Stan Meyers Secret, Preventing Electrolysis.

[SeaMonkey]

HMS, I feel you're making this a little more complicated than it is. Naudin's V1.0 set up without an insulated electrode provides the required charge exchange medium (the electrodes) so normal electrolysis occurs. Put simply the ions have some where to drop off and pick up charges to so become atoms of hydrogen and oxygen. Pulsing may make it slightly more efficient, but it's still normal electrolysis governed by Faraday's Laws.

To me, the mystery is how he got any gas evolved from the V1.1... that is if we are to trust what he says, and without a video of the V1.1 in action, I seriously have my doubts about it... and him.

If you are so sure about insulating an electrode, why dont you simply put aluminium foil around a pyrex container or a test tube to act as one electrode, and a ss tube in the water within the container as the other. After all, if you are completely insulating an electrode, it no longer needs to be as inert as SS or platinum, or nickel. You can use copper if you want, though aluminium foil is much the easier to mould.

If for some reason you feel the glass will be too thick, though here's an idea. Seal a sheet of aluminium foil between two sheets of clingfilm. Or if you want to make it more rigid, use see-through self-adhesive film... or even a laminator.

AC (RF) 'electrolysis' is definitely a 'doable' technique.

Kanzius demonstrated the process at 13.56 MHz but
whether anyone associated with him has continued
research to optimize the efficiency at some unknown
frequency is a mystery.

The experimenter's (inventor's) group over in Vancouver
BC claims to have re-discovered the secret to Meyer's
success. See Here


When coupling energy into a water solution capacitively
with insulated plates, the frequency which becomes
effective is likely to be very high. Puharich had success
with an amplitude modulated waveform of over 100 KHz
carrier frequency. Radio Frequency devices will be a
necessity.

[lamare]

There are two half reactions. One of them "eating" electrons, the other delivering electrons. So, in essence all it takes to form a complete reactions is a way to move the electron from where it is freed to where it is used in the other half reaction.

Normally, you would do that using an electrical wire, so to speak. However, there also is an effect known as "dielectric breakdown". This means that when a sufficiently strong electric field is applied to a dielectric, it stops being an insulator and becomes an inductor: Electrical breakdown - Wikipedia, the free encyclopedia

For water, this happens when field strengths in the order of 30 MV/m, or 30 kV/mm, or 3 kV/ 0.1mm are reached. ( Dielectric strength - Wikipedia, the free encyclopedia)

I should add that the required field strength for dielectric breakdown to occur in water does not necessarily have to be a static field strength. Given the fact that the electric field propagates at the speed of light, one can at least theoretically reach the required field strength with lower voltages, provided one switches fast enough. In other words, one can create a elecrical shockwave where at the wave front you have a very steep change in the field strength. If that would be the case, I would expect that Bedini kind of pulses are more effective than harmonic (sine wave) oscillators, because of the rapid switching provided by the sudden interruption of a current trough a coil.

I have done some calculations to estimate what kind of voltage rise times you would need if you want to create an electrostatic shockwave strong enough to reach the dielectric breakdown fieldstrength for water.

If we take B for the dielectric breakdown value and d for the distance across which this field would be present, then we would write: U = B . d

Since the traveled distance for the shockwave s = v . t = c . t, with c the speed of light, we can write:
U/t = (B . d) / t = B . c

So, we would need a voltage rise rate of about 30 MV/m times c, or 3*10^7 * 3*10^8 = 9*10^15. Or: 9 MV/ns. That's right, 9 MegaVolt per nano second. And that is not easy to do!

It may be possible using a spark gap, though. See: http://www.ece.unm.edu/summa/notes/SwN/SwN28.pdf - Fundamental physical considerations for ultrafast spark gap switching.

For the calculated configuration (E= 100 MV/m , load impedance 50 Ohm) the maximum rate of voltage rise is 6 * 10^15 V/s, which is close, so with careful design this may be achievable. But, as I said, certainly not easy to do.

FYI: I have continued to work on my article (Article:The Electret Effect - PESWiki). The pieces of the puzzle start to fall on their places. I went on a little sidetrack, since I suddenly understood how Edwin Gray's system works, so the actual electret / electrolysis stuff will have to wait..

