Eric Weeks wrote:
> I've been teaching about capacitors this spring to pre-med
> students so I have a response to this capacitor problem.
>
> I'm assuming the capacitor is charged to a fixed value, and then
> disconnected from the battery. In other words throughout this
> discussion Q is fixed, we have Q=CV, so if something changes
> C then likewise V changes. (If it was hooked up to a battery
> then V is fixed and Q could change as C changes.)
>
> Ignoring thermal motion, if you add in a dielectric material
> between two capacitor plates, the capacitance increases,
> thus the energy decreases, in other words work can be done. The
> reason for this is that the dielectric material polarizes, thus
> the + charges of one plate get to be very close to - charges
> within the dielectric, this is an energetically favorable system.
> So thus work could be done by lifting the dielectric material
> into the capacitor.

Eric, I think the case is obviously very difficult and we should not deal with it in a conventional way. Yes, "the energy decreases" but what does this mean? This energy has been defined on the assumption that all the acting forces are conservative, and if they are not?
>
>
> Thermal motion just means that the + charges of one plate aren't
> always close to - charges in the dielectric. So thus thermal
> motion changes the value of the dielectric constant of water,
> making it 80 rather than some much larger number.

Here we should be more specific. "Changes the value of the dielectric constant of water" may mean: A) Decreases the force of attraction between two opposite charges immersed in water so that, as we try to draw them appart, the counteracting force is 80 times weaker than in vacuum. B) Decreases the voltage between the plates so that, if we connect them with a wire, the current THROUGH THE WIRE will be 80 times smaller.
>
>
> So two statements from Pentcho that I strongly disagree with are:
>
> > If there were no thermal motion, the force of attraction
> > would slightly increase, due to polarization of water.
> >
> > So, in my view, the radical decrease in attraction can only
> > be due to thermal motion.
>
> If the dielectric is "perfectly polarizable" and has no thermal
> motion then the dielectric constant is infinite, any charge added
> to the capacitor plate is instantly compensated for by the
> opposite charge being induced on the dielectric. The only reason
> this does not happen is because of thermal motion: this is the
> opposite of what is stated in those quotes.

However water is very far from perfect polarization. Rather, the + and - poles are inseparable - as the + pole goes to the plate, it drags there the - pole as well.
>
>
> So to go through Pentcho's four steps I would start by writing
> down the two relevant equations.
>
> C = (kappa A / epsilon_0 d) relates the dielectric constant
> kappa (1 for air, 80 for water), the area of the plates A which
> is a constant, epsilon_0 which is a constant, and the separation
> between the plates d.
>
> Energy = (1/2)(Q^2/C)

Again, this is deduced on the assumption that only conservative forces are acting.
>
>
> > 1. As the capacitor is suspended over the pool (no contact with
> > water), the plates are slowly drawn together. Through a pulley,
> > the movement can be harnessed to lift a weight - work is GAINED.
>
> Yes. C increases as d decreases so the stored energy decreases,
> the stored energy can be converted into work (ie a pulley
> lifted).
>
> > 2. The capacitor is slowly let down and completely
> > immersed. Again, work is GAINED.
>
> Yes, C increases again as kappa increases.
>
> > 3. In water, the plates are slowly drawn apart until the
> > initial distance between them is restored. Work is SPENT,
> > but the work spent is 80 times smaller than the work gained
> > in step 1. (The movement is so slow that the friction is zero).
>
> Yes, C decreases so the energy stored increases, this is because
> it takes work to pull the plates apart. And I agree with the
> factor of 80.
>
> > 4. The capacitor is slowly lifted until the initial state of the
> > system (capacitor + pool + earth) is restored. Work is SPENT.
>
> C decreases again, by a factor of 80, so the stored energy
> increases and it takes a lot of work to do this, that's what the
> equations tell us. The physical cause of this is that the
> capacitor really likes having a dielectric inside, the dielectric
> allows for some + charges to be really close to - charges on the
> capacitor plate, and vice versa for the other plate.

These are forces acting horizontally whereas you move the capacitor vertically.

> To pull
> these charges apart takes a lot of work. Why does it take more
> work than when the two plates are close together? Well, with the
> dielectric in, + charges on one plate are really close to -
> charges on the dielectric, and farther from the - charges on the
> other plate. So when you remove the dielectric the + charges
> aren't happy, but they still are somewhat close to the - charges
> on the other plate. If the other plate is farther away to begin
> with, the + charges are less happy. By "happy" I mean that it's
> not energetically favorable.

If the capacitor is large enough, the field is independent of d. But the best way to resolve the problem is to look for the vertical component of the electrostatic interaction - the horizontal one is irrelevant. Let us assume that the capacitor is half-immersed - the situation close to the positive plate is someting like that:

I
I
I
I(-)(+) (-)(+) (water surface)
I
I

The vertical component of the attraction can only be due to interaction between the (-) pole of the left dipole and the region of the plate adjacent to and above it. Obviously, the number of such dipoles close to the plate, i.e. the degree of polarization, is essential. However the degree of polarization depends on the field, and, for a large capacitor, the field is independent of d. So, most probably, the "weight" of the capacitor is independent of the distance between the plates. At least this can easily be tested (I can't, unfortunately).

Pentcho

>
>
> So if the plates are close together, the + and - charges on
> the plate are almost as close as the + and - charges on the
> plate/dielectric. If the plates are far apart, there's a much
> bigger difference between having a dielectric in (and opposite
> charge very close to each plate) or having the dielectric out
> (and opposite charges very far away).
>
> > B) The net work gained in steps 1 and 3 is done at the expense
> > of net work spent in steps 2 and 4. This saves the second
> > law and implies the following. Roughly speaking, in step 2
> > the capacitor is "light" and little work is extracted from
> > letting it down, but in step 4 the capacitor is heavy and a
> > lot of work is spent for lifting it.
>
> Work is done removing the dielectric from the capacitor.
> Another way to put it is that the capacitor is attracted to
> the water, because opposite charges attract, and when the
> capacitor is in the water, the water polarizes to put those
> opposite charges next to the plate. This manifests itself as
> "the capacitor is heavy" but I would prefer to think of it as
> "lifting the capacitor against its own weight (constant) and
> against the electrostatic attraction to the polarized water".
>
> --Eric Weeks