Re: Confused by a derivation.

[Gary Turner <turner@MORNINGSIDE.EDU>]

I remember studying this for the first time and my initial reaction.
For an isolated plate, the field is Q/(2Ae), that is easy - the charge
arranges on both sides of the plate.
When an opposite plate is brought close, the charges 'all' move to one
side, doubling the field on that side to Q/(Ae).  But the other plate also
produces Q/(Ae), and E-fields add, so the total should be 2(Q/(Ae)).  The
field is zero inside the conductors because the two fields cancel out there.

The problem is that you cannot get an isolated plate to look like one
component of a parallel pair (without an external E-field).  This is one
example of where field line diagrams with field lines associated with
specific charges helps to visualize what is happening.  The field strength
between two parallel plates is only double the value for an isolated plate
because there are twice as many field lines (originating on twice as many
charges on the one side).  There are not four times as many field lines,
because those same field lines terminate on the opposite plate.  They do
not keep going, and there is not a similar set of lines associated with the
opposite plate. The field inside the conductor is not zero 'because the
fields cancel', but 'because there are no lines there at all'.



On Mon, 4 Feb 2002 03:14:46 -0500, John S. Denker <jsd@MONMOUTH.COM> wrote:

> Ludwik Kowalski wrote:
> > I am not very happy; perhaps somebody else will come
> > with an acceptable derivation.
>
> I don't understand what there is to be unhappy about.
>
> 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.