[Farrah Day]

This is all very interesting as it would seems we all have very different viewpoints and indeed different experience in this matter.

Sea Monkey, what John Kanzius did was definitely not electrolysis. The very word electrolysis infers electrodes - no electrodes were employed in Kanzius's discovery. And of course it has yet to be established just exactly what was occuring, though I'm tending to think along the line of it being a sodium burn rather than hydrogen - remember it was a saline solution. Professor Rustum Roy who took over the research on this from Kanzius told me it was actually a very simply reaction occuring, but would not elaborate. That was over a year ago and I've heard nothing since. I assumed that he intended to disclose the full details in a paper, but he is yet to update his web site or to the best of my knowledge, provide any further details on this phenomenom.

HMS, two electrodes in water with voltages applied across them, current flowing through the cct and gases being evolved - how could you not think this is electrolysis??

Farrah, you reminded me....I forgot to mention, and think this was mentioned earlier. If the dielectric material is too thick the circuit will not split water. The dielectric coating must be very thin......
Yesterday 11:42 PM

Why? What difference doe the thickness of the dielectric make, given that this is not what you want to break down? The fact that you intend to insulate an electrode will always mean that the impedance of the cell will be far greater than that of the cct... will it not?

Much of this simply does not make sense. Sure you will end up with a great water capacitor, but that is all. As I see it, you might as well be using a Leyden Jar.

Naudin's 'smaller than normal electrolysis bubbles' is I believe a non-issue as what other electrolyser is he comparing it to? It's all relative. Design and efficiency play a major part. A low voltage, low current electrolyser will see bubbles forming slowly and much larger than what you get from cells such as Naudin's, Meyer's and indeed mine - which I might add, gas bubble-wise looks visibly identical.

What I find rather worrying concerning Naudin's WFC, is that at the very top of his web site page, he depicts Meyer's version of how voltage splits the water molecule.

It's utter nonsense. Meyer does not even consider ions, he shows the water molecule being ripped apart completely, with the electrons being pulled off!! Where are the ions??

This is garbage on so many levels, but for some reason people are willing to accept this flawed science without question - without giving it a second thought!

We know that the water molecule, being a polar molecule will orient itself within an electric field, but we also know that the weak link will be one of the O-H bonds, and indeed they always break first so we get ions, H+ and OH-.

What would happen if you could pull the electrons off the water molecule as depicted by Meyer? You would be left with ions of H and O - no good, we need atoms. And what would become of the electrons?

It seems too many people trust implicitly what Meyer said, taking him at his word without for a moment considering the science. Whether this is because they are not educated enough to see the flaws, or are simply too lazy to research the science - or both - I don't know, but to put all your faith in what Meyer stated would be foolhardy in the extreme. People still quote various pieces from his Technical Brief, which is so full of flawed science and gibberish that it makes Harry Potter magic look more plausible.

Forgetting for a moment the conspiracy theorists, given that Meyer was ultimately exposed as a fraud, and we are indeed yet to see anyone successfully replicate his WFC and have it independently verified, we must surely, at the very least, err on the side of caution and scepticism.

Rather than taking Meyer at his word - and given what we know and don't know - it would be much safer and far more logical to assume Meyer to be a charlatan and a fraud, then be pleasantly surprised to later find out through your own research that he wasn't. This way you will be on much safer footing and not continually fall foul of psuedoscience.

I also feel that it would be foolish to simply try to replicate Meyer's WFC considering the all doubt about it ever working in the first place. Instead, it is far more logical to set up your own experiments in order to prove or disprove your own theories of operation, and take things from there. You don't for example need to fabricate or use a VIC, if you already have a EHT supply, some capacitors and a PWM. Improvise. If you know the result you are trying or hoping to achieve, forget Meyer, and work things through in your own way. That indeed is what I do.

Lamare, HMS, have you seen this: Water atomisation by high magnitude electrical impulses: A study.

[lamare]

What difference doe the thickness of the dielectric make, given that this is not what you want to break down? The fact that you intend to insulate an electrode will always mean that the impedance of the cell will be far greater than that of the cct... will it not?

Much of this simply does not make sense. Sure you will end up with a great water capacitor, but that is all. As I see it, you might as well be using a Leyden Jar.

