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Re: superposition



On Wed, 15 Jan 1997, John Mallinckrodt wrote:

Well... I'm afraid this simply won't work. With the conductive paper we
effectively set up boundary value problems--e.g., V is fixed on some
painted spots, grad V is perpendicular to the edges of all painted spots,
and grad V is parallel to the edges of the paper and to the edges of any
regions in which we scrape off the conductive coating. The paper solves
the PDE's for us. But superposing the fields that result from two
different sets of boundary conditions is not at all the same thing as
solving for the fields that result when we somehow "combine" boundary
conditions. (Frankly, I don't even understand how you would propose to do
the latter.)

Actually, if you do experiments and take data not too near the edges, it
works quite well (having done it). The departure from the non-bounded case
shows up only within about 2 cm of the edge of the paper. If that were not
so, they couldn't get anyone to buy these materials for the usual
field-mapping experiment. :-)

... and later, Donald said:

How about studying a conductive pattern consisting of just one small spot
of conductive paint, and a large circle of conducting paint centered on
it, as large as the paper will allow. The field should be radial, and easy
to deal with. Now investigate the field strength as a function of radial
distance from the center spot. It is approximately 1/r. The potential goes
as ln r. But not exactly. There's charge in transit from the center
outward. At any distance, r, the charge enclosed by a Gaussian surface is
that of the center electrode plus that in the paper with the circle of
radius r. How does that affect the field and potential variation with r.

Hmm, I can believe that the experimental results might not strongly
support a 1/r field strength variation, but I don't believe that "charge
in transit" could be successfully prosecuted as the culprit. As long as
the resistivity of the paper is uniform, the current density--and
therefore the field--must drop off as 1/r, mustn't they? Am I missing
something?

You are quite perceptive. The field comes out to be a 1/r field for
combined reasons: the cylindrical symmetry which you'd have even if the
had zero conductivity, and the charge density geometry which is also due
to the cylindrical symmetry. The result is a 1/4 field dependence.

My question in the last line (which should have had a '?') was a teaser. I
tried not to give away the result, by suggesting that the result might not
agree with 1/r dependence, when in fact, it does. The danger here is that
a student might get that result, and be quite happy about it, without ever
realizing that it's due to two causes.

I could imagine the paper being slightly warmer near the inner circle due
to the larger current density. This might lead to an elevated electric
field at small r, but I doubt that it is a large effect at commonly used
currents.

And that's an interesting hypothesis. We could check whether the slight
deviation from 1/r is in the right sense as to support that hypothesis.
Maybe, in a few weeks.

-- Donald

......................................................................
Dr. Donald E. Simanek Office: 717-893-2079
Prof. of Physics Internet: dsimanek@eagle.lhup.edu
Lock Haven University, Lock Haven, PA. 17745 CIS: 73147,2166
Home page: http://www.lhup.edu/~dsimanek FAX: 717-893-2047
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