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Re: messy electrostatics



To solve the general problem of the capacitance of two conductors you
could "simply"

1) Solve Laplace's equation in the region "between the plates" subject to
the boundary conditions V1 = V/2, V2=-V/2, and V_infinity=0 where V1 and
V2 are the electrostatic potentials of the plates and V_infinity is the
potential "at infinity". (Note that the V_infinity condition is not
necessary and may, in fact, constitute overspecification in cases where
the "region between the plates" does not extend to infinity such as that
of concentric spherical conductors.

2) Determine the charge densities on the plates from Gauss' Law (i.e.,
from the field strength it produces at the surface) and integrate them
over the plates to determine the charges on the plates. Call them Q11 and
Q21.

3) Perform steps 1 and 2 again with V1 = V and V2 = 0. Call the
resulting charges Q21 and Q22.

4) Find the values of C_c and C_b (what I might call the "common mode" and
"balanced mode" capacitances by solving the equations

V = (Q11 + Q12)/(2*C_c) + (Q11 - Q12)/(2*C_b)

and

V = (Q21 + Q22)/(2*C_c) + (Q21 - Q22)/(2*C_b)

5) The conventional "capacitance" of the plates is the value of C_b, i.e.,
the ratio of the magnitude of the charge on either plate to the magnitude
of the potential difference between the two plates when the plates are
equally and oppositely charged.

Note: The need to perform *two* Laplace analyses in the above procedure
arises from the fact that the potential difference between two arbitrary
conductors depends partially on the *net* charge they carry and partially
on the *difference* in the charges that they carry. In most practical
applications--those for which the plates are close together--the
dependence on the net charge is minimal.

There are companies that market software to do these calculations in
completely general three-dimensional cases. They use sophisticated
"finite element" and/or "boundary element" methods. If you have a spare
$50k they will be happy to talk to you!

John

On Fri, 13 Nov 1998, Ludwik Kowalski wrote:

Any two separated metallic pieces form a capacitor. How to find C?
Experimentally (assuming there are no other objects nearby) this can
done, at least in principle, by connecting the pieces to a power supply
of known d.o.p. and by measuring Q on one piece. Then C=Q/V.
It is true that measuring small Q can be complicated but that is not an
issue here.

I want to know how to calculate C without doing an experiment, for
example, to predict the value of Q. The problem was approximately
solved for many simple configurations, such as parallel plates, coaxial
cables and two spheres. But how to find C when a configuration is
not simple? For example, an aluminum foil on top of a balloon and
a suspended cylinder nearby. In other words, how to calculate C for
an arbitrary, but well defined, geometric configuration of two
metallic pieces?

The analytical approach is hopeless while a numerical integration
code should be possible. I do not know start writing it. Suppose we
subdivided the surface of each piece into 1,000,000 small elements;
and view C as a parallel connection of 1,000,000 capacitors. Then
C=dC1+dC2+dC3+ etc. How do we calculate individual dCi between
arbitrary oriented elementary planes whose sizes are different? And
how do we know which two pieces “go together” (are linked by the
same field lines)? A messy problem. Can it be solved?

Some say “what can be measured it can also be calculated”. Hmm.


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A. John Mallinckrodt http://www.csupomona.edu/~ajm
Professor of Physics mailto:ajm@csupomona.edu
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