Maybe you should watch the video posted here about a Leyden Jar being dissected:
http://www.energeticforum.com/renewa...html#post74102

Now this is really interesting. Where is the "charge" stored?
What would happen if you would fill the dissected - "charged"! - Jar dielectric with water?

You know, this practical demonstration goes totally against the textbook:

Chapter 19. Electrical Properties

Capacitors require dielectrics of high e that can function at high frequencies (small relaxation times). Many of the ceramics have these properties, like mica, glass, and porcelain). Polymers usually have lower e.

So according to the textbook, glass has a small relaxation time. So, according to the book, as soon as you remove the electric field that caused the polarization -- remove the capacitor plates -- the polarization should be gone in no time.

So, why oh why says the experiment otherwise?

That's because almost nobody has a clue about how this stuff really works, but it's beginning to make sense:

Article:The Electret Effect - PESWiki

There is an essential difference between the Newtonian analogy we use in electrical engineering (closed circuits) and the actual reality. The analogy of a capacitor in hydraulics (Newtonian analogy) is a piston moving back and forth in a closed cylinder wherein gas is pressurized. There is an essential difference, however. Imagine moving the piston inwards, pressurizing the gas, and put the thing on your workbench. The piston will immediately move back, because of the gas pressure. Now charge a capacitor and put it on your workbench. See the difference? The capacitor will just sit there, keeping it's charge. In other words: our hydraulic analogy is unstable, it 'wants' to release it's energy, while our actual electrical component is stable when 'pressurized'. It will only 'release' it's energy when something external is being done. It has to be disturbed, because the charges in a capacitor actually attract one another, which makes them like to stay where they are. So, when 'discharging' a capacitor, as a matter of fact, these attraction forces have to be overcome. And that does not release energy at all, it costs energy to do that. So, it actually takes the same amount of energy to charge a capacitor as the amount of energy it takes to discharge the capacitor. It is undoubtedly because of this that Steinmetz wrote, already in the beginning of the twentieth century:

"Unfortunately, to large extent in dealing with dielectric fields the prehistoric conception of the electrostatic charge (electron) on the conductor still exists, and by its use destroys the analogy between the two components of the electric field, the magnetic and the dielectric, and makes the consideration of dielectric fields unnecessarily complicated. There is obviously no more sense in thinking of the capacity current as current which charges the conductor with a quantity of electricity, than there is of speaking of the inductance voltage as charging the conductor with a quantity of magnetism. But the latter conception, together with the notion of a quantity of magnetism, etc., has vanished since Faraday's representation of the magnetic field by lines of force."

So, it may seem that the conservation law holds when considering electrical circuits in their 'prehistoric' analogy, in actual truth this is only the case because the interaction with the environment, the active vacuum, balance one another out. In reality twice the amount of work has been done than seems to having been done!

And you know what?
The charges (polar molecules) inside a dielectric, once polarized, also attract one another! This means that the "charge" will just sit there as long as you don't disturb the material. Once again, it seems like "discharging" the capacitor delivers energy, while in reality it costs energy. And that energy comes from the vacuum.

[HMS-776]

Farrah,

Thank you for that last post. Lots of good information.
Reading the dublin paper I definitely understand
what you are saying. Many different opinions here.

I am much more oriented with the electronics side of
things so I guess it's time to study more chemistry to better understand
the process regarding the ions.

[Farrah Day]

Hi Lamare, firstly I thought that dielectric relaxation was a measure of how long polar species took to orient themselve to an electric field, not how long they took to become disorient.

This is a very interesting demonstration by the MIT physics department. This gives major clues to the actual workings of capacitance.


make sure you watch the entire way through, the first half is nothing special.

YouTube - MIT Physics Demo -- Dissectible Capacitor

Quote:
Next we have a three-piece dissectible Leyden Jar consisting of two metal cups separated by a glass cup. When charged with the Wimshurst machine, we see by touching it with the shorting rod that it holds a large amount of charge. However, when disassembled, the metal cups can be brought into contact with each other and no spark will be generated. When the jar is reassembled it can then be discharged. This demonstrates that, in this situation, the charge actually resides on the surface of the glass (a dielectric), not on the metal.

