Everything is a capacitor
The wire is a capacitor
When attached to the 9V terminal, it'll charge up
Also the other terminal is a capacitor, and it's charge will adjust according to the new capacitance of the 9V terminal and connected wire.

Use VC = q

So I reckon yes, the wire is physically and detectably different from its 0V self.

Hi meemoe_uk, Bill M., et al,

The inventor, Joseph Hiddink (Toronto), was granted a US Patent called 'Capacitance Changer', that sort of kind of may apply to the thoughts of this thread...

Hiddink bought all of his neighbors a new television set, because he changed the physical configuration of a capacitor from two terminals to ONE terminal.

When that occurred, the ONE terminal being of a much lower capacitance, the voltage stored on the new one terminal went way high.

The voltage went so high that a lightning bolt struck his power pole, and he replaced all of his neighbor's TVs which were toasted, because he was such a nice guy.

He repeated the test in the country, parking in the shade of a tree in an open field. Again, from a blue sky, a lightning bolt struck the tree.

The lightning was attesting to the extreme voltage produced when the VC=q rule is applied. Apparently the super high voltage became a substitute for a leader bolt that attracts higher-altitude charges, but that's beside the point.

Two terminals have some capacitance, sure, but One has very, very little, and the voltage multiplied accordingly.

As a two terminal capacitor, Joseph used a plasma stream in a tube, and a foil wrapper. When the plasma stream was ignited (like a black light, without the phosphor coating of a fluorescent tube), a second voltage source applied several kiloVolts between the plasma stream in the light, and the outer foil wrapper, which formed a capacitance through the dielectric glass.

When the light was turned off, the absence of the plasma conductor left the foil holding all of the charge, which was multiplied by the ratio of the before and after capacitance values.

BTW -- Joseph said that when his patent was granted, that he received such a volume of hate mail that upset his wife to the point that she forbid him to continue his research. I suppose that says that most people are rattled by the thought of a capacitor value changing under an existing charge. Humans are funny (funny odd, not ha ha).

Joseph warns that any tests should be done in a grounded Faraday cage, for sure. And also that a neon bulb would show similar results if similar treatment is applied, but in a safer range of voltages --but still use a Faraday cage to avoid toasting equipment on the bench.

Doesn't the antique Whimthrust (sp?) machine amount to a rotating capacitance changer?

 

Take a 10pf and a 1000µF cap and put them in series, then connect the capacitor to the battery. The 10pf gets the bulk of the charge, but it is tiny.

I know this defies intuition, but if both caps start out discharged, they each wind up with equal charge. A less extreme example is easier to see.
Suppose you start with a 1uF cap and a 9uF cap in series, both discharged, and connect them across a 10V source (some series resistance will not change the results). The 1uF cap will wind up with 9V across it, and the 9uF cap will wind up with 1V across it.

Q=C*V
Q1=1uF*9V=9e-6 coulombs
Q9=9uF*1V=9e-6 coulombs.

This works for any two values of capacitance.
I haven't thought about how this applies to the original question. I'm just proofreading.

Now, if you had said that the 10pF cap winds up with the bulk of the voltage, I would agree. Perhaps that is what you meant to say.

As a side issue, the series capacitance is 0.9uF, and the voltage is 10V, making the total charge cross the series pair also equal to 9e-6 coulombs. Ain't that a weird stroke-o?

 

>
For most practical purposes, voltage is relative, but in theory it is absolute.
i.e. in most practice the only significant aspect to voltage is the potential difference between objects. But in theory you could count the number of electrons and protons in an object and calculate an absolute voltage, i.e. weigh up the charge of an object, then use VC = q to work out its absolute voltage.

Capacitance only exists between two objects which are separated by a dielectric. If you charge up two parallel plates to, say, 10 volts, remove the source, and then double the distance between the plates, the voltage will double, and the capacitance will be halved. Only Q is constant. You can't use Q=CV to calculate voltage, because C is not a constant property of an object. It is relative to another object which is separated from the first by a dielectric.
Absolute voltage does not exist. Voltage is always relative to another point.

 

Someone may have already mentioned this (I only skimmed the thread), but in the original question there is something that was completely overlooked.

