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

[jaydnul]

...plate of a different capacitor, why won't 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?

 

[jedishrfu]

The plates are separated enough and the electrons can't jump from one plate to the next hence no effective current flow. Capacitors hold charge.

Here's some more info on capacitors:

http://en.wikipedia.org/wiki/Capacitor

 

[Ratch]

lundyjb,

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

...plate of a different capacitor,

What do you mean by that statement? Do you mean that the charge from one plate will leak through the capacitor's dielectic, and arrive at the opposite plate (WRONG!)? Or do you mean that if there is an external conduction path from one plate to the opposite plate, the charge imbalance will equalize (RIGHT!)?


jedishrfu,

...no effective current flow

Current flow literally means "charge flow flow". You should just say "current" or "charge flow"

Capacitors hold charge

No, they do not hold charge. They store energy. A capacitor energized to 1000 volts has the same net charge as it did when it was energized to zero volts.

Ratch

 

[Nevertamed]

i understood the question, actually there would be current from one battery to the other but only for a short time, because there is no continuity (circuit not formed) between them, the electrons would pile up in the "receiving battery", if you want a continue current to exist you must set a path back from the receiving battery to the first one (both would be connected in series).

in an extreme scenario you could imagine that the receiving battery would take all electrons from the first battery (there's a finite amount of electrons), because the chemical reaction will always maintain each poles at a constant potencial level, but once electrons pile up in the receiving battery they will create a strong electric field that would reject and stop the incomming electrons.

 

[jedishrfu]

Thanks for your corrections, Ratch. You are correct, although it is common to say current flow.

With the charge comment, I was really referring to the one plate being more positive (lack of electrons) and the other being more negative (excess of electrons) but you are correct the net charge is zero for the capacitor.

 

[Ratch]

Nevertamed,

Wrong with respect to both a battery and capacitor. According to your reasoning, a bird would not be able to roost on a high voltage wire because of a supposed short transient current it would receive when it first touched the wire. It just doesn't happen that way.

A battery does not accumulate electrons on either pole, and a capacitor does not imbalance the charge on its plates unless there is a voltage difference between them. Just connecting one plate of a capacitor to a voltage source won't do anything, because there would be no voltage difference between the plates.

Ratch

 

[sophiecentaur]

Nevertamed,

Wrong with respect to both a battery and capacitor. According to your reasoning, a bird would not be able to roost on a high voltage wire because of a supposed short transient current it would receive when it first touched the wire. It just doesn't happen that way.

A battery does not accumulate electrons on either pole, and a capacitor does not imbalance the charge on its plates unless there is a voltage difference between them. Just connecting one plate of a capacitor to a voltage source won't do anything, because there would be no voltage difference between the plates.

Ratch

This actually does happen on the highest voltage transmission cables. Birds stick to Intermediate and Domestic voltage cables. (When I say "stick to", I don't mean glue-like)

I don't know how they know to avoid them. Perhaps the local field strength causes them to tingle and so they keep clear of such attractive perches.

 

[Ratch]

sophiecentaur,

This actually does happen on the highest voltage transmission cables. Birds stick to Intermediate and Domestic voltage cables. (When I say "stick to", I don't mean glue-like)

That has to be caused by high voltage induction (affect without touching due to high electric fields), not conduction. I see plenty of birds roosting on domestic power lines without any ill effects.

Ratch

 

[Nevertamed]

Nevertamed,

Wrong with respect to both a battery and capacitor. According to your reasoning, a bird would not be able to roost on a high voltage wire because of a supposed short transient current it would receive when it first touched the wire. It just doesn't happen that way.

A battery does not accumulate electrons on either pole, and a capacitor does not imbalance the charge on its plates unless there is a voltage difference between them. Just connecting one plate of a capacitor to a voltage source won't do anything, because there would be no voltage difference between the plates.

Ratch

Actually there's a little current through the bird, since there is finite resistance in the wire, and finite resistance in the bird, it still forms a parralel circuit. of course the wire's resistance would be meaningless compared with the bird's, and we all know the current always "choose" the easiest path, only a tiny amount of current goes through the bird

 

[sophiecentaur]

sophiecentaur,



That has to be caused by high voltage induction (affect without touching due to high electric fields), not conduction. I see plenty of birds roosting on domestic power lines without any ill effects.

Ratch

Yes - we all do. It must be a matter of Volts per metre and the capacity and length of the bird's body which limits the Voltage to which they a tolerant. I was referring to High Voltage cables - 132kV and above.

