Re: Electrostatics problem

["John S. Denker" <jsd@MONMOUTH.COM>]

Hi Folks --

At 05:44 AM 3/1/99 +0100, Miguel wrote:
> ... conclude that no inner cavity in a conductor will
> contain a electrostatic field. But still one can see such statements
> telling the students that the 'inner' (let's call it so) surface of
> an open cylinder won't have no charges. One of our V.der Graff machines
> (is it the one from Edmunt S.?) comes with the typical guide on electrostatic
> experiments which say that.

There is a slick way to solve this problem (and a host of similar
problems).  It is both quantitative and intuitive.

The trick is to convert this seemingly electrostatic problem into an
electrodynamic problem.  Near the end of the calculation we let the
frequency tend to zero and recover the electrostatic result.

1) For simplicity, replace your circular tube with one that has a
rectangular cross section.

2) Call it a waveguide.  Solve the Maxwell equations therein, as discussed
below.

3) As a warmup, consider the mode where the field has:
  x) a standing sine-wave profile in one direction,
  y) no dependence on the second direction, and
  z) at frequencies above cutoff, a running sine wave in the long direction.

4) But we are really interested in the behavior below the cutoff frequency,
in which case the field has:
  x) a standing sine-wave profile in one direction,
  y) no dependence on the second direction, and
  z) below cutoff, a dying exponential in the long direction.

5) Couple to this mode using your van de Graaff generator.  Imagine turning
the field on and off sinusoidally at 0.1 Hz, which is below cutoff (by a
goodly margin :-) and look at the field inside the tube near the peak of
the cycle, when it is intuitively clear that there are no significant
dynamic effects.

6) Conclude that the applied field always causes a field inside the tube,
but that the magnitude of the field inside the tube falls off exponentially
as you get away from the open end(s).

==========================================

Now you may be wondering why it would be *easier* to do an electrodynamics
problem than the corresponding electrostatics problem.  I think the
following may have something to do with it:

1) We must be careful when dealing with infinite quantities.  In
electrostatics, conductors make me nervous because they have infinite
conductivity.  Therefore I have a hard time imagining the conductor being
"almost" there or "partially" there, which in turn means I can't
interpolate between the conductor-present and conductor-absent cases.  OTOH
in electrodynamics, even at arbitrarily low frequencies, the conductor
always has a little bit of inductance.  Therefore I can imagine, if I want,
having a small voltage drop across the conductor, without having things
blow up.

2) It's obvious that you can have electric fields inside a conducting tube.
 I can *see* through the tube, for crying out loud.  Therefore the alleged
no-field result can only hold "in the limit" and I want to know "in WHAT
limit".

Cheers --- jsd