Re: Why does electrostatic attraction in water decrease?

[Pentcho Valev <pvalev@BAS.BG>]

"John S. Denker" wrote:

> Pentcho Valev wrote:
> > Perhaps [conductivity] is not the essential problem.
>
> I agree.
>
> > to define the energy in an electric field, one considers a
> > set of charges as they are moved, isothermally, from infinity to their
> > places in the system.
>
> I wouldn't "define" the energy that way, but you can
> _measure_ the energy that way, subject to restrictions
> discussed below.
>
> > If this assembling is infinitely slow (to avoid
> > friction), the work of assembly is identified with the energy of the
> > system.
>
> OK, keeping in mind that the energy is "in" the fields
> not "in" the electrons.
>
> > An essential condition is that all of the forces are CONSERVATIVE
>
> That means we must neglect magnetic effects;  this is
> one of the restrictions alluded to above.
>
> > - if, for instance, two opposite charges are drawn apart at some stage,
> > the energy of the system must increase at the expense of the work you
> > spend but NOT at the expense of heat absorbed from the surroundings.
>
> OK.  But making a fuss about thermal effects is probably not
> the essential issue either.  One could take a microscopic
> view and include the work (defined as F dot dx) done by the
> environment, including it on the same footing as the macroscopic
> forces.
>
> > This
> > IS the case when you draw two opposite charges apart in a vacuum, but does
> > NOT seem to be the case when you draw them apart in a dielectric liquid
> > (e.g. water).
>
> Really?  What's the evidence of that?

No direct evidence I must admit.

> > In the latter case, conductivity is irrelevant - the
> > essential factor is the increased liquid pressure between the charges that
> > pushes them apart.
>
> Really?  What's the evidence of that?

You have not seen Panofsky's book I quoted in one of my previous postings. It is
written on pp. 115-116: "Thus the decrease in force that is experienced between
two charges when they are immersed in a dielectric liquid can be understood only
by considering the effect of the pressure of the liquid on the charges
themselves. In accordance with the philosophy of the action-at-a-distance
theory, no change in the purely electrical interaction between the charges takes
place."

> The effect of a dielectric can't be summarized as a pressure
> in any reasonable way;  consider this:
>   -- Scenario 1:  I bring a charge to point A and another charge
> to point B (bringing them in from infinity as described above)
> in the absence of any dielectric.
>   -- Scenario 2:  Same, except that there is a slab of _solid_
> dielectric material in some of the space between A and B.
>
> There's no pressure on the charges in either scenario, but
> the energies and forces are quite different.

Yes, but the case has nothing to do with the one in which the charges move in a
liquid dielectric. If, for instance, they are opposite, in your Scenario 2 you
will obtain some more work than in your Scenario 1, due to the polarization of
the solid dielectric. If, in Scenario 3, the charges go the same route but are
in water all along, you will obtain 80 times less work than in scenario 1.

Pentcho