 

[jim hardy]

Sophie is correct.

Birds tend to avoid power lines > about 30KV.
I'd guess it tickles their feet.

It'd be the minute capacitive current that charges their body alternately to +/- line peak voltage at line frequency , 50 or 60 hz.

When working such lines hot, linemen wear a special suit that forms a Faraday cage around them.

http://www.hubbellpowersystems.com/lineman/accessories/suit.asp

 

[Nevertamed]

Nevertamed,

Just connecting one plate of a capacitor to a voltage source won't do anything, because there would be no voltage difference between the plates.

Ratch

the thread starter stated " If the positive plate on a charged capacitor will pass current to the negative plate of another capacitor" which is true, because there would be a voltaje between both plates regarless of being part of different capacitors. however there would be only a short current as the voltaje fade away

 

[Nevertamed]

Nevertamed,



A battery does not accumulate electrons on either pole



Ratch

batteries create a constant imbalance of charge, to maintain a DC voltage

 

[sophiecentaur]

Just connecting one plate of a capacitor to a voltage source won't do anything, because there would be no voltage difference between the plates.

Ratch

In fact, there is a small but finite Capacitance between the unconnected end of the capacitor and ground. This will lead to a finite but small charge flow, via the main Capacitor when it's connected. You can regard it as two capacitors in series - one large and one tiny one.

 

[Ratch]

Nevertamed,

the thread starter stated " If the positive plate on a charged capacitor will pass current to the negative plate of another capacitor" which is true, because there would be a voltaje between both plates regarless of being part of different capacitors. however there would be only a short current as the voltaje fade away

Remember, caps do not get charged, they get energized. The positive plate is deficient in electrons. You refer to two capacitors and four plates. So which plates will there be a voltage between? Where is the conduction path for the "short current"? I don't know how to answer your message because I don't know to what you are referencing.

batteries create a constant imbalance of charge, to maintain a DC voltage

What does that mean? How much charge for what voltage? If it were a cap, I could multiply the voltage by the capacitance and give you an answer. But a battery is a electrochemical device, and does not use a electrostatic field to maintain its voltage like a cap does. Both a battery and a cap sustain a voltage, but they do it in very different ways. Besides, a battery is an active device, whereas a cap is not.

sophiecentaur,

In fact, there is a small but finite Capacitance between the unconnected end of the capacitor and ground. This will lead to a finite but small charge flow, via the main Capacitor when it's connected. You can regard it as two capacitors in series - one large and one tiny one.

Your description of the above really confuses me. First of all, ground is nothing special, just a common connection point. The capacitance between one end of the capacitor and the other end is the value of the capacitor. Where and how does the second capacitor come into play? Inquiring minds would like to know.

Ratch

 

[sophiecentaur]

sophiecentaur,

Your description of the above really confuses me. First of all, ground is nothing special, just a common connection point. The capacitance between one end of the capacitor and the other end is the value of the capacitor. Where and how does the second capacitor come into play? Inquiring minds would like to know.
Ratch

OK then. Replace 'ground' with 'the other terminal of the battery'. The effect will then even less but there are still, effectively, two capacitors in series with the battery. (And another small one directly across the battery terminals.) If the battery is supplying an emf, then this network of capacitors will charge up according to Q=CV.
Explanations in terms of equivalent lumped components are often helpful - antenna theory is often usefully approached this way.


btw If you want to be fussy about nomenclature, then I think it would be more desctiptive to say that Capacitors become Polarised, if you don't like 'charged'. (I have not read of the term "Energised" in this context - it is used more int the context of Batteries, I think but I don't think it is defined very rigorously, though people do talk of 'energising a coil' when you switch on the current, I suppose). That just means that there is a displacement (imbalance) of net charge from one side to the other. With no dielectric, the charge needs to be 'taken off' one side and 'put into' the other. When there is a dielectric, the additional charge (giving it higher Capacitance) is due to charges being easily displaced within the dielectric material as its molecules become polarised. You need to move more charges in this case for a given Voltage.

 

[Nevertamed]

Nevertamed,



Remember, caps do not get charged, they get energized. The positive plate is deficient in electrons. You refer to two capacitors and four plates. So which plates will there be a voltage between? Where is the conduction path for the "short current"? I don't know how to answer your message because I don't know to what you are referencing.



What does that mean? How much charge for what voltage? If it were a cap, I could multiply the voltage by the capacitance and give you an answer. But a battery is a electrochemical device, and does not use a electrostatic field to maintain its voltage like a cap does. Both a battery and a cap sustain a voltage, but they do it in very different ways. Besides, a battery is an active device, whereas a cap is not.



