Thanks for the replies. I'm a little late in replying myself, but if
I might be permitted to summarize the idea and add a few new insights:
Recall that the experiment is two concentric metal rings standing in
a shallow tray of water and connected across 5 V at 300 Hz.
I was trying to understand why water inside the inner ring is an
equipotential by considering the electric field everywhere in space.
In particular, I had the erroneous idea that the electric field is
zero (or at least negligible) outside the water, and this led me
astray.
A better way to understand why V = const is to use Ohm's law either
in its macroscopic form delta V = IR or microscopic form E = J/sigma.
Normally I think of these equations in the following way: you set up
a certain delta V or E and this drives a resulting I or J. But just
like our F = ma discussions, sometimes it is easier to think of a as
primary and deduce F from it.
In the present case, we argue based on symmetry that I or J must be
zero inside the inner ring because there is no source/sink inside it.
Consequently the potential must be constant there.
In fact, we can argue that J everywhere in the water is tangential to
the water surfaces because any normal component has nowhere to go.
Ions cannot jump out of the water. The water thus acts like a sheet
wire. It does not confine electric field; rather it confines current.
Consequently the electric field inside the water has the same radial
symmetry as if the two rings were infinite cylinders.
This does not however imply that the electric field outside the water
is zero. Outside the water we have both J=0 and sigma=0, so Ohm's law
can't tell us anything about E.
However, we can draw some conclusions about E outside. What keeps the
current from following the fringe field lines that would exist if the
water were not present? Evidently some surface charge must arise on
the top and bottom surfaces of the water. The same occurs in a wire
as otherwise the current would not follow bends in the wire! But
E_normal = 0 inside the water. Hence E_normal = charge density /
epsilon_0 just outside it. (Darn it, why do we use sigma both for
conductivity and surface charge density?) Presumably the sign of the
surface charge density is the same as the sign of the charge on the
nearest cylinder in order to cancel the fringe fields inside the
water. Thus there will also be an E_tangential outside the water. In
fact, since the plates stick part way out of the water, I don't see
any reason that this component of E should be much weaker just
outside the water than inside. Further away outside, E will probably
follow the curving field lines you would get for two rings with no
water present.
In no sense then is E confined to the water, quite unlike B in the
iron core of a linking transformer say.
Is this a fair summary of the whole picture? Carl
--
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/