Contrary to what the poster quotes, I personally don't believe this demonstrates that the charge actually resides on the surface of the glass, it simply shows that the molecules within the glass remain polarised until imposed upon by another electric field or fields.

As I mentioned before that is a great video demonstration, but even this can be misinterpretated.

One thing that is a bit of a misnomer is the charging of a capacitor. A capacitor does not actually charge up as such. Think about it logically, if you assume that one plate of a capacitor is charging up with electrons, what is happening to the other plate? It is of course discharging of electrons.

So what is it? Is the capacitor charging or discharging? Of course it is doing both, so clearly you do not charge up a capacitor, you simply create a situation whereby the charges separate.

One thing for sure, the science is never quite as simple as it may at first seem!

[lamare]

Hi Farrah,

As I mentioned before that is a great video demonstration, but even this can be misinterpretated.

One thing that is a bit of a misnomer is the charging of a capacitor. A capacitor does not actually charge up as such. Think about it logically, if you assume that one plate of a capacitor is charging up with electrons, what is happening to the other plate? It is of course discharging of electrons.

So what is it? Is the capacitor charging or discharging? Of course it is doing both, so clearly you do not charge up a capacitor, you simply create a situation whereby the charges separate.

One thing for sure, the science is never quite as simple as it may at first seem!

You're definiately right about that!

I just noticed something very peculiar. I looked at Baudin's "charge stepping" effect and suddenly the figure Stan has in his memo's came to mind. Now there's an interesting difference. In Stans figure we see a complete discharge, where in Baudin's we only see charge building up. Of course, this could be an error in Stans figures, but I also looked up his picture where he has the outside of the capacitor covered with Delrin.

Since the whole capacitor is submerged in the water and we may assume he was using water with a considerable amount of ions, in other words: conducting water, you have a very interesting capacitor design, because on the outside of the outer tube you have the basically the same thing as in an electrolytic capacitor. So, it could be that the dielectric at the outside gets polarized, especially since the inner capacitor is charged with high voltages, and because of the relaxation time of the dielectric, this might cause interesting imbalance effects inside the capacitor.

I don't know, but it is an intriguing tought, because Stans "step charging" figure suggests something strengthens the charging process. The pulses get stronger while the cap seems to fully discharge. This hard to explain if that would be caused by polarization of the water. That's a fluid, which is depolarized before you can even think of blinking an eye. So, me thinks somehow having a dielectric layer on the outside of the outer tube, which is the positive in Stans system IIRC, does make a difference if this whole construct is submerged in the water.

[SeaMonkey]

This is all very interesting as it would seems we all have very different viewpoints and indeed different experience in this matter.

Sea Monkey, what John Kanzius did was definitely not electrolysis. The very word electrolysis infers electrodes - no electrodes were employed in Kanzius' discovery. And of course it has yet to be established just exactly what was occurring, though I'm tending to think along the line of it being a sodium burn rather than hydrogen - remember it was a saline solution. Professor Rustum Roy who took over the research on this from Kanzius told me it was actually a very simply reaction occurring, but would not elaborate. That was over a year ago and I've heard nothing since. I assumed that he intended to disclose the full details in a paper, but he is yet to update his web site or to the best of my knowledge, provide any further details on this phenomenon.

[/url]

Yes, electrolysis as we understand it conventionally
does require electrodes in order that the oxidation
and reduction reactions may take place at the
surfaces.

Within an intense alternating electromagnetic field
there will be 'eddy currents' within the conductive
electrolyte solution and other ionic "resonance"
effects and "standing wave" possibilities.

Current flow within a conductive medium may be
stimulated either by direct connection (electrodes)
or by electromagnetic induction.

"Water Electrodes?"

The proper frequency (or spectrum of frequencies)
to most effectively dis-associate the gaseous atoms
is yet to be fully "revealed."

Puharich, Boyce and Kanzius have given us clues!

We should also recognize that the magnitude of
potential involved need not be high.

[lamare]

I don't know, but it is an intriguing tought, because Stans "step charging" figure suggests something strengthens the charging process. The pulses get stronger while the cap seems to fully discharge. This hard to explain if that would be caused by polarization of the water. That's a fluid, which is depolarized before you can even think of blinking an eye. So, me thinks somehow having a dielectric layer on the outside of the outer tube, which is the positive in Stans system IIRC, does make a difference if this whole construct is submerged in the water.

