Re: Vacuum - The Final Frontier

From: bz (bz+sp_at_ch100-5.chem.lsu.edu)
Date: 03/29/05 

Date: Tue, 29 Mar 2005 22:57:33 +0000 (UTC)

John C. Polasek <jpolasek@cfl.rr.com> wrote in
news:fbcj41lgc84jjbetc1pobq2f7jootoaqdl@4ax.com:

> On Tue, 29 Mar 2005 18:17:29 +0000 (UTC), bz
> <bz+sp@ch100-5.chem.lsu.edu> wrote:
>
>>"Jack" <chemphysicsdude@yahoo.com> wrote in
>>news:1112050186.340517.160720 @o13g2000cwo.googlegroups.com:
>>
> snip
>>>
>>
>>I have been reading his paper "The World behind the Quantum Vacuum"
>>and I think I have seen a problem. He bases his whole idea on the
>>allegation that when you have two parallel plates forming a capacitor
>>with a dielectric between the plates and you charge the capacitor,
>>remove the charging source, remove the dielectric, the plates will be
>>left with no charge.
>>
>>[quote]
>>The plates are simply electrodes that enable full contact with the
>>dielectric during charging and discharging. Once the capacitor is
>>charged, and disconnected, the plates can be put aside (of course not in
>>a vacuum dielectric since the vacuum cannot be grasped). The protrusion
>>in Fig. 1 is figurative only, as these "excess" electrons have already
>>made their way down the wire. After disconnection, the plates are left
>>neutral and the dielectric remains under stress since there is no path
>>for reverse current. [unquote]
>>
>>I suspect that statement is demonstrably wrong.
>>
>>He cites the dis-assembleable lyden jar as proof.
>>
>>I say that, first of all, the lyden jar has different geometry and
>>second, they say 'there is no spark' when the metal parts are touched
>>together after disassembly. That does not prove that there was no charge
>>left, just that the charge was too small to show a visible spark.
>>
>>I cite David Halliday and Robert Resnick, Fundamentals of Physics. 1970
>>edition. There is a very interesting experiment described in on page 496
>>
>>"Let us use a parallel-plate capacitor, carrying a fixed charge q and
>>not connected to a battery, to provide a uniform external electric Field
>>E_0 into which we place a dielectric slab.... ...the slab, as a whole,
>>although remaining electrically neutral becomes polarized.... There is
>>no trasfer of charge over macroscopic distances.... If a dielectric is
>>placed in an electric field, induced surface charges appear which tend
>>to weaken the original field within the dielectric. This weakening of
>>the electic field reveals itself as a reduction in potential difference
>>between the plates of [the] charged isolated capacitor...."
>
> This is a puerile argument. H&R use CGS electric logic in which the
> vacuum is totally impotent and that only an electric tension exists in
> the gap. Then how does my Fig. deliver 135 amperes of RF current?

I don't see CGS or SI or any other 'logic' in their observation that the
dielectric introduction reduces the potential difference.

As for 135 amperes of RF current, circulating currents in a parallel resonant
circuit or the current through a series resonant cicuit can be high.

I don't know what 'my Fig' is, but a 1750 PF capacitor will resonante at 10
MHz when connected to a 1.447e-6 Henry coil. If we have a high Quality (Q~
900) circuit, with a total DC resistance of oh, say .01 ohm (silver
conductors, wide ones could do much better) then 135 amps would produce
I^2 R, heating or 182 watts of heat.

Almost twice as much heat as a 100 watt light bulb would produce.

X_c and X_L would each be about 9 ohms.

In parallel resonance, the circuit would appear like a very high impedence at
10 MHz. A high voltage would appear across the circuit, about 900 times the
voltage across the resistive element which would drop 1.35 volts, so you
would have about 1200 volts across the circuit at the current you give.

What frequency did you use and what inductance and what are the resistances
of the circuit elements?

In any case, it is not difficult with a low power transmitter to get 182
watts at 10 MHz. Your capacitor spec. doesn't list the internal series
resistance (power factor is another way the series resistance is reported),
but I would guess it it much less than 0.01 ohms so we can probably get the
current you cite with much lower power.

> Yes the PD of a charged vacuum capacitor will be reduced upon

PD, potiential difference?

> introduction of a dielectric by the value K >1, where pemittivity of
> vacuum = 1
> V = g*D/1*eps0 or V_k = g*D/k*eps0.
> With H&R there is no eps0 nor D = 1*eps0*V/g. eps0 is not accounted as
> 1 or any value; it's nonexistent in their view.

K&R has it as 8.85418e-12 coul^2/nt-m^2 which is the same as 8.8... Farad/m
and is very close to the 8.854187817e-12 Farad/m that mathcad lists.

