Re: Brain FRY Problem # 3.1

On Tue, 2 Oct 2007, Edward Green wrote:

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:

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?

If you want to worry about the contact angle where the liquid meets the
plates and/or surface tension, then it'll get a bit more complicated. For
the parallel plate capacitor, the contact angle is the only thing that
will make the surface non-flat, and then, in that case, the surface
tension will mean that the pressure isn't the same on both sides.

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.

Since the voltage across the dielectric filled portion is the same as the
voltage across the air filled portion, E must be the same in both cases.
This means that the effective charge density (ie actual charge density -
effective surface charge on the dielectric) must be the same in both
cases. As long as the air-dielectic interfac is normal to the plates, E
remains normal to the plates and the same everywhere. Magic, eh?

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 I'm impressed by is just how few approximations we need to make to
get a very simple solution to the parallel plate version, just the usual
macroscopic EM assumptions, assume the contact angle to be 90 degrees,
and ignore edge effects at the vertical sides.

The coax version might make a nice problem for students - calculate the
effect of the non-flat top of the liquid on the field. Rotationally
symmetric, so it's a 2D problem. I should check first that there isn't a
simple analytical solution.

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

Ah yes, I've been meaning to comment on the Stokes flow, chirality, and
drag torque. About what point are the moments taken? The point must be a
point about which the object has the necessary symmetry properties. The
unbalanced dumbell doesn't rotate due to drag torque, but due to a
gravitational torque about this point. I need to think about this a bit
more.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.