Ratch

actually when they get charged they get energized (they get energy from a external source and store it as an electric field between the plates), about the 2 caps and the 4 plates:

we have the first cap: with one plate full of electrons and the other one lacking (after being charged or energized if you like it)

we have a second cap: with one plate full of electrons and the other one lacking (after being charged or energized)

if you set a path (an hypothetical wire) from the first capacitor's full of electrons plate to the second's capacitor lack of electrons plate; there will be a short current until this imbalance disappear (there wuld be a voltage between them, despite the fact the plates involved belong to different caps)

about your second quote:

In a way, a capacitor is a little like a battery. Although i know they work in completely different ways, capacitors and batteries both store electrical energy. A battery has two terminals. Inside the battery, chemical reactions produce electrons on one terminal and absorb electrons on the other terminal, it maintains this imbalance of charge ( charge separation) as the acid keep working in order to offer a constant voltage

 

[sophiecentaur]

I don't understand the need to use a term, other than 'charged' do describe what happens to a battery. Everyone but everyone knows what it means and it's used in every book I have ever read on the subject. The qualifying word 'differentiallly', which might be added in front is omitted because it is such an accepted term.

Yes, of course Capacitors store Energy but the Energy they 'store' is not the same as the Charge as it depends upon the Capacitance value - the two quantities, Charge and Energy are not synonymous. It is the Charge that is used in the context of most circuit calculations but - horses for courses.

The answer to the original question lies with the parasitic Capacitances involved.

 

[Nevertamed]

i like and always use "charge"

 

[Ratch]

sophiecentaur,

OK then. Replace 'ground' with 'the other terminal of the battery'. The effect will then even less but there are still, effectively, two capacitors in series with the battery. (And another small one directly across the battery terminals.) If the battery is supplying an emf, then this network of capacitors will charge up according to Q=CV.
Explanations in terms of equivalent lumped components are often helpful - antenna theory is often usefully approached this way.

I still cannot understand what you are saying because I cannot visualize the circuit you are talking about. Perhaps a schematic?

btw If you want to be fussy about nomenclature, then I think it would be more desctiptive to say that Capacitors become Polarised, if you don't like 'charged'. (I have not read of the term "Energised" in this context - it is used more int the context of Batteries, I think but I don't think it is defined very rigorously, though people do talk of 'energising a coil' when you switch on the current, I suppose).

I don't think "polarization" is a good word to use for energy storage. It sounds too much like a polarized capacitor such as an electrolytic. Energized simply means "charged with energy".

When there is a dielectric,...

A capacitor always has a dielectric.

the additional charge (giving it higher Capacitance) is due to charges being easily displaced within the dielectric material as its molecules become polarised. You need to move more charges in this case for a given Voltage.

Just another way of saying the capacitor is more energized. E=1/2*C*E^2 or E = 1/2*(Q^2)/C

 

[Ratch]

Nevertamed,

we have the first cap: with one plate full of electrons and the other one lacking (after being charged or energized if you like it)

we have a second cap: with one plate full of electrons and the other one lacking (after being charged or energized)

if you set a path (an hypothetical wire) from the first capacitor's full of electrons plate to the second's capacitor lack of electrons plate; there will be a short current until this imbalance disappear (there wuld be a voltage between them, despite the fact the plates involved belong to different caps)

I disagree. There will be no current between two floating caps. Try energizing two caps, and put a ammeter between the two plates to see if there is any current.

In a way, a capacitor is a little like a battery. Although i know they work in completely different ways, capacitors and batteries both store electrical energy. A battery has two terminals. Inside the battery, chemical reactions produce electrons on one terminal and absorb electrons on the other terminal, it maintains this imbalance of charge ( charge separation) as the acid keep working in order to offer a constant voltage

If they work in different ways, then they are not like each other. A battery is an active element, and a cap is not. A battery produces a voltage by chemical means, not by storing energy by accumulating electrons like a cap does. If there is not a conductomg path for the terminals of a battery, the electrons will not accumulate.

Ratch

 

[Ratch]

sophiecentaur,

I don't understand the need to use a term, other than 'charged' do describe what happens to a battery. Everyone but everyone knows what it means and it's used in every book I have ever read on the subject. The qualifying word 'differentiallly', which might be added in front is omitted because it is such an accepted term.

Because it is a wrong descriptive, even if everyone thinks they know what it means.