>
>>I contend that by symetrical considerations, removing the dielectric
>>from a charged capacitor would seem to be unlikely to leave the
>>electrodes in an uncharged state because that would require that when an
>>uncharged dielectric is place between the plates of an isolated
>>capacitor, the charge on the plates (and the voltage) would drop to
>>zero.

> If you know anything of field theory and conservation of charge try
> this:

If I am going to try it, why would I need to know anything about field theory
and conservation of charge?

assuming a switch has just been closed....
> 1st, at the battery, electrons issue as particles,

somewhat simplified, but ok. Very high current [can] flow at first. Limited
by the total dc resistance of the wires and internal resistance of the
battery and capacitor. This current would continue to flow if there was an
internal short circuit in the capacitor.

> then along the wire
> as current that time-integrates to charge

somewhat simplified, but ok. I would say that the electron flow into the
capacitor decreases as the plate charges until the potential difference
between the plates equals the voltage output of the battery. At this point,
the switch may be opened. The plates will continue to display a voltage
difference equal to the battery voltage until the charge is removed. Note
cosmic rays will eventually discharge the capacitor. Any leakage resistance
path between the plates will discharge the capacitor. Inserting a dielectric
will partially discharge the capacitor while storing energy in the
diaelectric. I am unsure what happens under these conditions if the
dielectric is again removed.

> , and once on the plates the
> charge is represented as the areal displacement charge
> D = Q/A coulombs/m^2 we know Q and we know A
> For continuity, D has to carry right across the vacuum gap which I
> show can be accomplished with charge density rho, by integrating
> Maxwell's divergence equation (Eq. 4 op cit) as follows:
> D = rho*delta
> where delta is a minute polarization displacement of the pairs so.
> D = E*eps0
> But this is also true inside the thickness of the plates:

There is no electric field within the thickness of a conductor.
"The electric field inside a conductor carrying a static charge is zero." K&R
p492
Also, see how the Van de Graf generator works, charges deposited inside a
conductive sphere go to the outside of the sphere.
There is zero, zip, nada, no, none, nil static charge _within_ a conductive
body, including the capacitor plates.

If your theory depends on charge within the conductor, it has problems.

> D = E* (eps0*infinity)
> In other words it takes zero value of E to maintain D coulombs/m^2 on

Wrong. It takes zero current flow into the capacitor to maintain the charge.
The voltage difference between the plates is observable. If your theory
depends on the E field going away when the battery is disconnected, it has
problems.

> the plates. There is no energy invested in the plates as the product
> of D times (E = ) zero is zero.

There is energy stored as an electrostatic field between the plates, due to
the imbalance in charges between the two plates. One is deficient in
electrons, the other has a surplus of electrons.

> All the energy is in the gap, in the vacuum and it consists of
> mechanical energy of electrons and positrons spring loaded out of
> their normal positions by the value delta.

I disagree. All the energy is in the gap, in the vacuum and it consists of an
electric field between the two plates as one plate has an excess of electrons
and the other has a deficency.

> The electric energy .5 E*D
> can be shown to be the same as .5K*delta^2 K defined in Eq. 11 as
> 2.6e14N/m.

That doesn't mean that there is any relationship to reality.

>
> CGS electrics as exemplified by H&R is parlor physics where D = E. To
> be sure, you will look long and hard to find any mention of vacuum
> capacitors or permittivity in at least the old texts I have to hand.

Vacuum capacitors have been known for as long as there have been vacuum
tubes.
K&$ has problems involving electrostatic deflected CRT tubes. They calculate
the beam deflection due to the electric field. p441&442.

K&R index: Permitivity constant, 426, 560 and speed of light 645.

>
>>In any case, it should be an easy experment to perform.
>>
>>I contend that if removal of the dielectric doesn't leave the plates
>>entirely uncharged THEN the idea that the capacitor stores its energy
>>ENTIRELY within the space between the plates is false.
>
> The vacuum maintains D throughout the gap by virtue of field E and the
> plates maintain the same value D using a field intensity of zero.

The field intensity is NOT zero. A test charge introduced between the plates
shows that. A volt meter measuring the EMF from one plate to the other will
also show that.

> Metals can be accounted as having infinite permittivity. It's dD/dE.

that would make the force between charges in a metal zero.
The metal would fall apart. F=(1/(eta0 4 pi))(q1 q2)/d^2

>
>>That would imply that Coulomb and Faraday were right and Polasek is
>>wrong.
>>
>>Does anyone have an electrometer, a voltage source, a way to get signal
>>in and out of a good vacuum, something to use for a dielectric, and a
>>way to remove the dielectric from between the plates without breaking
>>the vacuum? 

-- 
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an 
infinite set.
bz+sp@ch100-5.chem.lsu.edu   remove ch100-5 to avoid spam trap

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