Re: Electrostatic shielding

[Leigh Palmer <palmer@SFU.CA>]

> On Wed, 31 Jan 2001, Lemmerhirt, Fred wrote:
>
> >  Now when the
> >  point charge moves way off center, how do we know that only the negative
> >  charge on the inner surface redistributes itself to compensate?  I can't
> >  seem to come up  with the condition that requires the outer  positive charge
> >  to remain uniformly distributed.
>
> Aha!  I suddenly see Leigh's point.  I'm blinded by empirical behavior of
> hollow conductive objects:  when you place a charged object into a
> conductive shell, the charge "effectively" migrates to the outer surface
> of the shell, and then redistributes itself regardless of the charge
> distribution within the shell.  The field outside the shell tells us how
> much charge we've placed inside.
>
> On the other hand, since the field within the metal of the hollow shell is
> zero, there is no continuity between the flux inside and the flux outside.
>
> Hmmm.  Perhaps the zero flux within the metal of the shell is maintained
> in a manner similar to the zero g-field within a hollow spherical shell of
> mass.  It's a set of very strong forces which sum to zero.
>
> When we place a charged object inside a conductive shell, it induces an
> opposite charge on the inner surface.  The reason for this is clear.  But
> HOW does it also induce an alike charge on the outer surface, since there
> can be no communication between the inside and the outside through the
> thick layer of metal?  Yet there MUST be communication, otherwise the
> charge on the outside surface wouldn't "know" that it's supposed to stay
> there, and it would recombine with the charge on the inner surface.
> During the time when the charged object is being inserted (through a hole
> in the shell?), that's when the transient currents and the "communication"
> must take place; when the charges on the inner and outer surface come to
> an agreement of how to distribute themselves and yet maintain zero field
> within the metal.


Now we're getting somewhere! Let's continue with some fundamental
electrostatics.

The electric field at any point in space - regardless of the presence
or absence of conductors, dielectrics, eccentrics, etc. - is given by
the superposition of electric fields of all the electric charges in the
universe. That is fundamental; we can't get away from it. (Remember, we
are discussing only electrostatics here. You guys are making me very
aware of crossing my t's and dotting my j's.)

We can apply this law to Fred's question very easily. In the concentric
case, consider the electric field due to the central charge and the
inner surface charge at a point within the conductor. By symmetry we
see that *this contribution* to the total electric field is zero. *The
contribution of all the other charges in the universe* to the field at
this point must also be zero if the sum of the fields is to be zero.
("conductor" => E==0 for those who didn't understand the shorthand when
I said it before.) If the rest of the universe consists of charge
distributed on the outer spherical surface of the conductor then we
know what that distribution is: it is uniform over the surface. (This
solution is unique only so far as it specifies the relative charge
distribution - the total charge on the outer surface may have arbitrary
magnitude and sign.

Now we move the point charge away from center. As we do so, the charge
distribution on the inner surface adjusts. Surface currents bring the
total field back to the condition where there is no component of the
electric field at the surface which is parallel to the surface. (If
such a component existed, a current would flow - if I were permitted to
use that vulgar redundancy in this picky company.) During this entire
process the electric field (or lack thereof) due to the rest of the
universe has absolutely no influence on what goes on within the shell.
Gauss's law holds at all times. and E==0 within the conductor at all
times. Therefore the electric field contribution due to all interior
charges at any point within the conductor is identically zero at all
times.

It is interesting to note that this "behavior" is dependent upon the
inverse square nature of the electric field of a point charge. The
undetectability of violations of the Faraday cage effect have been
used to set very small upper limits on the possible deviation of the
exponent in Coulomb's law from -2.

Writing while watching my back is very tiring, but it is probably good
pedagogical exercise. I hope it is helpful to some.

Leigh


Problems worthy of attack
prove their worth by hitting back.

Piet Hein

(He also said "What doesn't kill me makes me stronger", probably in
Danish, but Nietsche had said the same thing earlier, in German.)