## If the positive plate on a charged capacitor 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?
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 The plates are separated enough and the electrons cant 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

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

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

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.
 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.
 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

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 Quote by 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
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.

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

 Quote by 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
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

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 Quote by Ratch 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.
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 Quote by Ratch 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

 Quote by Ratch Nevertamed, A battery does not accumulate electrons on either pole Ratch
batteries create a constant imbalance of charge, to maintain a DC voltage

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 Quote by Ratch 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.

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

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 Quote by Ratch 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.

 Quote by Ratch 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

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