Re: Why does electrostatic attraction in water decrease?

[Bob Sciamanda <trebor@VELOCITY.NET>]

Read Panofsky more carefully.  His example of a parallel plate capacitor
immersed in a liquid dielectric says:
1) The electrostatic force of attraction between the charged plates is
still Coulomb's law, the same as it was without the dielectric (assuming
the same charges).
2) The E field between the plates increases the mechanical pressure in the
liquid between the plates.  Outside the plates, this pressure is not
increased because here the field is zero - or much weaker, in a real case.
3) This increased inside pressure is what weakens the net force of
attraction between the plates.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "John S. Denker" <jsd@MONMOUTH.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, April 24, 2002 12:31 PM
Subject: Re: Why does electrostatic attraction in water decrease?


Pentcho Valev wrote:
> Panofsky's book ... 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."

That's obvious nonsense, for several reasons:

 1) Pressure is a scalar.  To the extent that there is a pressure, it
pushes on both sides of the charged particle equally.  No net force,
therefore no work, therefore no contribution to the energy budget.

 2) It's silly to emphasize the long-range effects of the charged
objects without discussing the equally-long-range effects of the
displacement charges within the dielectric.

> >   -- 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.

The water "in the middle"  (location yyy) tends to attract the
charges inward, increasing the energy of assembly.  The water
"behind" the charges (locations xxx and zzz) tends to attract
the water outwards, decreasing the energy of assembly.  If you
do the numbers, I bet it all works out.

        xxx     yyy     xxx
        xxx     yyy     xxx
        xxx     yyy     xxx
        xxx  +  yyy  -  xxx
        xxx     yyy     xxx
        xxx     yyy     xxx
        xxx     yyy     xxx