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Re: [Phys-L] amusing electrostatics exercise

I find the following exercise to be interesting without being overly

Let's start with a slightly messy real-world application, and then
simplify it. My motivation was that I wanted an explanation of how
an electrophorus works. I found a whole bunch of bogus explanations.
There may be some good explanations out there, but they seem to be
hopelessly outnumbered by bogus explanations. This version is typical:

In particular, the diagram shows an unbalanced negative charge on
the dielectric ... but it does not show the corresponding positive
charge. There must be some electric field lines that end on the
dielectric, but it is impossible to guess how those lines are
arranged because we have no idea where the counterelectrode is.

For today, let's not attack the full real-world problem. Instead,
let's do a warm-up exercise. Consider the situation in the following

There is a room with conducting walls (shown in blue). The floor
of the room is X by X. The height of the room is Y. Centered over
the middle of the floor is a thin metallic disk of radius r (shown
in red). The disk is horizontal at a height h above the floor.
There is a charge Q on the disk ... and correspondingly a charge -Q
on the walls of the room. The disk is not very large, and the room
is much wider than it is tall, so that X ≫ Y ≫ r.

Your mission, should you decide to accept it, is to find an algebraic
expression for the voltage on the disk, as a function of h. Voltage
is measured relative to the walls, which we take to be effectively
"chassis ground".

The disk has an insulating handle, so you can change h arbitrarily
without changing Q.

Make whatever reasonable simplifying assumptions you like. (Replacing
the conducting walls with non-conducting walls would not be considered

As a specific numerical instance, set
X = 10 m
Y = 2 m
r = 0.1 m
h = 1e-5 m initially

Charge up the disk to 1.5 volts using a battery, and then disconnect
the battery. Raise the disk (maintaining constant Q during phase of
the process). What happens to the voltage?

The usual jsd puzzle rules apply: Everything I have said is (to the
best of my knowledge) true and non-misleading, although some details
may be irrelevant. OTOH I have not told you everything I know. I
have not told you the answer or even the method of solution ... and
more importantly, the problem may be underspecified.

Again, the point is that after doing this warm-up exercise you should
have a fighting chance of figuring out how an electrophorus works.

Wow, that's absolutely delightful. Please, please consider writing this all
up for The Physics Teacher once all is said and done here on PHYS-L.

I'll take a stab at starting your exercise. (As usual, an advantage of the
Digest mode is I don't get biased by seeing what anyone else tries until

Initially there's -Q on the floor right below the +Q plate and the
potential difference increases linearly as V = Q*h/A*eps0 where A = pi*r^2.
Physically, this voltage increase happens because I have to do work to pull
the positively charged plate away from the negatively charged floor.

However, as the plate moves away from the floor, the negative charges there
start moving away from the region directly below the plate and start going
up the walls. Electric field lines spread out and we have an electric
dipole, with an image charge beneath the floor.

Eventually, when the plate reaches half height, h = Y/2, then V must level
off by symmetry. Thereafter, V will decrease as h increases because it gets
attracted to the greater amount of charge on the ceiling than the floor.

So one might expect something like a second-order Taylor series, resulting
in the parabola:

1 - V/V0 = (1-2h/Y)^2

where V0, the maximum potential at h=Y/2, is Q*Y/4*A*eps0. (The factor of 4
was chosen to properly match dV/dh at h=0.) -Carl

Carl E Mungan, Assoc Prof of Physics 410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363