is a good diagnostic. See

http://www.monmouth.com/~jsd/physics/laplace.html

for more on this.

Since I set potential boundary conditions in my technique charge

is not conserved. I don't see how you accommodated to azimuthal

symmetry in your relaxation calculation, John. It looks like you

have used the rectangular form of the Laplacian everywhere, even

on and near the axis. That will surely lead to funniness. Why do

you not use the cylindrical Laplacian? The calculation is still

two dimensional when you do so.

I see your comment on "conditional formatting", a term I'd never

seen before. I'm really creaky on Excel, having progressed from

v.1.0 -> v.3.0 -> Excel 98 without having really ever read a

manual*. I will backslide and learn about conditional formatting.

Yesterday I found a zoom function in one of the menus. I can

zoom out to 25% and see the some of the equipotentials quite

easily, directly in the page as displayed on my 17" monitor.

Printing the page makes them even clearer.

Leigh

* In our religion (orthodox Macish) we are taught that reading

manuals is sinful, thus most Mac users are illiterate, preferring

the Columbus approach (called by us physicists "suck it and see")

to using a new application.

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