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.

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.

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.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)

> 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. 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.

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