Re: Brain FRY Problem # 3.1

On Sep 27, 4:03 pm, "Timo A. Nieminen" <t...@xxxxxxxxxxxxxxxxx> wrote:
On Thu, 27 Sep 2007, Edward Green wrote:

<...>

If a fixed potential V pushes charge Q from one plate 1 to plate 2 of
a capacitor, the energy reduction associated with charge movement is
VQ. If, following introduction of a dielectric, the same potential V
can now push charge nQ, the energy reduction is now nVQ. So the extra
energy reduction is (n-1)VQ. Have I made a mistake so far?

Looks like the right path.

Really? I had grave doubts about that argument as soon as I posted
it, though I haven't had the opportunity to return to the net before.

Maybe its OK with a little more hand waving, though. Forget about
"energy stored in the field", however nice that idea sounds. Focus on
the work done by the external driving force, which is indeed VQ and
nVQ respectively: we don't even have to assume the system is
conservative. Further hand-waving involves the idea "energy loss is
good", which is a short cut for an entropic argument. It's entropy
change that drives energy transport.

"Understanding" may be looking at something the same ***-eyed way
long enough that it looks straight.

Since you don't really want a Q in your
solution, you can replace it with CV, and since you don't want a C in your
solution, you can replace it with the expression relating the geometry
(with the original dielectric constant, since your Q is the original
charge) to the capacitance.

Certainly I was hoping Q would drop out, and that we would wind up
with an expression purely in terms of V,n and the volume of
dieletric... in other words, and expression for the implied energy
density dissipated by introduction of the dielectric. But, since this
is not a parallel plate capacitor, will we run into trouble with the
actual field geometry? One might expect the liquid to be sucked up
farther along the central wire by the greater field strength.

The infinite (or long, or moderately long
length) of the entire capacitor doesn't matter, you're only dealing with a
length h.

Using parallel plates, this is done as an undergrad demo or lab. I don't
know off-hand how accurately it gives you the dielectric constant.

Four extensions to the problem:

(a) What approximations are being made?

(b) Solve the problem considering the field energy directly, without using
the capacitance or surface charge.

That was my first idea, though perhaps only implicitly evoking the
field energy. I took the attitude "we don't know nothing about
dipoles, but we know this magical stuff allows more charge to move
through the same geometry at the same potential".

V is enough to give you E(r).

(c) Since each capacitor cylinder is an equipotential, E doesn't vary
along the length of the capacitor. If we consider a small volume element
of the liquid at the top, it has an induced dipole moment. What force acts
on a dipole in an electric field? Isn't it proportional to the gradient of
the field? Since the dipole moment is also proportional to the field,
shouldn't the force be proportional to grad(E)? But E is independent of z,
so why is there an upwards force? Why does a dielectric liquid climb up a
parallel plate capacitor?

Somehow that was cunningly worded to reach just the wrong conclusion:
which I suppose is fine, because we often think in ways cunningly
designed to reach the wrong conclusion. However, that is by accident,
and you are setting up the trap delibrately!

Energetic arguments don't care about all these vulgar details, so are
preferred: it's enough that the introduction of dielectric allows a
dissipation of energy by the external driving forces.
Maybe we are in fact assuming the process is reversible, though?

If you want to know the details (well, I assume you already do),
perhaps even in the parallel plate case the fluid really creeps up the
walls, so the field at the boundary is not perpendicular to the
plates. But then this should work with a fixed slab of dielectric
drawn into the gap? There is some edge effect nonetheless: the abrupt
disappearance of dipoles at the boundary crooks the field.

(d) Following on from (c), assuming that E=E(r), neglecting surface
tension etc, what is the profile of the surface of the liquid in a
cylindrical capacitor?

Well, I already guessed that: it's drawn up the central wire. Perhaps
we already know enough to make this analytic... unfortunately, I'm
feeling rather dull these days, and haven't had the time to think
about this deeply, etc., etc., ... :-/

.