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Re: Confused by a derivation.



I don't understand your explanations, but if you're happy, I'll quit - I
would only repeat myself anyway.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "Ludwik Kowalski" <kowalskiL@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Sunday, February 03, 2002 6:45 PM
Subject: Re: Confused by a derivation.


Bob Sciamanda wrote:

An idealized infinite plane sheet of charge produces (by itself) a
field
of sigma/2epsilon everywhere. This is true whether this charge layer
is
on a conductor, on an insulator, just isolated charge, or whatever.

Yes.

You still are insisting that if this charge layer is on a conductor
then
it, by itself, has the magical property of reducing its field to zero
on
one side, and doubling its field's value to sigma/epsilon on the
other.

No, I did not want to say so. I was referring to a situation in which
net charges on the original plate migrated to one side of the plate. The
field near that side of the plate doubles because all field lines (and
not only 1/2 of them) are now emerging from that side. The field on
the other side is zero because the local sigma there is zero.

Your b) assumption is untrue.

In part b) I wanted to say that the correct formula for C would result
if the assumption were true. The reason I am not satisfied with the
common derivation is that it solves a different problem. Conducting
plates are present but the derivation ignores them. Replace metallic
plates with plastic sheets and you have a situation in which turning
one plate with respect to another has no effect on local distribution
of sigma. Charges on conducting sheets do not behave this way.

I think I found an explanation I was looking for; I know why only
one metallic plate contributes to the net E in a narrow gap and not
two plates (see the wrong picture below). It has to do with the fact
that two plates are linked by one flux. The field lines originating
on the +1000 V plate end on the -1000 V plates. It would be wrong
to expect one flux to make more that one contribution to E.

MMMMMMMMMMMMMMMMMMMMMM__________ +1000V
MMMMMMMMMMMMMMMMMMMMMM
UPPER METALLIC PLATE (no field inside)
++++++++++++++++++++
| | | | | | | | | contributes +sigma/eps-o
v v v v v v v v v


| | | | | | | | | contributes -sigma/eps_o
v v v v v v v v v
- - - - - - - - - - - - - - - - - - - -
LOWER METALLIC PLATE (no field inside)
MMMMMMMMMMMMMMMMMMMMMM__________ - 1000 V
MMMMMMMMMMMMMMMMMMMMMM

What was wrong with this reasoning? I WAS WRONGLY EXPECTING
ONE FLUX TO MAKE TWO CONTRIBUTIONS TO E.
Ludwik Kowalski


Thanks to Bob, JohnD and JohnM. To get the correct formula
for C one can do two things:

a) Add two contributions 0.5*sigma/eps_o, as if there were
two flat non-conductive layers of sigma. That is the
approach used in most textbooks.

b) Assume that only one metallic plate contributes to the
total field equal to sigma/eps_o. There must be a way
to justify this somehow.

We know that (a) leads to a correct formula for C. Let me
speculate along the line of (b). Suppose a single plate is
connected to +1000 . The field lines starting on both
surfaces go to infinity. I bring the second plate to form a
narrow gap. By induction this floating plate is polarized
while charges on the positive plate are redistributed, as
stated by JohnD. The field in the gap is now due to three
layers of charge but these are not independent. The field
due to +sigma, like near any metallic surface is sigma/eps_o.
But the fields due to charges on the floating plate (-sigma
and +sigma) cancel. Thus only one layer actually contributes
to the total field inside the gap when one plate is floating.

I wish I could argue that the same must be true when the
second plate is no longer floating, that is when it is connected
to a negative terminal of the power supply. In this case we
have two layers on two metallic surfaces. These two layers
are not independent; they influence each other. All field lines
originating on the positive plate terminate on the negative plate.
I suspect that this can lead to a good explanation of why each
plate contributes only 1/2 of what it would contribute if it were
alone. I know it is true (because it give the correct formula for
C) but I can not explain it. Perhaps somebody will.
Ludwik Kowalski