This begins to look promising. Puharich says the same thing:

http://www.puharich.nl/Files/Patent.pdf

"There is an `Open Circuit` reversible threshold effect that occurs in Component III due to water polarization effects that lead to half wave rectification and the appearance of positive unipolar pulses; and There are electrode polarization effects in Component II which are a prelude to true electrolysis of water as evidenced by oxygen and hydrogen gas bubble formation.

A principal effect that occurs in Stage B, Phase 3, in Component II, the thermodynamic device, is that the two electrodes undergo stages of polarization. It has been observed in extensive experiments with different kinds of fluids in the cell of Component II , i.e., distilled water, sea water, tap water, Ringers solution, dilute suspensions of animal and human blood cells, that the inner surface of the outer ring electrode at 3' in FIG. 3 (the electrode that is in contact with the fluid) becomes negatively charged.

What is this "Component II"?

The thermodynamic device is fabricated of metals and ceramic in the geometric form of coaxial cylinder made up of a centered hollow tubular electrode which is surrounded by a larger tubular steel cylinder, said two electrodes comprising the coaxial electrode system which forms the load of the output of the electrical function generator, Component I. Said center hollow tubular electrode carries water, and is separated from the outer cylindrical electrode by a porous ceramic vitreous material. Between the outer surface of the insulating ceramic vitreous material, and the inner surface of the outer cylindrical electrode exists a space to contain the water to be electrolysed. This water cell space comprises the third component (Component III) of the invention. It contains two lengths of tubular pyrex glass, shown in FIGS. 2 and 3. The metal electrode surfaces of the two electrodes which are in contact with the water are coated with a nickel alloy.

[lamare]

Hi Lamare, firstly I thought that dielectric relaxation was a measure of how long polar species took to orient themselve to an electric field, not how long they took to become disorient.

Should be more or less the same.

Contrary to what the poster quotes, I personally don't believe this demonstrates that the charge actually resides on the surface of the glass, it simply shows that the molecules within the glass remain polarised until imposed upon by another electric field or fields.

As I mentioned before that is a great video demonstration, but even this can be misinterpretated.

One thing that is a bit of a misnomer is the charging of a capacitor. A capacitor does not actually charge up as such. Think about it logically, if you assume that one plate of a capacitor is charging up with electrons, what is happening to the other plate? It is of course discharging of electrons.

So what is it? Is the capacitor charging or discharging? Of course it is doing both, so clearly you do not charge up a capacitor, you simply create a situation whereby the charges separate.

This article gives a pretty good idea how capacitors are "charged". The actual amount of "charge" on the capacitor does not change by one bit:

ELECTRICITY MISCONCEPTIONS: Capacitor

When we "charge" a conventional metal-plate capacitor, the power supply pushes electrons into one plate, and the fields from these extra electrons reach across the gap between the plates, forcing an equal number of electrons to flow out of the other plate and into the power supply. This creates opposite areas of imbalanced charge: one plate has less electrons and excess protons, and the other plate has more electrons than protons. Each plate does store charge.

However, if we consider the capacitor as a whole, no electrons have been put into the capacitor. None have been removed. The same number of electrons are in a "charged" capacitor as in a capacitor which has been totally "discharged." Yes, a certain amount of charge has been forced to flow momentarily during "charging," and a rising potential difference has appeared. But the current is directed THROUGH the capacitor, and the incoming electrons force other electrons to leave at the same time. Every bit of charge that's injected into one terminal must be forced out of the other terminal at the same time. The amount of charge inside the capacitor never changes. The net charge on each plate is cancelled by the opposite charge on the other plate. Capacitors are never "charged" with electric charge!

