Re: Brain FRY Problem # 3.1

On Oct 2, 1:34 am, "Timo A. Nieminen" <t...@xxxxxxxxxxxxxxxxx> wrote:

On Mon, 1 Oct 2007, Edward Green wrote:
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.

The parallel plate version is a standard problem (although a not-so-common
one). Your approach looked pretty much like the standard approach. I
didn't check the details, but (n-1)VQ might be correct. It's close,
anyway.

I'll have a look and see if I can find something nice about it to
send to you. There's been at least one paper in Am. J. Phys. looking at
the complexities that the simple energy argument leaves out.

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.

See below!

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".

E.D is all you need. Even simpler, you're finding the difference between
E.D with air, and E.D with the liquid. Thus, you'll get a k-1 term (i.e.,
your n-1). Equate with gravitational PE and you're done. Easy for the
parallel plate problem, a little calculus for the coax problem.

(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!

Not at all. No trap. There really is no macroscopic EM force on the liquid
at the surface.

See below.

So, what happens? With no surface force on the top surface, the pressures
immediately above and below the surface must be equal (to atmospheric
pressure). If the liquid has risen to a height h, the pressure at the
external liquid level must be rho*g*h greater than outside the capacitor
at the same level - the capacitor increases pressure!

Comment: the assumption of equal pressure across the interface
assumes a flat meniscus. Probably a detailed treatment with a
dielectric fluid will not result in a flat mensicus?

I've also thought of a simpler model: consider a slab of idealy
uniformly polarized dielectric in the gap of an ideal parallel plate
capacitor: the dielectric fills the width of the gap (plate to
plate), but not the full area.

What happens near the air/dielectric interface?

The electrostatic problem reduces to two parallel plate capacitors
*** joined at one edge, with different charge desnities (because of
the effective surface charge density of the dielectric nulls out some
of the plate charge at their interface.

Either such capcitor considered alone shows edge effect (field leaking
out or bowing near the free edge), but, since the effective charge
densities are different, these effects don't simply null out at the
boundary: the field from the higher effective plate charge density
predominates near the boundary, and E actually points partially into
the air gap.

Of course for the force on the dielectric, we need the gradient...

Now, what was my point going to be again?

Why? How? It's not obvious (published for the first time in 1973 (J. P.
Gordon, PRA 8, 14-21, 1973) that I'm aware of; if any other reader knows
of an older publication, I'm interested in knowing about it). It's all in
the grad(|E|^2) [forgive the typo above!] force. Basically, |E|^2 is
effectively a potential giving the force. Pressure is also a potential
giving a force. In hydrostatic equilibrium, they must cancel. Therefore,
there is an increase in pressure proportional to |E|^2. Also proportional
to k-1.

For the parallel plate case, that's all there is to it.

(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., ... :-/

This one is largely just to apply the pressure increase above. If you like
playing with numbers, try it. Having the dielectric creep up the centre
cylinder might well distort the field - this could be tricky. For a simple
approximation, ignore it.

This is basically why I was talking about a parallel plate capacitor
above, to avoid this difficulty.

Are you interested in having a go at deriving the pressure increase for
yourself? It's an interesting exercise you can do either microscopically
or macroscopically. If you want to do the microscopic version, just assume
you have an ideal gas - keep it simple. There's some nice stuff in
Einstein's 1905 Brownian motion paper. I have a couple of relevant papers
on arxiv (with W. Singer), but iirc most of the theoretical details aren't
in them, although the stuff about optical forces might be useful (easy to
translate to the static case, if you keep in mind that the optical force
is a time-average of the instantaneous force). Anyway, I think it makes
for a nice connection between the micro and macro classical EM theories.
It's all in Gordon if you just want the answer.


Thanks as always for your scholarly efforts.

Once again, a very textbook looking problem seems to lead into some
fairly subtle questions, some of which may not have been resolved
until recently, or remain controversial. This is not the first time
this has happened!

What was the last thing we were talking about... ah yes, the Stokes
fluid...

.

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