Yes, of course Capacitors store Energy but the Energy they 'store' is not the same as the Charge as it depends upon the Capacitance value - the two quantities, Charge and Energy are not synonymous. It is the Charge that is used in the context of most circuit calculations but - horses for courses.

Well, that is true. Energy and charge are two different things. Caps don't store any charge, but they do store energy.

The answer to the original question lies with the parasitic Capacitances involved.

Parasitic capacity is usually very small and only of interest at high frequencies. Are you sure you want to get involved with this?

Ratch

 

[Ratch]

Nevertamed,

i like and always use "charge"

Yes, and NASA likes to see their astronauts "walk" in space, too. But tell me. Did you ever ask NASA what would happen if their tether broke. Would the astronaut walk away from his space vehicle?

Ratch

 

[sophiecentaur]

sophiecentaur,
A capacitor always has a dielectric.

How about these? In practice, of course, there is never 'nothing' between the plates of a capacitor but on these jobbies, ε_r is reckoned to be as near as dammit to unity. A 'dielectric' is a material that polarises - not a vacuum.
This MIT link defines Polarisation as charge separation, in any form so my use of the word is justifiable, I think.

I disagree. There will be no current between two floating caps. Try energizing two caps, and put a ammeter between the two plates to see if there is any current.

I will try a diagram, to show you what I mean about the Capacitors involved when I have time - Power Point is OK but slower than writing and ranting for me. You object to the used of the word 'Parasitic' on the grounds that they are only relevant at HF. But that's the whole point. They are small but if they work at HF then they have a finite effect at all frequencies and at DC.
This is hardly worth answering, it's so obvious. I was clearly talking about what would happen at the start of this process - when you add the battery or start to move the (original) wire in a field. If you acknowledge that there will be a charge separation / difference then some current was flowing at one time.

You have clearly nailed this particular flag to your mast and are sticking to it but can you quote me this as a generally used term? 'Correctness' is something that needs a bit more support than just what you are saying. But, again, is this relevant to the main discussion? We can usually describe the World in more than one way. Energy is a pretty good way to describe and predict things (I am a big fan, in general) in Physics but Forces, Masses, Charges and Impedances are often used more conveniently. Perhaps this is why we learn and then use Q=CV so often - despite the fact that V is Energy per Unit Charge.

Energized simply means "charged with energy".

A bit of tautology, (or is it self-referencing?) here, I think. From what I can find, the term 'energised' seems to be used for circuits in which a current is flowing continuously but the word is used very freely and imprecisely so I don't think it is any better than 'Charged', aamof.

 

[Nevertamed]

Nevertamed,
I disagree. There will be no current between two floating caps. Try energizing two caps, and put a ammeter between the two plates to see if there is any current.

Ratch

they are not just 2 floating caps, they are two points with a imbalance of charge between them, one fully negative charged plate (from the first capacitor) and one fully positive charge plate (from the second capacitor), forget the fact both plates belong to different capacitors, they are still 2 points with a voltage present, if you set a path there will be a current until the charge depletes.

 

[Nevertamed]

Nevertamed,





If they work in different ways, then they are not like each other. A battery is an active element, and a cap is not. A battery produces a voltage by chemical means, not by storing energy by accumulating electrons like a cap does. If there is not a conductomg path for the terminals of a battery, the electrons will not accumulate.

Ratch

i didnt say caps and batts are identical, i said "in a way, caps are a lil like a battery" i meant they both have similarities

 

[Nevertamed]

Nevertamed,
Yes, and NASA likes to see their astronauts "walk" in space, too. But tell me. Did you ever ask NASA what would happen if their tether broke. Would the astronaut walk away from his space vehicle?

Ratch

maybe they like speaking metaphorically like lots of people...

 

[sophiecentaur]

sophiecentaur,
I still cannot understand what you are saying because I cannot visualize the circuit you are talking about. Perhaps a schematic?

The 'equivalent circuit' I am proposing is on the attachment and you can read my original ideas with that circuit in mind.
C1 is the capacity between the inner and outer parts of the battery. They are there whether or not there happens to be any chemical in there. The emf and internal resistance r of the battery are shown and so is the finite Capacitance (C2) between the added wire and the case / negative of the battery or ground.
When the emf is turned on (pouring in the electrolyte, perhaps), charge will flow as the arrows show and the two Capacitors will 'charge up'. This means that there is an initial current.
This diagram instantly transfers to the original question because you can leave out the battery and you still have the induced emf and the Capacitor C2. By doubling the length of the wire (the OP) you are increasing C2. My only query about this is how to determine the effective capacity change.
I googled 'equivalent circuit capacitor antenna' and got more hits than I had time to read. They establish the principle that I'm getting at. The fact that antennae are mostly used for AC is not a hinderance because the situation for the DC condition is only arrived at after a change - not DC - in the situation.

