Re: Confused by a derivation.

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

MY DILEMMA WAS SOLVED. IT HAPPENED AS I WAS
EXPLAINING WHAT CONFUSED ME. LET ME POST
WHAT I WROTE, THE CONTENT IS NOT DIFFERENT
FROM WHAT BOB POSTED TODAY (AND WHAT JOHND
WAS SAYING IN THE MESSAGE SHOWN BELOW)
Ludwik Kowalski
**********************************************
Referring to this:

> I am not very happy; perhaps somebody else
> will come with an acceptable derivation.

JohnD, wrote:

> If somebody sees a problem, please explain. …..
> I don't understand what there is to be unhappy about.

Let me try to better (?) explain my concern. The problem is
to calculate E in the narrow gap between the identical parallel
plates, one with the net charges +Q and ?Q.

1) Here is how the problem is solved in many textbooks.
    The electric field of a single isolated plane of charge,
     on each side, is sigma/(2*eps_o); this comes from
     Gauss’s Law. We have two such planes. They reinforce
     each other in the gap and cancel each other outside the
     gap. Therefore the net E=sigma/eps_o.

2) What is wrong with the above? Nothing. Except it is
    a different problem. It is not at all obvious that the field
    due to two isolated planes of charge is the same as the
    field due to charges residing on conducting plates.

3) I agree with JohnD that a single metallic plate carrying
    a charge +Q (distributed uniformly on both sides) will
    have the field sigma/(2*eps_o) on each side. This is
    exactly the same as the field from a single isolated
    plane carrying the same total charge +Q.

    But the similarity ends when the plane carrying the net
    charge ?Q is brought to create a narrow gap. If the layers
    of sigma are immobilized then we have a superposition
    of two fields. Nothing is changed in the distributions of
    +Q and ?Q.

4) For metallic plates, however, the distributions of +Q and
    ?Q are no longer the same as they were before the gap was
    created. Instead of being distributed on both sides of
    conductive plates the charges are distributed only on
    sides facing each other.

5) Applying Gauss’s law we can show that the uniform
    field created by +Q is +sigma’/eps_o while that created
    by ?Q is ?sigma’/eps_o. By the principle of superposition
    the net field in the gap should be E’=2*sigma’/eps_o.
    How does this compare with E=sigma/eps_o due to the
    isolated layers of +Q and ?Q?

    E and E’ would be identical if sigma=2*sigma’. Is this
    justifiable? Yes, it is. That is what JohnD and Bob were
    saying. Hmm, I have just convinced myself that two
    problems have identical solutions.

THANKS TO ALL WHO RESPONDED.

John Denker wrote:

1) There is a slightly unfortunate ambiguity in the definition
of area:  If you buy some aluminum foil that is 2 feet wide
and 25 feet long, you've got 50 square feet of aluminum foil.
That's enough to cover 50 square feet of real estate.  But if
you want to paint it (both sides) you need to buy 100 square
feet worth of paint, so in this unusual sense your foil has
100 (not 50) square feet of "area" -- but when talking about
the two-sided area of thin objects you must be super-explicit
that you are referring to the two-sided area.

2) Every field line ends on a charge.

2a) For one plate of a parallel-plate capacitor, all the
field lines go out one side.  This is required by overall
charge neutrality (we're assuming the capacitor is being
operated as a 2-terminal device in accordance with
Kerchhoff's "law") so it remains true even for non-flat
capacitors (e.g. a coaxial cylindrical capacitor).

2b) For a single isolated flat plate, half the lines go
out one side and half go out the other.  This should be
obvious by symmetry for a flat plate.  It's not obvious
(and not true :-) for non-flat plates.

The maximum field (per unit charge) is 2x greater in
case (2a) than in case (2b).  I don't see a problem here.

If somebody sees a problem, please explain.