Re: mapping out equipotentials

[John Denker <jsd@MONMOUTH.COM>]

At 09:48 AM 1/31/01 -0500, Carl E. Mungan wrote:

> However, a secondary part is to measure the potential at various
> points *inside* the inner cylinder. Suppose for the sake of argument
> it is the 5 V electrode. What one finds is 5 V everywhere inside to
> within a few mV I believe.
>
> The question is: WHY DO YOU FIND THIS?

Ohm's law.  I=0 implies delta V=0

Even Ohm's law doesn't hold exactly, there is a generalized law that
guarantees delta V=0 when I=0 and there's some nonzero conductivity.

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


> Answer #1: Because the water inside is a conductor and a conductor is
> an equipotential.

If and only if it carries no current.


> Answer #2: Because the inner cylinder shields the cavity inside it.
>
> Counter-argument: A hard but soluble problem is to find the electric
> field everywhere on the plane inside a uniform ring of charge. The
> answer is only zero right at the center.

A ring isn't a shield.  Something counts as a shield if and only if it
intercepts all the field lines.  A cylinder with end-caps will do.  A
really long cylinder is a good approximation, because exponentially little
escapes out the distant ends.

> Answer #3: The water somehow changes the symmetry of the problem so
> that the ring is actually like an infinite cylinder, for which the
> electric field is zero inside.

Ohm's law again.  Resistors in parallel.  You have the smallish but nonzero
conductivity of the water in parallel with the verrrry much smaller
conductivity of the air.  Guess which one dominates.

> If you agree with this explanation, then I venture to say if I
> suspended the ring in the center of an aquarium (to change the water
> from 2D to 3D) and repeated the experiment I would no longer find an
> equipotential inside the inner cylinder, right?

Right.