 

[Ratch]

sophiecentaur,

How about these?

I thought you would bring them up. It doesn't change anything, however.

In practice, of course, there is never 'nothing' between the plates of a capacitor but on these jobbies, εr is reckoned to be as near as dammit to unity. A 'dielectric' is a material that polarises - not a vacuum.

It does not matter whether a dielectric is a material or not. What matters is whether it has the properity of a dielectric, specifically, whether it has a dielectric constant. A vacuum does. Its dielectric constant is called the "permittivity of free space".

This MIT link defines Polarisation as charge separation, in any form so my use of the word is justifiable, I think.

Yes, caps that are energized also have to be polarized. But, you did not addresses my concern, which is that saying a cap is polarized can easily be mistaken for designating the cap as an electrolytic, even if it is not.

I will try a diagram, to show you what I mean about the Capacitors involved when I have time - Power Point is OK but slower than writing and ranting for me. You object to the used of the word 'Parasitic' on the grounds that they are only relevant at HF. But that's the whole point. They are small but if they work at HF then they have a finite effect at all frequencies and at DC.
This is hardly worth answering, it's so obvious. I was clearly talking about what would happen at the start of this process - when you add the battery or start to move the (original) wire in a field. If you acknowledge that there will be a charge separation / difference then some current was flowing at one time.

I don't think something as elaborate as PP is needed. A simple schematic with all the salient points should suffice. A conduction path is needed at all times for charge to move. That is what I will be looking for.

You have clearly nailed this particular flag to your mast and are sticking to it but can you quote me this as a generally used term? 'Correctness' is something that needs a bit more support than just what you are saying. But, again, is this relevant to the main discussion? We can usually describe the World in more than one way. Energy is a pretty good way to describe and predict things (I am a big fan, in general) in Physics but Forces, Masses, Charges and Impedances are often used more conveniently. Perhaps this is why we learn and then use Q=CV so often - despite the fact that V is Energy per Unit Charge.

What you say in the above paragraph is true, but I am finding it hard to determine what is the point or conclusion.

A bit of tautology, (or is it self-referencing?) here, I think. From what I can find, the term 'energised' seems to be used for circuits in which a current is flowing continuously but the word is used very freely and imprecisely so I don't think it is any better than 'Charged', aamof.

If "energized" means "imbued with energy", would that make a better definition?

Ratch

 

[sophiecentaur]

If "energized" means "imbued with energy", would that make a better definition?

Ratch

This can only be a matter of preference. After all, would you say, in everyday life, that you "climbed a hill" or that you "Increased your Gravitational Potential"? Both would be correct and acceptable in the appropriate context. If we are talking about Charge (which I though we were), then why not allow our experiment to "charge a capacitor"? If we are talking about supercapacitors and road transport then it's arguable that 'energising' them would be appropriate - but did we ever in this World, talk about Energising a Battery? "Energiser" is a trade name for Ever Ready High Power batteries but that, I think, is aimed at describing what they actually do to your toys and not what Ever Ready do to their batteries.

Anyway, I did that diagram (above post). Hope it makes sense for you.

OH - I just spotted what you wrote about permittivity of free space etc.. Dielectric constant is used to say how the effective permittivity is affected by adding a dielectric, surely. If space has a dielectric constant of Unity, then the quantity is totally redundant. But. again, is this really relevant to the price of fish and chips?

 

[Ratch]

Nevertamed,

they are not just 2 floating caps, they are two points with a imbalance of charge between them, one fully negative charged plate (from the first capacitor) and one fully positive charge plate (from the second capacitor), forget the fact both plates belong to different capacitors, they are still 2 points with a voltage present, if you set a path there will be a current until the charge depletes

I cannot visualize what you are describing without calling the the second capacitor floating. A schemat is needed.

i didnt say caps and batts are identical, i said "in a way, caps are a lil like a battery" i meant they both have similarities

I think the differences outnumber the similarities.

maybe they like speaking metaphorically like lots of people...

