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Re: messy electrostatics

Responding to this message (Fri., 13 Nov. 1998),

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 can also be calculated”. Hmm.

John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU> wrote:

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!

In a private message XXX wrote:

Such a calculation can be done (at least in principle) be relaxation.
(See e.g. E.M. Purcell, electricity and Magnetism problem 3.29 in
the first edition, 3.30 in the 2nd edition) Once the potential has be
approximated on the grid, numerical derivatives give the field
strength, and Gauss's Law will provide an approximate value for
the charge density on the surface of the conductors.

===============================================

Inspired by these replies, I “invented” my own relaxation method.
Monte Carlo experts certainly used a similar approach but why
should this take away a pleasure of pretending to be an inventor?
Students are expected to reinvent the wheel. Here is the description
of my algorithm. I do not have time to write the code now but I
plan to do this next weekend. Critical comments and suggestions
will be appreciated.

1) We know that a solution exists; charges +Q and -Q will be distributed

somehow on two insulated metallic objects. We also know that
solution is unique; otherwise the experimental data would not be
reproducible. And charges are on the surfaces, not inside the
metallic
pieces (Gauss).

2) Subdivide each Q into N equal charges dQ so that SUM[dQi]=+Q
and SUM[dQj]=-Q. Distribute discrete charges randomly. (In such
way that all dQi stay on the positive electrode while all dQj stay
on
the negative one.) This gives a possible descrete distribution of
mutually interacting 2*N charges. Calculate the electrostatic
potential energy of this distribution.

U=k*SUM[dQ(i)*dQ(j)/r(i,j)] k=Coulomb’s law factor.

The sum covers all possible pairs of subcharges dQ so that some of
the terms are positive (both dQ are “like” charges on the same
piece
of metal) and some are negative (for the “unlike” pairs on
different
pieces). The computer code remembers the last value of U, U_last.
Choose another random configuration and calculate the next value
of U, U_now. If U_now is smaller than U_last then the new
configuration is more acceptable and U_last is replaced by U_now.
Otherwise, the code ignores the change and makes another random
discrete distribution of charges. The procedure is repeated as many

times as necessary till “a more acceptable” is found.

3) The above is done over and over. It is clear that any sequence of
“more acceptable” configurations converges to a desirable “final”
configuration. If N is sufficiently large, then the “final” is a
good
approximation of the “true” (continuos) distributions of Q1 and
Q2 on metallic surfaces, provided the number of steps is very
large.
We know that the “true” distribution (on both pieces) is the one for

which the electric potential energy is minimized. And this helps
us to find where each dQi and each dQj must be.

4) The tedious part, for metallic pieces of arbitrary shape, will be
the
analytical description of surfaces. But in some cases, this can be
quite simple, as for two nearby spheres. The random part is trivial

if a reliable generator of random numbers is available. The way of
going from step n to step n+1 is irrelevant, it does not have to be

random. This offers many options for speeding up calculations
according to specific shapes of electrodes.

5) What should be done next? We need C=Q/V. The value of Q is
known, it was chosen arbitrarily, for example 1 mC, from the
beginning. To find V the code must calculate the work done, W,
when a positive probe charge dQ (also arbitrary, but dQ<<Q) is
moved from the negative electrode to the positive one, along
any path. Once this is done (using Coulomb’s law to calculate
net forces from well positioned charges dQi and dQj) the value
of V is simply W/dQ.

Ludwik Kowalski