So, you are right, the charge is not really stored in the glass, but remains on the plates. Here you see what happens normally when you dissect a capacitor:

Dissectible Capacitor Apparatus (condensing electroscope) for electrostatics demonstrations


And Wikipedia has it right, too:

Leyden jar - Wikipedia, the free encyclopedia

When not properly explained, this demonstration promotes the myth that capacitors store their charge inside their dielectric. This erroneous belief was taught in schools throughout the 1800s, and is still sometimes encountered. However this phenomenon is a special effect caused by the high voltage on the Leyden jar.[7] In the dissectible Leyden jar, charge is transferred to the surface of the glass cup by corona discharge when the jar is disassembled; this is the source of the residual charge after the jar is reassembled. Handling the cup while disassembled does not provide enough contact to remove all the surface charge. Soda glass is hygroscopic and forms a partially conductive coating on its surface, which holds the charge.[7] Addenbrook (1922) found that in a dissectible jar made of paraffin, or glass baked to remove moisture, the charge remained on the metal plates.[8] Zeleny (1944) confirmed these results and observed the corona charge transfer.[9] In capacitors generally, the charge is not stored in the dielectric, but on the inside surfaces of the plates, as can be seen from the fact that capacitors can function with a vacuum between their plates.[10]

Still, it is very intersting, because what all this clearly shows is that the polarized dielectric enacts a considerable attraction force upon the charge carriers in the plates. Now apparantly it depends on the characteristics of the material used what happens when you dissect a capacitor. Under certain conditions, apparantly, these attraction forces are so powerful that charge is actually ripped of the metal and resides on the surface of the dielectric.

It is interesting to think about this further, because when you think about the negative plate, where electrons should be transferred from the metal plate to the dielectric surface, that should be pretty easy. At the positive plate, there is a shortage of electrons in the metal. If this would also be transferred to the dielectric, then electrons would actually have to be ripped from the insulator and transferred to the metal plate. I doubt if the latter could actually take place.

Therefore, this little detail in a story how to dissect a Leyden Jar is interesting:
Electrostatics

Pick up the insulating cup by its bottom, and offer it to the nearest student, telling him or her to put a hand inside to see if there's any charge in there. [...] Ask the class to be 'very, very quiet' as the student (probably with great caution) inserts her fingers into the insulating cup. The student feels the charge, and others can hear the 'crackling' sound, but the student feels nothing even slightly painful, just a pleasant Coulomb tickling. Point out that there's charge on the outside of the insulating cup also.

Is the latter statement actually true?
If yes, then wouldn't you also "feel the charge" and "hear a crackling" sound when you would pickup the cup?

He doesn't mention that, so this suggests that there is a difference between the in- and outside of the cup...

And if that is correct, then you can only "feel the charge" at the inside, *if* you have charged the jar such that the inner cup is the negative pole of the capacitor. So, if you happen to have a Leyden Jar laying around and have nothing better to do: this would be interesting to check out.

Anyway, I think what this experiment teaches us is that the charges inside a capacitor really do attract one another with considerable force and that therefore Bearden is right. When you discharge a capacitor, the capacitor does not release energy at all. It actually takes energy to discharge the capacitor and it is only because of the interaction with the active vacuum that this energy is available.

And it is because of the interaction with the vacuum that energy seems to be released by the capacitor, while in actual fact the vacuum delivers twice that amount of energy. Once to overcome the attraction forces in the capacitor, and once to push the charge carriers trough the wire to the other plate....

[Farrah Day]

Very interesting stuff, Lamare.

Quote:
Originally Posted by Farrah Day
Hi Lamare, firstly I thought that dielectric relaxation was a measure of how long polar species took to orient themselve to an electric field, not how long they took to become disorient.

Should be more or less the same.

What I'm getting at here is that unless influenced by further electric fields, why would the polarised molecules disorient? Over time, yes, I would expect the molecules to disorient due to all the electric fields surrounding, but not immediately. So the relaxation time is not a measure of how long the molecules take to disorient, but rather how long they take to orient when exposed to an electric field.

So, you are right, the charge is not really stored in the glass, but remains on the plates. Here you see what happens normally when you dissect a capacitor

No Lamare, that's not what I'm saying.

Capacitors take a bit of thinking through, and not everyone will be in agreement - but one thing for sure, not everyone can be right.

This is how I see it:

If charges remained on the surface of the glass dielectric when it was taken apart, surely the handling of it would have dischaged them. However, the glass still being polarised, over time will surely attract ions from the surrounding air, which if then touched will discharge through the person touching them.

There are no charges within or on the glass dielectric, it is simply has it's molecules organised - polarised. The same thing as when you brush a piece of iron continuously with a magnet to magnetise it. You are simply polarising it.

