Re: Flat conductors (was I need help).

["John S. Denker" <jsd@MONMOUTH.COM>]

Ludwik Kowalski wrote:
...
> Can a situation be nearly electrostatic?

Sure.

> A very small current
> is always present in any electrostatic setup; right? Does it
> change distributions of E outside conductors significantly
> (with respect to an ideal distribution)? Probably not.

Saying "probably" sounds like a guess.
I _was_ going to say stop guessing, work it out.
But then I tried working it out myself.  It's a nasty
tricky little problem to work out.  So I'll just tell
you my approximate results.  First of all, to keep things
from getting unreasonably complicated, I assumed that the
resistor paper was lying a distance L above a conducting
"ground plane".  This gives me a definite capacitance per
unit area:

        rrrrrrrrrrrrrrrrrrrrrrrrr   (resistor paper)
         |
        (L)
         |
        ggggggggggggggggggggggggg   (conducting ground plane)


Then using
 -- capcitor equation (voltage related to charge)
 -- definition of field (field related to voltage)
 -- Ohm's law (current related to field)
 -- continuity (current related to charge)
I derived something that looks like
 ++ the diffusion equation, plus
 ++ terms that enforce the boundary conditions
    (keeping the charges from leaping off the paper)

Specifically, a disturbance in the charge pattern with
wavelength lambda will die out exponentially with a time
constant tau, where
        tau = (eps0 lambda^2) / (k L)
where 1/k is the resistivity of the paper (ohms per square).
For the experiment in question, this should be a pretty
short time, because eps0 is so small.
  http://physics.nist.gov/cgi-bin/cuu/Value?ep0

> Do you agree that if rho were very large, such as in sulfur
> in a vacuum, then the two silver dots separated by 10 cm
> would produce a dipole field in 3D (but not too close to a
> small silver dot)?

Huh?  There is a dipole field in any case, with or without
the resistor paper.  You just can't measure it with an ordinary
voltmeter.  You could measure it with other instruments, but
a garden-variety voltmeter just shorts out the field unless
you robustify it with the resistor paper.  In the absence of
the voltmeter, the dipole field is always there.  The resistor
paper doesn't disturb it (except near the four edges of the
paper).  The natural dipole field has no dE/dz in the plane
of symmetry, so putting the paper in that plane has no effect.

> In our carbon-impregnated paper the field seems to be
> very different from that of a dipole, it is like a 2D field
> of two long cylinders.

Pray tell, what's the difference, in the plane of symmetry?
Assume we're not near the four edges of the paper, and/or
assume the cylindrical rods are encased in a long four-sided
tubular chassis, so we have the analogous boundary conditions.

> Clearly something changes when
> rho is changed from something like 10^-15 to 0.32 ohm*m.
> Would the transition be gradual or sudden? Should one
> expect something intermediate between the 3D and the 2D
> field for some range of rho values? I think so. Why not?

Gradual.

1) There would be a gradual increase in the timescale for
charges to diffuse out to the boundaries of the paper and
set up the boundary conditions.

2) Much more importantly, there would be a gradual deterioration
in your ability to measure the field.  The disturbing effect
of the voltmeter would gradually increase.

If you use a crummy voltmeter, you will get all sorts of
goofy answers, even with ordinary resistor paper, let
alone with any higher-resistivity substance.  That's
especially true in regions far from the sources, i.e.
where the fields are relatively small.