In my previous posting, I hinted at the existence of non-conservative forces acting as a dielectric liquid is placed in an electrostatic field. But I gave no further explanations. Here is some elaboration.

Let us assume that a constant-charge parallel-plate capacitor with vertical plates is suspended over a pool of water. The magnitude of the force of attraction beween the plates is F. Then the capacitor is let down and immersed in the pool - the force of attraction is greatly reduced and becomes F/80. What can be the molecular mechanism behind that? The situation in water can be illustrated by the following picture:

+P (-)(+) (-)(+) (-)(+)........(-)(+) P-

where +P and P- are the positive and negative plates and (-)(+) are water dipoles. 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. For instance, thermal motion can force the second dipole on the left to rotate:

+P (-)(+) (+)(-) (-)(+)........(-)(+) P-

Clearly, the rotation results in a local electrostatic "push", and the sum of all such "pushes" amounts to a kind of thermal pressure acting against the force of attraction. If this pressure does exist, it would be NON-CONSERVTIVE - as the plates are drawn apart in water, the pressure will do work at the expense of heat absorbed (somewhat analogous to work a gas does on expansion).

This picture suggests another perpetuum mobile with the same molecular mechanism. Four steps:

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.

2. The capacitor is slowly let down and completely immersed. Again, work is GAINED.

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).

4. The capacitor is slowly lifted until the initial state of the system (capacitor + pool + earth) is restored. Work is SPENT.

Now if only steps 1 and 3 are taken into account, the net work gained is great. At the expense of what energy is this work done? There are two possibilities:

A) The net work gained in steps 1 and 3 is done at the expense of heat absorbed from the surroundings. Then the second law is of course violated.

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.

Pentcho