Re: Flat conductors (was I need help).

[Brian Whatcott <inet@INTELLISYS.NET>]

I'll take a shot at responding - interspersing my
responses with your text.

At 04:39 PM 2/26/02, you wrote:
> 1) Correction: the current at 300 volts was 11 mA, not 5.5 mA,
> as posted about two hours ago. This is probably irrelevant, as
> far our problem is concerned. But it shows that the new sheet
> has exactly the same R as the one I used before. The sheet is
> very linear (in terms of the mA versus volts) up to at least
> 300 V.


I am uncomfortable at dissipating 3 watts over 30 cm2 or so.


> 2) There is one question I keep asking but nobody answers.
>
> In air, or on an acetate sheet, two small silver dots separated
> by a distance much larger than their diameters, would produce
> a field of a 3D dipole (except very close to circles). The electric
> field lines would not be confined to a single 2D plane because
> all meridian planes are identical. An electric current, of the
> order of pA would be flowing, more or less along the E lines
> in the 3D space.
>
> If the resistance of the acetate sheet were progressively lowered
> then more and more current would flow through the sheet and
> less and less above and below it. Eventually we would have a
> flat conductor carrying practically all the electric current. The
> distribution of E is no longer like for a 3D dipole, it is like
> that of a 2D dipole (two infinitely long cylinders perpendicular
> to the 2D sheet). It looks like if the electric current, in process
> of changing its path, also changed the configuration of E. The
> current is saying to the field "I become 2D and you must do
> the same." Why is it so?


Where the current is planar the electric voltage distribution in the
plane due to that current is predictable.
But in the surrounding insulator, there is still a field
("field" = force available to remotely move a charged object)

In this space, the field is not like the bipolar field due to two
isolated charged spots, and not like the field due to two charged
cylinders in 3 space. That's because in this air space, a charged
particle would see a  planar potential gradient spreading from
one spot and converging on another which goes with a different
3D field distribution than any of the ones already depicted.

> The only way I can rationalize this is to assume that static
> surface charges associated with the steady state flow are
> responsible for what really happens. How else can this be
> explained? Static charges, by the way, were discussed here
> recently in a thread whose name was "Chabay/Sherwood."
> Ludwik Kowalskil


Static charges associated with a charge flow seems to me to
 be an unhelpful or  misleading mind picture in this case.
       You can measure point voltages, both on the paper and
(at least conceptually) in the air to derive a picture of electric fields.



Brian Whatcott
  Altus OK                      Eureka!