Re: Vacuum - The Final Frontier

From: John C. Polasek (jpolasek_at_cfl.rr.com)
Date: 03/30/05 

Date: Tue, 29 Mar 2005 22:28:26 -0500

On Tue, 29 Mar 2005 22:57:33 +0000 (UTC), bz
<bz+sp@ch100-5.chem.lsu.edu> wrote:

>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.

CGS' constitutive eqn is D = E + 4piP. With D = Q/A staying fixed,
then adding P with a dielectric has to reduce E = V/g, that's how.
SI' constitutive eqn is D = eps0*E + P. It recognizes that D and E are
different creatures, one the forcing function (E) and D the resultant.

>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.

Not to be picky but your L and C resonate at 3.13 Mhz not 10 Mhz. Also
the L/C ratio means the "final" sees an impedance of 29 ohms. Is that
reasonable? With 29 ohms and 100 amps would you see dissipation of
29*10,000 watts inside the final stage? Enough of that.

chop
>> 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.
The voltage is restored. The dielectric would be a temporary booster,
requiring fewer volts when present to maintain D = Q/A.
>> , 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.
That's what I said and do say below.
>"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.
Then you can't read.
>> D = E* (eps0*infinity)
>> In other words it takes zero value of E to maintain D coulombs/m^2 on
D = Q/A gets a free ride on the plates which require zero voltage
drop. Metal acts as infinite dielectric constant.
>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.
OK when you start talking that way I see I have wasted a lot of time.
>> 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.
Yes, if K&R shows how to design a capacitor all you need is eps0 and
dimensions in meters. But in theoretical physics, permittivity appears
to be a hot potato, not much discussed, is all I meant to say.
>>
>>>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.
It is in the metal of the plates.
> 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?
John Polasek

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