Re: Confused by a derivation.

["RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>]

A lovely and well presented arguement!  Carl are you going to post it on
your web-page with diagrams?

I particularly like the elegant use of superposition and uniqueness of
solutions that you present in (3).  While I think intro students will be
able to follow steps (1) and (2) I'm less confident about step (3); though I
do think it is worth their while to try to follow it and to see it, if for
no other reasons then that they may finally get it when they hit
intermediate E&M.  In principle it is not beyond there knowledge base, but
the uniqueness arguement ends up being rather sophisticated for that level
of student, somewhat similar, in subtlety, to the classic problem of a
spherical cavity in uniform "spherical" distribution of charge.

Joel

> -----Original Message-----
> From: Carl E. Mungan [mailto:mungan@USNA.EDU]
> Sent: Tuesday, February 05, 2002 1:51 PM
> To: PHYS-L@lists.nau.edu
> Subject: Re: Confused by a derivation.
>
>
> Michael Edmiston wrote:
>
> > I am looking at Tipler, 4th Ed. which is
> > what I am teaching from right now.
> >
> > <snip>  But on page 708 he also says... "However, the positive
> > surface charge density on the outside surface is not
> disturbed - it is still
> > uniform - because it is shielded from the cavity by the conductor."
>
> In case anyone does not have Tipler handy, what the text is referring
> to here is a spherical shell which is overall charge neutral, but
> inside of which there is a point charge Q at an off-center location.
> The induced charge on the inside surface of the shell is -Q and is
> nonuniform, being more dense nearer the charge, and the charge on the
> outside of the shell must therefore be +Q but it *is* uniform. Why?
>
> I have the exact sentence Michael refers to highlighted with a
> notation next to it that it is a non sequitur. It simply does not
> follow and cannot be proven from anything Tipler has said up to that
> point or that a student could hope to guess.
>
> In detail, here is a correct argument for why the outside surface
> charge density is uniform:
>
> 1. Consider the following configuration A: a point charge +Q inside
> the cavity and a net charge of -Q on the shell. What will happen? All
> -Q will go to the inner surface and distribute itself (nonuniformly)
> in just such a way as to make E zero inside the metal of the shell.
> If we draw a Gaussian surface within the shell, it encloses the point
> charge and inner surface, which must add up to zero charge since E is
> zero everywhere on the Gaussian surface.
>
> 2. Consider the following configuration B: no charge in the cavity
> and a net charge +Q on the shell. What will happen? All +Q will go to
> the outer surface and distribute itself (uniformly) in just such a
> way as to make E zero inside the metal. Again, a Gaussian surface in
> the metal must enclose no charge. (Furthermore, we cannot even put -q
> and +q at different places on the inner surface, even though this
> also results in zero enclosed charge. The reason is that if we did
> this, there would have to be field lines in the cavity, and if we
> then integrated along a field line, we would find a potential
> difference between these places on the surface, contradicting the
> fact that the metal is an equipotential.)
>
> 3. Superpose configurations A and B. This is the desired
> configuration of +Q in the cavity and zero net charge on the shell.
> By uniqueness (of solutions to Laplace's equation), the resulting
> charge distribution is the only possible charge distribution. But
> this is just -Q nonuniform inside and +Q uniform outside. QED
>
> That is, if the charge on the outside were *not* uniformly spread
> over the surface, then the shell would *not* be an equipotential.
> Proof by contradiction. Throwing terms like "shielding" around is
> completely misleading, as Joel already pointed out. The student only
> has to ask: What about the -Q charge (or any other charges) at
> "infinity" (which must be there if the universe is to be charge
> neutral and on which the field lines must ultimately end), doesn't it
> mess up the "shielding"? I think this is why the inside of  a cavity
> is shielded from fields outside the cavity (in the usual Faraday cage
> arrangement), but vice-versa is not true. Comments from anyone else?
> --
> Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
>        U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
> mungan@usna.edu    http://physics.usna.edu/physics/faculty/mungan/