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Re: Confused by a derivation.



Bob,

We're getting close. Here is where we still have points of difference.

(1) Charge density on the inner surface of the plate.

(1a)Usually we are given the "total charge on the capacitor," call this Q.
Students must first realize this does not mean +1/2Q on one plate and -1/2Q
on the other plate, it means +Q on one plate and -Q on the other plate. Nor
does this mean the capacitor has zero charge. This is just the definition
of what we mean when we say the capacitor has total charge of Q. That
language means +Q on one plate and -Q on the other.

(1b) Say the plate dimension is 1x1x0.001 (meters) The total plate surface
area in this case is a tad bit over 2 square meters. Is the surface charge
density Q/2? No, it is Q/1 on the inside and 0 on the outside. This is not
just a definition. This is the information the student must know in
addition to Gauss' Law. I think this is what Ludwick was initially
asking... can we show this with Gauss' Law.

The bottom line for point (1) is that I strongly disagree when you say "You
do not need to know that all of the charge is on the inner surface." Unless
you are sampling the surface with a proof plane and a calibrated faraday
ice-pail (or have some other method of measuring surface charge density)
then the information you have is the total charge on the capacitor and the
dimensions of the plates. If these are indeed the data you have, then you
indeed must know that all the charge is on the inner surfaces of the plates.

On the other hand, if an angel or other trustworthy person tells you what
the charge density is on the inner surface, then I totally agree that you do
not need to know that all the charge is on the inner surface.

(2) Knowing that E is sigma/epsilon near the surface of a conductor is a
Gauss' Law result, but it is not what I would call a solution of the
capacitor problem using a direct evaluation of Gauss' Law. It is using the
result from a previously solved problem.

Here is how I would like to see a student solve for E in a capacitor using
Gauss' Law.

(2a) Draw a right cylinder with end caps parallel to the plates of the
capacitor. The radius of the circular ends is r and the length/height of
the cyclinder is h.

(2b) Place one end cap within the gap and the other end cap in the gap.

(2c) Total charge contained in the gaussian surface is pi*r^2*sigma where
sigma is Q/A of inner surface. Therefore total charge in the surface is
pi*r^2*Q/A. [Here is where the student has to know that we are talking
about Q as the total capacitor charge and A is the inner surface area or one
plate/]

(2d) Integral(E dot dA) = q(in)/(epsilon) = (pi*r^2*Q/A)/(epsilon) =
(pi*r^2*Q)/(A*epsilon)

Integral (E dot dA) = sum of (E dot A)for left cap, right cap, curved walls.

E dot A curved walls = zero [ because E parallel to wall, i.e. perpendicular
to surface vector]
E dot A for end within conductor = zero [ because E = zero inside
capacitor ]
E dot A for end in gap is not zero.

Therefore E dot A for end in gap is (r^2*Q)/(A*epsilon)

But E dot A is also EA = E*pi*r^2 [because E perpendicalar to surface, i.e.
parallel to surface vector]

Therefore E*pi*r^2 = (pi*r^2*Q)/(A*epsilon)

Therefore E = Q/(A*epsilon) (A = area of inner surface of one plate)

* * * * *

That process is the process I want students to be able to go through when I
tell them.... "Use Gauss' Law to find the electric field at..."


Michael D. Edmiston, Ph.D. Phone/voice-mail: 419-358-3270
Professor of Chemistry & Physics FAX: 419-358-3323
Chairman, Science Department E-Mail edmiston@bluffton.edu
Bluffton College
280 West College Avenue
Bluffton, OH 45817