They are supposed to be a technically oriented organization, not an entertainment production company. They could have come up with something much better like "exterior excursion", or like they did in the beginning, "extra-vehicular activity' (EVA), both of which are correct. It is a shame that the lame-stream media just swallowed it, and gave them a pass. Ratch

 

[Ratch]

sophiecentaur,

I googled 'equivalent circuit capacitor antenna' and got more hits than I had time to read. They establish the principle that I'm getting at. The fact that antennae are mostly used for AC is not a hinderance because the situation for the DC condition is only arrived at after a change - not DC - in the situation.

I sure am glad I can see a schematic. I never would have understood the circuit without it. I can't argue with you because I don't quite know what you are proposing. Antenna circuits desire resonance, which means some inductance, too. This is probably a good time to end this thread, because the explanations are too involved and complex to benefit anyone.

Ratch

 

[Ratch]

sophiecentaur,

Both would be correct and acceptable in the appropriate context.

That is the point, both are correct.

If we are talking about Charge (which I though we were), then why not allow our experiment to "charge a capacitor"?

That is the point, caps do not get charged.

If we are talking about supercapacitors and road transport then it's arguable that 'energising' them would be appropriate

Energizing a capacitor is always correct.

but did we ever in this World, talk about Energising a Battery?

No, but it would be a correct description of an action.

"Energiser" is a trade name for Ever Ready High Power batteries but that, I think, is aimed at describing what they actually do to your toys and not what Ever Ready do to their batteries.

Yes, you are correct.

Anyway, I did that diagram (above post). Hope it makes sense for you.

Nice diagram. I guess you are saying that there are a couple of capacitors that will be energized when the battery becomes active. At least I know where the caps are located and their relationship in the circuit.

OH - I just spotted what you wrote about permittivity of free space etc.. Dielectric constant is used to say how the effective permittivity is affected by adding a dielectric, surely. If space has a dielectric constant of Unity, then the quantity is totally redundant. But. again, is this really relevant to the price of fish and chips?

It shows that everything has dielectric constant, and the dielectric constant of materials is relative to the permittivity of free space.

Ratch

 

[sophiecentaur]

sophiecentaur,
I sure am glad I can see a schematic. I never would have understood the circuit without it. I can't argue with you because I don't quite know what you are proposing. Antenna circuits desire resonance, which means some inductance, too. This is probably a good time to end this thread, because the explanations are too involved and complex to benefit anyone.

Ratch

Any wideband antenna cannot be highly resonant. The best you can do is to get as wideband a match as possible. Matching is the only reason to aim at resonance and the resonant half wave dipole is only the start in the art of antenna design. Many receiving antenna operate far from resonance - your portable VHF sound receiver antenna (a simple version of an unbalanced radiator, which would never be satisfactory for transmission aamof) being a typical example.
In any case, I only introduced the antenna idea as an example of reducing a circuit to its 'equivalent' components (with or without inductance). If you are not aware of this as a general way of approaching problems then it could be worth your while to follow it up. It is a very powerful tool, used in many instances.

This may be a 'personal view' as my original idea was to suggest a possible approach to the original problem that was posed. I have not seen any alternative proposed on this thread so it at least suggests a way through.

 

[justsomeguy]

I have not seen any alternative proposed on this thread so it at least suggests a way through.

The answer was given a page or so back, though not in direct reply to the OP. His question was:

...plate of a different capacitor, why won't 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?

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.

 

[Ratch]

sophiecentaur,

Any wideband antenna cannot be highly resonant. The best you can do is to get as wideband a match as possible. Matching is the only reason to aim at resonance and the resonant half wave dipole is only the start in the art of antenna design. Many receiving antenna operate far from resonance - your portable VHF sound receiver antenna (a simple version of an unbalanced radiator, which would never be satisfactory for transmission aamof) being a typical example.
In any case, I only introduced the antenna idea as an example of reducing a circuit to its 'equivalent' components (with or without inductance). If you are not aware of this as a general way of approaching problems then it could be worth your while to follow it up. It is a very powerful tool, used in many instances.

This may be a 'personal view' as my original idea was to suggest a possible approach to the original problem that was posed. I have not seen any alternative proposed on this thread so it at least suggests a way through.

How did you get involved is describing a half wave resonant dipole VHF antenna? I am sure the OP has no idea what you are talking about. I see it as a solution in search of a problem.

Ratch

 

[sophiecentaur]

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]

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]

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.

love it

 

[sophiecentaur]

love it

Handbags at dawn?

 

[justsomeguy]

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.

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.

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]

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

 

[jaydnul]

"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]

"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 won't 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]

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]

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]

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]

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]

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.

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]

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?