Furthermore, with reference to the MIT video, if the charges resided on the copper electrodes, as soon as he clinked them together they would have discharged.

When the copper electrodes are taken off the glass dielectric, there is no longer any influence from the polarised glass dielectric, so the copper being a conductor quickly stabilises as the electrons move throughout the metal to re-establish neutrality.

Once the copper electrodes are again placed over the glass, which is still polarised, then there again will be electron movement with in the copper creating a charge imbalance across the width of the electrode.

To me this all seems quite logical and not at all mysterious... that is until someone asks a question that my logic can't put an answer to. But we'll see.

Fascinating stuff, nevertheless.

[sebosfato]

Sorry guys but you are not understanding how a capacitor really works...

I have studied this on a lecture of physics thus my source is at least trustworthy...

So its like this...

I have to come back to how electrostatics were discovered to explain this to you... Do you remember, that you can electrify an object with attrition and by induction?

It works by creating an imbalance of electrons between two objects, because one object is more likely to hold or accept more electrons than the other.

Take glass for example: if rub a piece of glass in your hair you will capture electrons from the glass and hold than in your hair, as a result the glass will become positively charged... If you approach this glass without touching to a metal, this positive charge will attract the electrons inside the metal to the side where the positive charged object is close to... If now you just touch the metal with a ground wire and than disconnect and after this you move away the piece of glass you will remain with a negatively charged plate... This is induction.

Ok

Did you understood that the glass have electrons? Great!

Now What is The capacitance.

Capacitance is the measure of the amount of energy that you need to apply to polarize a certain dielectric up to a certain degree... Thus WHen you apply a voltage on two electrodes with glass on its inside you will start creating a side of the glass witch will be missing electrons and another in witch there is an excess. I physics they talk about band gap... So the so called dielectric breakdown is nothing more the point where you open all the holes possible and the object" start to conduct electricity... Thus It would discharge all the energy all in once.


So more easily. When you apply a voltage on a capacitor you are creating a physical electrostatic force in both sides making the dielectric to acquire positive charge on one side and negative on the other side. That because you make "force" the electrons inside the glass to move from the negative to the positive, as the material is very resistant you will not be able to keep this electrons flowing thus it will create a electrostatic imbalance on the material. If you rise this charge up to the limit where you make it conductive you will have dielectric breakdown.-


So The charge remain inside the glass. And the amount of energy stored depends on the how polarized (organized) you could make the glass dipoles....

If you take out the glass from inside of the capacitor you will see that the electrodes remain uncharged. Is the electrostatic force of the glass over the electrons of the electrodes that makes they want to move if you close the circuit...

Hope that help you understand

[Farrah Day]

Sorry guys but you are not understanding how a capacitor really works...

Seb, I totally agree that at least one of us does not understand how a capacitor and dielectrics really work... but I'm not convinced it's me!

Glass being a very good insulator means that it does not have very many unbound electrons... if any. Therefore it cannot be electrons moving so much as molecules aligning. A quality of a dielectric is for it's molecules to polarise without electron current flowing!

I don't think any charge as such remains in the glass, the molecules within the glass simply remain polarised - there's a big difference.

Hence, I ultimately feel that much of what you said is wrong. That said, I'm sure many other people will agree with your every word.

[lamare]

There are some interesting documents about Bedini's batteries available at:

http://www.energeticforum.com/attach...ions-dmr07.pdf
http://www.energeticforum.com/attach...ons2-dmr08.pdf

First article also available at: http://www.panaceauniversity.org/OTG...0-%20DMR07.pdf

The reaction that possibly takes place at the anode of Bedini's batteries is this one, according to the first paper:
PbSO4 + H2O + 2e -> 2PbO. PbSO3 + H2SO4 +H +O

If I understand this correctly, then this reaction can take place at one electrode, without the need for the electron to be tranferred between two half reactions. If that is the case, then there is a significant difference between the reactions in a WFC and in batteries.

[Farrah Day]

I lay awake in bed last night pondering a flaw in my logic concerning the MIT Leyden Jar experiment... something is still not quite adding up, and I'm uncomfortable with it.

However, as we're somewhat going off-topic, and because I don't wish to clog HMS's thread up with further multiple posts on this subject, I'll start a fresh thread.