From: "Christopher R. Carlen"
Subject: Re: What's E-field in a capacitor?
Date: Wed, 06 Nov 2002 18:12:51 -0800
Organization: Sandia National Laboratories, Albuquerque, NM USA
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John Popelish wrote:
> Since the E field is in units of volts per distance, for a parallel
> plate capacitor, the E field inside the capacitor, far from the edges
> must approach the capacitor voltage divided by the plate separation
> distance. Compare that basic definitial concept to your analysis.
Well, according to that:
E = V/d.
E = Q/(C d), and since C=p A/d then
E = Q/((p A/d) d) = Q/(p A).
Whereas I had said E=2Q/(p A) so you are wrong of course.
Heh heh. Just kidding. :-)
I knew my conclusion was wrong, but I want to understand deeply why "you
can't do it like that."
The answer that seems to be most useful, is that Gauss's Law yields the
result that E=Q/(p A). The Gauss' Law result depends on symmetry
considerations, which are that for the parallel plate cap., the bulk of
E is within the plates, and uniform over the plate area.
Once having the result that E=Q/(p A), based on the considerations of
the field structure between the plates, one cannot then go building a
case that the field from each individual plate is E=Q/(p A), and then
say that the field is the sum of the contributions from two plates.
Thus, the crux of the problem seems to be in the way in which I am
interpreting the boundary conditions for E at a conductor/dielectric
interface placed in an external E field, which also using Gauss' Law
yields E=Q/(p A).
This seems to be indicating that the field very close to a charged plate
equals E=Q/(p A). However, the boundary conditions analysis is based
on the assumption that the plate is placed in an external E-field, and
so the field conditions at the surface of the plate are structurally
equivalent to those within the parallel plate cap. Hence the consistency.
However, for a charged plate with uniform charge density, the E-field is
known to be E=Q/(2 p A) at an infinitesimal distance from the plate, or
at any distance from an infinitely large plate (two ways of saying the
same thing), from Coulomb's Law. This must be so since the charged
plate by itself has half of it's field pointing in one direction, and
the other half of its field pointing in the opposite direction.
Thus, two such plates placed parallel to one another and oppositely
charged, would produce a field that is the sum of the contributions from
the two individual plates, which would be E=Q/(p A), agreeing with what
we know from Gauss' Law, as well as from the definition of the
capacitance and E-field in terms of potential times distance.
Whew! I'm glad we have gotten *that* settled!
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA