Re: Confused by a derivation.

[Bernard Cleyet <anngeorg@PACBELL.NET>]

Do not neglect to disconnect the 1 kV source before you juxtapose the
second plate!  (Otherwise the charge on the plate will increase
enormously.)

bc  who worked energy / work / force problems with caps (with variable
gaps), dielectric sheets, and batts. 'till he was blue in the face back in
1958.

P.s. If the plates have a>> d,  how can the fields cancell (2nd plate)
except with in the plate?  I pray I've not forgotten everything since
'58/59.



Ludwik Kowalski wrote:

> 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
>
> John Mallinckrodt wrote:
>
> > On Sun, 3 Feb 2002, Ludwik Kowalski wrote:
> >
> > > The field inside the conducting plate is zero. For that reason
> > > my expectation would be:
> > >
> > >           MMMMMMMMMMMMMMMMMMMMMM__________
> > >           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__________
> > >           MMMMMMMMMMMMMMMMMMMMMM
> > >
> > > What is wrong with this reasoning?
> > > Ludwik Kowalski
> >
> > There is nothing wrong with it except, perhaps, for the word
> > "contributes" (which seems to imply that the field outside one
> > plate should be *added* to that outside the other) and, also
> > perhaps, the slightly enigmatic presence of the + and - signs.
> >
> > The bottom line any way you look at it is that the field between
> > the plates has magnitude sigma/eps_o and elsewhere is zero.  This
> > is the only solution that BOTH satisfies Gauss' law AND yields
> > zero field within the conductors.
> >
> > John Mallinckrodt      mailto:ajm@csupomona.edu
> > Cal Poly Pomona          http://www.csupomona.edu/~ajm