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Re: how many volts ?




Hi Ludwik,

"Ludwik" == LUDWIK KOWALSKI <kowalskil@alpha.montclair.edu> writes:

Ludwik> On March 25, 1997 Bob Sciamnda wrote:
>> ............. The relation can obviously be turned into:
>>
>> |Q| = C*|V| , where the C_i,j are the "coefficients of
>> capacitance".
>>
Ludwik> How can this be applied to our problem?


Well... I think I may have missed too much of this thread to help (by
subscribing to too many email lists I have developed a serious itchy
trigger er, delete finger) I like to think of the C_i,j as the inverse
of the matrix that gives:

V = k G*Q (where k is the usual 9e9 Nm^2/C^2)

i.e., the 'G' is a sort of 'geometry matrix'. Most of the elements of
G can be approximated as '1/r_i,j' where r_i,j is just the distance
between the ith and jth conductors. The tricky ones are nearest
neighbors, and those can be gotten by breaking each neighbor into
little pieces, assuming a certain distribution over the little piece,
and then adding up 1/r for all the little pieces. This also works for
the diagonal elements of G. We could then break your plates up into
'chunks' (a set of N conducting chunks for the left plate, and another
set of N for the right plate). Once we have the geometry straight we
can get 'G' for the set of chunks and from that 'C'. Finally.. adjust
the values of 'V' until we get the right total charge...

Clear as mud.

;->

I usually have my E&M folks work out the charge distribution on a
square plate this way.. since I felt I learned a lot by working it out
myself (since my advisor (Mike Moloney) asked me to! Thanks Mike!)
when I was an undergraduate.

-steve