Let's say that I have a battery that has a very large area for a negative terminal. Now I position a "wire" that happens to have the same area and shape extremely close to the negative terminal without touching them. This, of course, forms a capacitor that is already connected to the negative terminal of the battery (since one plate of the cap IS the negative terminal of the battery). I then connect the "wire" to the positive terminal. This will result in the capacitor charging according to the voltage and capacitance resulting in an excess of electrons on the battery side and a deficit on the "wire" side of the capacitor. I then disconnect the "wire" from the positive terminal, leaving the "wire" close to the negative terminal thus leaving the capacitor charged. Now I want to move the "wire" away from the battery. Yes, it will remain charged (by having fewer electrons than it normally wood). But it will also require energy to move it away from the battery because they are electrically attracted.

But the big thing that was missed is that the battery will be left with a net negative charge on it. When the next "wire" is brought near the battery, a charge separation will be induced in the "wire" even before you connect it to the positive terminal. The end result is that on each succeeding iteration you will pull less charge off than the time before and you will quickly reach steady state in which subsequent iterations have no effect.


long rod as the negative terminal. I now wrap a wire in such a way that it forms a

 

Someone may have already mentioned this (I only skimmed the thread), but in the original question there is something that was completely overlooked.

Let's say that I have a battery that has a very large area for a negative terminal. Now I position a "wire" that happens to have the same area and shape extremely close to the negative terminal without touching them. This, of course, forms a capacitor that is already connected to the negative terminal of the battery (since one plate of the cap IS the negative terminal of the battery). I then connect the "wire" to the positive terminal. This will result in the capacitor charging according to the voltage and capacitance resulting in an excess of electrons on the battery side and a deficit on the "wire" side of the capacitor. I then disconnect the "wire" from the positive terminal, leaving the "wire" close to the negative terminal thus leaving the capacitor charged. Now I want to move the "wire" away from the battery. Yes, it will remain charged (by having fewer electrons than it normally wood). But it will also require energy to move it away from the battery because they are electrically attracted.

But the big thing that was missed is that the battery will be left with a net negative charge on it. When the next "wire" is brought near the battery, a charge separation will be induced in the "wire" even before you connect it to the positive terminal. The end result is that on each succeeding iteration you will pull less charge off than the time before and you will quickly reach steady state in which subsequent iterations have no effect.

You lost me. I don't think you are saying that the battery will discharge. What are you saying?

 

It has a lot in common with a Kelvin electrostatic generator. Look into that and then let's discuss further.

 

It has a lot in common with a Kelvin electrostatic generator. Look into that and then let's discuss further.

I understand that principle.
If I understand your "process", you are successively charging one plate of a capacitor by connecting it to the positive terminal of a battery (the other plate is connected to the negative battery terminal), removing it, and then doing the same with another plate, ad nauseum. Is that correct? If so, how does that relate to the Kelvin electrostatic generator?

 

In the Kelvin generator, you have a capacitor that is formed between the ring and the water and that is charged. Then the water drop breaks free (disconnecting the wire) while still charged and takes that charge away with it someplace else.

The result is a growing charge separation that builds until it eventually produces a limiting mechanism, either a spark or it pulls the drop to the ring or it flings the drops away entirely.

 

In the Kelvin generator, you have a capacitor that is formed between the ring and the water and that is charged. Then the water drop breaks free (disconnecting the wire) while still charged and takes that charge away with it someplace else.

The result is a growing charge separation that builds until it eventually produces a limiting mechanism, either a spark or it pulls the drop to the ring or it flings the drops away entirely.

As I said, I understand the Kelvin generator. I was hoping you would explain what the equilibrium mechanism is in your thought experiment with the battery.

 

As I said, I understand the Kelvin generator. I was hoping you would explain what the equilibrium mechanism is in your thought experiment with the battery.

Start off with two uncharged plates and then connect a battery between them to establish a voltage V between them.

A certain amount of charge, Q, will be taken off one plate and put on the other. You then disconnect the battery and remove the plate and ship it to Kansas. You now have a plate that has a net charge on it, half of which resides on each side.

When you move the next uncharged plate into position, there is already an electric field in place and so a charge distribution is induced in the other plate with plus half the charge of the fixed on one side and minus one-half the charge on the other, thus there is a voltage between them that is V/2.

When you connect the battery, charge is moved from one plate to the other in order to establish V again, but this time only half of the charge has to be moved because the induced charge provides the other half. Thus when you disconnect the second plate and ship it to Oregon, it only takes Q/2 with it leaving the fixed plate with a net charge of 1.5Q.

The total charge supplied by the battery up to this point is 1.5Q.

Repeat the process again. Each cycle will remove half of the charge that was removed in the previous cycle. Thus the total charge removed will converge to 2Q, which will also be the net charge on the fixed plate. Once this has been "reached", connecting the battery will move no charge and when you remove the plate it will carry away no net charge with it.

Let's assume that you could get a net charge of, say, 6Q on there. Now what happens when you bring in an uncharged plate? It will induce a charge of +/-3Q on opposite sides of the plate and the voltage will be 3V. What happens when you connect the battery? It will move 2Q of charge in the OTHER direction in order to REDUCE the voltage down to V. Thus, when you remove the plate, it will carry away -2Q of charge and leave the fixed plate with 4Q of net charge. Repeat the process over and over and you will again converge on a net charge of 2Q on the fixed plate.

Now, I have assumed that charge can only reside on the fixed plate. In practice, it would spread over the entire system, including the surface of the battery, and you would be able remove more than 2Q of charge, but it would still converge to some fixed multiple of Q. Even if that multiple turns out to be 142.73, you will quickly reach the equilibrium point and not be able to draw an arbitrary amount of charge from the battery and drain it (unless 142.73Q happens to be greater than the capacity of the battery).

 

Well, I had to read it 47 times, but it finally soaked in to my dense skull. Thanks for taking the time to write such a detailed explanation.

 

No problem. I'm glad you stuck with it and even happier that you agree with my experiment (at least I assume you do).

Now that you do understand it, do you agree that it has a reasonable linkage to the Kelvin generator, or do you think I'm stretching it too much?

 

No problem. I'm glad you stuck with it and even happier that you agree with my experiment (at least I assume you do).

Now that you do understand it, do you agree that it has a reasonable linkage to the Kelvin generator, or do you think I'm stretching it too much?

I don't see that your experiment has much in common with the Kelvin generator, unless you mail the negatively charged bucket of water to Kansas and the positively charged one to Oregon. However, in your experiment those plates both had the same charge polarity, so I guess that comparison doesn't hold water either.

 

You can stack the plates you remove on top of each other a couple feet away. The connection is that you induce a charge separation and then you physically separate part of the charge and remove it leaving the rest of it with a growing net charge.

 

I think you are asking if, with two pieces of copper wire, one piece on the + terminal and one piece on the - terminal, (using current flow convention) the wire connected to the - terminal has a surplus of electrons while the wire connected to the + terminal has a deficit of electrons, is that right?

Ultimately, using a super powerful microscope that can see electrons (imagination here gentlemen...), you would see that this is NOT the case. In a battery, the energy that comes out the terminals is a result of a galvanic force, or the same force that is involved in oxidation. Because of electric field laws, sulfuric acid's force against lead oxide is balanced by the field force of the number of electrons held internally in the PbO due to the presence of the oxide. Same thing on the lead plate re: sulfuric acid, its associative force is balanced by a dearth of those electrons. You can transport electrons from the PbO side to the Pb side using a wire, or circuit, and the galvanic action takes place. In a way, first the electron moves from PbO to Pb, THEN H2SO4 + Pb + PbO can break down into the lower potential energy form of PbS and H2O, because the electric field force preventing this enthalpic action was neutralized by the circuit.

In a capacitor, the number of electrons doesn't change one plate re: the other, but merely the "want" to move. It's as if each q takes on a delta_q and this delta_q is what moves down the line at the speed of light *while the q bounces down the line at 1m per hour.*

I hope I helped, and that I'm not too far off base...

EDIT: addition between asterisks above

 

No, there really is a net charge involved. If you take a parallel plate capacitor, charge it up, isolate it, then pull the plates apart one of the plates really does have a net excess of electrons and the other plate has a net deficit of electrons. Talking about the charge being the result of electrons "wanting" to move is metaphysical claptrap.

 

No, there really is a net charge involved. If you take a parallel plate capacitor, charge it up, isolate it, then pull the plates apart one of the plates really does have a net excess of electrons and the other plate has a net deficit of electrons. Talking about the charge being the result of electrons "wanting" to move is metaphysical claptrap.

My mistake. That is correct.