Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: Charged disk; was electrostatic ...



I could not resist to write a little program and it confirmed
Leigh's expectation (close to 4 not 16), or according to 1/r,
as in Bob's message below. What I did was a real brut force
Monte Carlo approach. 400 electrons were randomly
distributed on the surface 500 times. Electrostatic potential
energy was calculated for each of these random configurations.
The configuration which had the lowest energy was saved and
electrons on each hemisphere were counted. The statistic
was poor but good enough to distinguish between the 4
and 16 ratios.

It seems to that minimizing the energy is good in principle
but not in practice because the minimum is very flat. In other
words nearly all random distributions have nearly the same
potential energy. Unfortunately I have no time to play with
this problem. Perhaps one could be more productive in
trying to minimize electric fields at several points inside the
pear (theoretically E should be zero at avery point but getting
this by chance is very unlikely, especially with only 500 trials).
It took about 2 hrs to make 500 trials with 400 electrons on
my 180 Mhz machine. Just think how many pairs of electrons
must be considered to calculate each energy. It would be a
little longer with calculations of E, I think.
Ludwik Kowalski

Bob Sciamanda wrote:

Separate spheres widely.
Connect them with a long wire.
Charge the system.
Remove the wire.
Now you have two spheres at the same potential and sufficiently isolated
from each other to allow a closed form calculation giving:

Q1/R1^2 = {Q2/R2^2}*{R2/R1}

IOW the ratio of the surface charge densities is the inverse of the ratio
of the radii. Check my math :)

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor