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If the positive plate on a charged capacitor will pass current to the negative...

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sophiecentaur
#37
Dec14-12, 05:38 PM
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You brought up the resonance idea, seemingly in an attempt to discredit a useful approach to a problem. I just pointed out that equivalent circuits are used all over and cited antennae as an example. Why did you use an irrelevant idea to reject a perfectly valid point? Is there anything wrong with using Capacitance to explain what happens when you connect charge polarised objects? If you don't understand it then don't reject the idea per se. Get informed rather than getting cross.
sophiecentaur
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Dec14-12, 05:55 PM
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Quote Quote by justsomeguy View Post
The answer was given a page or so back, though not in direct reply to the OP. His question was:

The answer is a simple "yes" to his final question. Batteries generate electricity via chemical reaction, they do not store it. Electrons do not travel from one plate in the capacitor to the other as they do between battery terminals -- except when the dielectric has failed.

If you just connect two dissimilar terminals of two batteries together and leave the other two terminals hanging, no reaction takes place. No reaction means no ions or free electrons are created, which means no voltage differential and no current.
This question doesn't need to apply just to batteries. It applies to any two sources of emf connected in series. Introducing the chemical generation of ions just muddies the water and misses the essentials of the problem. There is no answer unless the absolute potentials of one terminal of each battery is specified. If you assume that they are both the same voltage and their negative terminals are connected to earth (a nominal zero potential) then, when one battery is removed and its negative terminal is connected to the positive terminal of the other, charge will flow and that charge will be determined by the capacity of the 'upper' battery to Earth, because the 'lower' battery will maintain its emf wrt ground by passing just enough charge so that Q=CV. The C2, in my earlier diagram is the equivalent of the capacity of 'everything' that becomes connected to the lower battery positive terminal.
Nevertamed
#39
Dec14-12, 06:09 PM
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Quote Quote by sophiecentaur View Post
You brought up the resonance idea, seemingly in an attempt to discredit a useful approach to a problem. I just pointed out that equivalent circuits are used all over and cited antennae as an example. Why did you use an irrelevant idea to reject a perfectly valid point? Is there anything wrong with using Capacitance to explain what happens when you connect charge polarised objects? If you don't understand it then don't reject the idea per se. Get informed rather than getting cross.
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sophiecentaur
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Dec14-12, 06:12 PM
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Quote Quote by Nevertamed View Post
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justsomeguy
#41
Dec14-12, 06:31 PM
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Quote Quote by sophiecentaur View Post
This question doesn't need to apply just to batteries. It applies to any two sources of emf connected in series. Introducing the chemical generation of ions just muddies the water and misses the essentials of the problem.
I thought the "problem" was the OP asking why you can measure (and use) voltage between the leg of one capacitor to another even though each one has a leg disconnected, but why you cannot do this with batteries? This is very much the essential of the problem the OP asked, and the answer given is correct.


Quote Quote by sophiecentaur View Post
There is no answer unless the absolute potentials of one terminal of each battery is specified.
If there is no reaction taking place, as there cannot be when one of the terminals is not connected, then the potential is zero.

Quote Quote by sophiecentaur View Post
If you assume that they are both the same voltage and their negative terminals are connected to earth (a nominal zero potential) then, when one battery is removed and its negative terminal is connected to the positive terminal of the other, charge will flow and that charge will be determined by the capacity of the 'upper' battery to Earth, because the 'lower' battery will maintain its emf wrt ground by passing just enough charge so that Q=CV. The C2, in my earlier diagram is the equivalent of the capacity of 'everything' that becomes connected to the lower battery positive terminal.
You need to doublecheck this I think. What do you suppose is generating the electrons in the 'upper' battery? If no reaction is taking place, then no ions are being created, and thus no free electrons. It's like suggesting there will be voltage output from a generator that isn't even running as long as you connect it the right way.
Ratch
#42
Dec14-12, 07:52 PM
P: 315
sophiecentaur,

You brought up the resonance idea, seemingly in an attempt to discredit a useful approach to a problem.
You brought up the antenna idea, seeming in an attempt to pettifog the problem.

I just pointed out that equivalent circuits are used all over and cited antennae as an example.
Don't you think we all know that?

Why did you use an irrelevant idea to reject a perfectly valid point?
You mean inductance? I was not rejecting anything, I was saying that it had to be considered.

Is there anything wrong with using Capacitance to explain what happens when you connect charge polarised objects?
Nothing at all, provided you submit a clear concise explanation that the OP can understand of how it applies to the question he asked.

If you don't understand it then don't reject the idea per se.
It is your idea, so it is up to you to explain it in a way that the OP and I can understand it.

If you don't understand it then don't reject the idea per se. Get informed rather than getting cross.
So that is where you come in. Inform me.

Ratch
Jd0g33
#43
Dec14-12, 08:06 PM
P: 315
"But a battery is a electrochemical device, and does not use a electrostatic field to maintain its voltage like a cap does."

That was exactly what i was looking for. Thanks Ratch
Nevertamed
#44
Dec14-12, 09:48 PM
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Quote Quote by lundyjb View Post
"But a battery is a electrochemical device, and does not use a electrostatic field to maintain its voltage like a cap does."

That was exactly what i was looking for. Thanks Ratch
If the positive plate on a charged capicitor will pass current to the negative...plate of a different capacitor, why wont the positive end of a battery (lets say AA) pass current to the negative plate of a different AA battery. Does it have something to do with the chemical reaction that happens inside the battery?

A: YES


P.S. ha ha ha would be shorter that way xD
Ratch
#45
Dec14-12, 10:12 PM
P: 315
Nevertamed,

If the positive plate on a charged capicitor will pass current to the negative...plate of a different capacitor,
Caps are not charged, they are energized. There will have to be a conduction path for charge to move anywhere. You seem to be describing a floating capacitor, so there is no conduction path. You need to produce a schematic so we know what you are avering.

Does it have something to do with the chemical reaction that happens inside the battery?

A: YES
The chemical reaction of a battery is quite different than the differential equation that describes the voltage/current relationship of a cap.

Ratch
sophiecentaur
#46
Dec15-12, 11:04 AM
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Quote Quote by Ratch View Post
Nevertamed,



Caps are not charged, they are energized. There will have to be a conduction path for charge to move anywhere. You seem to be describing a floating capacitor, so there is no conduction path. You need to produce a schematic so we know what you are avering.



The chemical reaction of a battery is quite different than the differential equation that describes the voltage/current relationship of a cap.

Ratch
Why do you keep on about this? This idea seems to be entirely your own. Can you support it with a reference? If you can't then why not let it drop?
When current has stopped flowing and, assuming there is no self-discharge in a battery, there is no difference in essence between the excess charges on the terminals and the charges on the terminals of a Capacitor. When a small enough charge flows elsewhere, you can assume the PD across either component remains the same.
This is why I have been suggesting the use of equivalent components. As you claim to be familiar with the idea then why not accept the approach as a start to understanding the problem?
You say that I need to explain myself but my simple diagram should be enough to make the point. Any structure / component will exhibit Capacitance - either wrt infinity, a local Earth or another structure / component. If you don't like the term "parasitic" then what would you prefer? You can put these parasitic Capacitances onto an equivalent circuit and that gives you a chance of solving the problem with the usual circuit analysis tools at your disposal.
It would help to have the OP clarified with a diagram but I thought the strong implication was that we start with two batteries or capacitors that had been charged from the same source - so there would initially be 0V PD between the two positive terminals and the two negative terminals.

I was thinking about your mention that the Inductances would also need to be considered. This would be true if you wanted to describe the changing situation when the re-connection was made but, as we are discussing the final state of things, with no current flowing, Inductance is hardly relevant.
sophiecentaur
#47
Dec15-12, 11:19 AM
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Quote Quote by justsomeguy View Post
You need to doublecheck this I think. What do you suppose is generating the electrons in the 'upper' battery? If no reaction is taking place, then no ions are being created, and thus no free electrons. It's like suggesting there will be voltage output from a generator that isn't even running as long as you connect it the right way.
Could you explain what you mean here? There is an emf across a battery even when not connected to a load. That must imply an imbalance of charges. The chemical reaction stops because charges are not being removed via a connected circuit and the potentials balance out within the cell. There will be an 'real' Capacitance across the plates and terminals of the battery which will be 'storing' this unbalanced charge. The Capacitance, itself will be small because the separation is relatively great - but still finite (in the order of a few pF). The fact that a lot of current can flow when the circuit is connected, without much reduction in PD, gives a very high equivalent Capacitance. But the response to an instantaneous load is not instant (in the order of ms) due to the time for the chemical processes to get going so there is not a simple equivalent Capacitance value.

Your analogy with a generator would be more accurate if you were to compare the generator on load and off load but still running at constant speed.
Ratch
#48
Dec15-12, 12:42 PM
P: 315
sophiecentaur,

Why do you keep on about this?
Because energizing in the truth and charging is not.

This idea seems to be entirely your own. Can you support it with a reference? If you can't then why not let it drop?
I support it with an explanation which no one has refuted. "Charge" is the wrong descriptive.

When current has stopped flowing and, assuming there is no self-discharge in a battery, there is no difference in essence between the excess charges on the terminals and the charges on the terminals of a Capacitor. When a small enough charge flows elsewhere, you can assume the PD across either component remains the same.
If all you are talking about is an unchanging voltage and no charge flow, then yes, they would be equivalent. But a cap's voltage is because of a charge difference between its plates. A battery's voltage is because of valance electron displacement of its two metal terminals due to chemical reaction, and not electron crowding and separation due to a dielectric as in a capacitor. If we use a miniature DC generator, then there would be a third way to make a voltage. Neither of those ways are equivalent to each other except in narrow circumstances.

This is why I have been suggesting the use of equivalent components. As you claim to be familiar with the idea then why not accept the approach as a start to understanding the problem?
As I explained above, equivalents are only valid in certain circumstances.

You say that I need to explain myself but my simple diagram should be enough to make the point. Any structure / component will exhibit Capacitance - either wrt infinity, a local Earth or another structure / component. If you don't like the term "parasitic" then what would you prefer? You can put these parasitic Capacitances onto an equivalent circuit and that gives you a chance of solving the problem with the usual circuit analysis tools at your disposal.
It would help to have the OP clarified with a diagram but I thought the strong implication was that we start with two batteries or capacitors that had been charged from the same source - so there would initially be 0V PD between the two positive terminals and the two negative terminals
Yes, capacitance is everywhere, but we usually ignore it because it is insignificant in most cases. I don't think the OP was thinking of any other capacitance other than the two caps he was asking about.

I was thinking about your mention that the Inductances would also need to be considered. This would be true if you wanted to describe the changing situation when the re-connection was made but, as we are discussing the final state of things, with no current flowing, Inductance is hardly relevant.
And in the final state of things, with no voltage changing, neither is capacitance.

Ratch
justsomeguy
#49
Dec15-12, 12:43 PM
P: 166
Quote Quote by sophiecentaur View Post
Could you explain what you mean here? There is an emf across a battery even when not connected to a load. That must imply an imbalance of charges. The chemical reaction stops because charges are not being removed via a connected circuit and the potentials balance out within the cell.
This is half the story. It's not just that electrons are not being removed, but that no more are being freed either. There are a finite number available from when the reaction was last stopped.

Quote Quote by sophiecentaur View Post
There will be an 'real' Capacitance across the plates and terminals of the battery which will be 'storing' this unbalanced charge. The Capacitance, itself will be small because the separation is relatively great - but still finite (in the order of a few pF). The fact that a lot of current can flow when the circuit is connected, without much reduction in PD, gives a very high equivalent Capacitance. But the response to an instantaneous load is not instant (in the order of ms) due to the time for the chemical processes to get going so there is not a simple equivalent Capacitance value.
No, there is no capacitance in an ideal battery. In the real world there is a tiny bit of charge left on the plates after the reaction is stopped, but it's negligible and easily drained. Once that's done, there is no more current flow despite the chemistry of the battery being in a state to produce much more. You have not supported your claim that:

(air) --- +[B1]- --- +[B2]- --- (ground)

Will result in real current flow for any duration. Once the plate of the imaginary capacitor is drained, there is no more current flow, despite there being plenty of reactants left in the battery to produce more.

Your analogy with a generator would be more accurate if you were to compare the generator on load and off load but still running at constant speed.
No, it was more accurate the way I described it. The chemical reaction is the generator. As long as there are no reactions taking place, there is no power generated. That reaction simply cannot take place if one of the terminals is disconnected.

A battery is not a capacitor. There is not a vast reserve of free electrons you can draw on as long as you wish without the chemistry taking place. The chemistry cannot take place as long as one of the terminals is disconnected. It's a furnace and the flow of fuel is interrupted. There may be some residual heat in the system that you can use, but that's it.
sophiecentaur
#50
Dec15-12, 06:01 PM
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Quote Quote by Ratch View Post
sophiecentaur,



Because energizing in the truth and charging is not.



I support it with an explanation which no one has refuted. "Charge" is the wrong descriptive.
But no reference or even a quote with your personal terminology being used elsewhere? That's hardly PF style for asserting "truth".

If all you are talking about is an unchanging voltage and no charge flow, then yes, they would be equivalent. But a cap's voltage is because of a charge difference between its plates. A battery's voltage is because of valance electron displacement of its two metal terminals due to chemical reaction, and not electron crowding and separation due to a dielectric as in a capacitor. If we use a miniature DC generator, then there would be a third way to make a voltage. Neither of those ways are equivalent to each other except in narrow circumstances.
But the chemical reaction in a battery doesn't continue all the time. It stops when the potential builds up (no load) and the chemical potential is equalised. Batteries have a long shelf life because the reaction is only there when charge is allowed to flow and the Potential drops to permit it. What is the difference for the plates of a charged capacitor and the plates of a battery? Electrons are built up on one and depleted on the other in both cases (due to electric fields). In a DC generator, the PD is caused by EM induction. Could you explain the difference in detail, as far as the charges on either side of the emf source are concerned?

As I explained above, equivalents are only valid in certain circumstances.

Yes, capacitance is everywhere, but we usually ignore it because it is insignificant in most cases. I don't think the OP was thinking of any other capacitance other than the two caps he was asking about.
You ignore a component when its value is small enough to ignore it in the circumstance. In the case of the two batteries (or two capacitors) the capacity is what it is and the charge the is displaced is given by Q=CV. I assume you are familiar with that expression. The relevant capacity is small - a few pF and so is the charge imbalance (a few pC). Without using vague terms like "floating cap" (which is something that happens when you falll in the water ), can you explain it in better terms than that? I was suggesting that the OP could actually get an answer by considering the other capacitances involved in his experiment. Many explanations on PF involve introducing additional variables and mechanisms. I wish your objections could include some formulae or figures. It would give them some weight (PF style again).

And in the final state of things, with no voltage changing, neither is capacitance.

Ratch
What voltage change do you refer to? Which capacitance is changing, too? If you put a DVM across two previously charged capacitors in series (or two batteries) what voltage would you expect to measure?
Ratch
#51
Dec15-12, 06:47 PM
P: 315
sophiecentaur,

But no reference or even a quote with your personal terminology being used elsewhere? That's hardly PF style for asserting "truth".
A concensus does not determine the truth. Reason and facts do. Everyone I discussed this with cannot refute me, yet they prefer to "go with the flow" even though they know it is descriptively wrong. As is "current flow" also wrong.

But the chemical reaction in a battery doesn't continue all the time. It stops when the potential builds up (no load) and the chemical potential is equalised. Batteries have a long shelf life because the reaction is only there when charge is allowed to flow and the Potential drops to permit it.
Yes, agreed. Let's see where this takes us.

What is the difference for the plates of a charged capacitor and the plates of a battery? Electrons are built up on one and depleted on the other in both cases (due to electric fields).
I thought I made it clear in post #48 where I said "A battery's voltage is because of valance electron displacement of its two metal terminals due to chemical reaction, and not electron crowding and separation due to a dielectric as in a capacitor." So no, it is not the same in both cases.

In a DC generator, the PD is caused by EM induction. Could you explain the difference in detail, as far as the charges on either side of the emf source are concerned?
The DC generator has no dielectric, therefore, its voltage will be only by EM induction as you stated, and not by charge separation and crowding or spacing out of electrons.

You ignore a component when its value is small enough to ignore it in the circumstance. In the case of the two batteries (or two capacitors) the capacity is what it is and the charge the is displaced is given by Q=CV. I assume you are familiar with that expression. The relevant capacity is small - a few pF and so is the charge imbalance (a few pC).
I agree with the above.

Without using vague terms like "floating cap" (which is something that happens when you falll in the water ), can you explain it in better terms than that?
A floating cap is a topology, not a verb. It means that one of it terminals is left "floating", i.e. it is not connected to anything.

I was suggesting that the OP could actually get an answer by considering the other capacitances involved in his experiment. Many explanations on PF involve introducing additional variables and mechanisms. I wish your objections could include some formulae or figures. It would give them some weight (PF style again).
The OP asked for a qualitative explanation, not a quantative one.

What voltage change do you refer to? Which capacitance is changing, too? If you put a DVM across two previously charged capacitors in series (or two batteries) what voltage would you expect to measure?
I was answering your previous statement where you said "with no current flowing, Inductance is hardly relevant."

I answered "And in the final state of things, with no voltage changing, neither is capacitance." So in the proper context, you should have interpretated it as I saying capacitance is not relevant if voltage is not changing. You instead interpreted it as I as saying that capacitance changed.

Ratch
sophiecentaur
#52
Dec16-12, 05:08 AM
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I am afraid I'll wait to change my description of what we do to capacitors until I see it in wider use. Good luck with your educational crusade.
As for the rest of your post, it contains nothing but assertions and no explanations so it doesn't take us any further.
To be fair to both of us, the original scenario is a bit vague so we may be talking at cross purposes about all this.
jim hardy
#53
Dec16-12, 10:11 AM
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methinks there's accepted terminology in any field and for discussion to be meaniingful, one needs to use it.

That's why study in any field begins with vocabulary,

and a good textbook has a definition of terms at beginning of each chapter.
Ratch
#54
Dec16-12, 12:09 PM
P: 315
jim hardy,

methinks there's accepted terminology in any field and for discussion to be meaniingful, one needs to use it.

That's why study in any field begins with vocabulary,

and a good textbook has a definition of terms at beginning of each chapter.
That is all fine and good. But how does that apply to what we were discussing?

Ratch


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