I believe your assumption is incorrect.

When you connect a long wire to the negative terminal of a battery, the wire is only polarized.....there is no charge transfer.

The reason that there is no charge transfer is because of the positive pole. The positive pole keeps the same number of electrons on the negative pole.

If you touch the wire to a mono-pole charge.......i.e.......a charged balloon or a charged sphere......then you can charge the wire.

Thank you for the answer, BR-549.
Polarization of the wire would redistribute the free electrons along the wire. Thus the density of the electrons would be lower at the end attached to the battery, and it would be higher at the opposite end.
This brings another question: why the potential difference between any point of the wire and the positive terminus of the battery be about exactly the same?
Also, can you please support your answer with formulas or references or prove it any other way?

 

Let me throw out a similar thought experiment: Suppose a properly protected linesman is working on a 1MV DC power line. He cuts out a 5 meter section and throws in down to you, and you're standing on the ground with your bare feet in the mud. Do you feel anything? Ignore air conduction.

I propose you will not feel a thing because the wire does not carry enough charge to detect. Love to hear other's analyses.

The question now is not how much a charge the wire would have. The question is would the wire have any charge at all.

 

Now I'm sure it's BillyMayo, or his clone. Now he doesn't have a theory, he is as baffled as we are, and he's teaching several of us how this works...all at the same time.

Thanks for the reminder, Santa Claus
It's been a long time since I was well and properly trolled.

I am not trolling you. It is you who replied to my thread. And you still keep posting in my thread, but do not provide answers in your posts. So why are you keep posting if you do not have ideas?
I do not know what BillyMayo has done to you, I have no acquaintances with such a name. Do not blame me for his deeds.

 

So what happens if you connect a battery to a capacitor, then disconnect the capacitor from the battery, and then separate the two plates of the capacitor (each still connected to the dangling wire that previously connected them to the battery)? Gee, you have two conductors that are charged, one positively and the other negatively. How is this any different than connecting a long wire to a terminal of a battery (hint -- it isn't, the wire is merely one plate of a capacitor with the other plate being, if nothing else, the positive terminal of the battery).

There is no doubt that a capacitor can be charged in a closed circuit. There is no doubt, that a wire can carry a charge. There is doubt, however, that a capacitor or a wire can be charged in an open circuit.

 

There is no doubt that a capacitor can be charged in a closed circuit. There is no doubt, that a wire can carry a charge. There is doubt, however, that a capacitor or a wire can be charged in an open circuit.

Again, connect a wire to a battery and you have a made a capacitor that is connected to a battery. There is NO difference. It IS a capacitor.

 

What are you looking for that constitutes a "solution" -- more to the point, what did my response not cover that you are looking for?

Hi, WBahn, I just completed reading your blog post. It is curious.
Though, I believe that your conclusion: "We can now appreciate that the number of ions that it takes is so small that there is effectively no accumulated charge on the terminals and that the timescale on which it happens is so small that it is effectively instantaneous", - is not completely correct.
But let's rather talk about my questions.
Let me ask you a direct question:
Do you believe, that if we connect an open circuit wire to a negative electrode of a battery, electrons from the electrode will flow into the wire, and that would cause the battery to "pump" more electrons into the negative electrode?

 

Again, connect a wire to a battery and you have a made a capacitor that is connected to a battery. There is NO difference. It IS a capacitor.

There is a difference: a capacitor when charged is neutral as a whole, a wire, on the other hand, when charged is not neutral.
Do you believe a capacitor can be charged in an open circuit?

 

The reason people have so much trouble understanding these things is because people forget that charge is a SURFACE AREA entity.

The difference between a single open wire connected to a battery and a capacitor connected across a battery is the difference between polarization and charging.

The only reason you have voltage on a battery terminal is because the terminal has a limited surface area.

As you increase the length of the wires connected to the terminals....the voltage goes down.

If you increase the length far enough....you will have zero voltage at the end of the wires.

If you continue to add wire......the voltage at the terminals will reach zero also.

We have completely diluted and dissipated the charge potential with surface area.

To make a potential or voltage, we have to have accumulation....to have accumulation...we must have limited surface area.

Also remember that there is a difference between a dipole source and a monopole source.

 

I am not trolling you. It is you who replied to my thread.

Anyone that has been married will recognize this logic.

 

Hi, WBahn, I just completed reading your blog post. It is curious.
Though, I believe that your conclusion: "We can now appreciate that the number of ions that it takes is so small that there is effectively no accumulated charge on the terminals and that the timescale on which it happens is so small that it is effectively instantaneous", - is not completely correct.
But let's rather talk about my questions.
Let me ask you a direct question:
Do you believe, that if we connect an open circuit wire to a negative electrode of a battery, electrons from the electrode will flow into the wire, and that would cause the battery to "pump" more electrons into the negative electrode?

Asked and answered.

 

The reason people have so much trouble understanding these things is because people forget that charge is a SURFACE AREA entity.
The difference between a single open wire connected to a battery and a capacitor connected across a battery is the difference between polarization and charging.
The only reason you have voltage on a battery terminal is because the terminal has a limited surface area.
As you increase the length of the wires connected to the terminals....the voltage goes down.
If you increase the length far enough....you will have zero voltage at the end of the wires.
If you continue to add wire......the voltage at the terminals will reach zero also.
We have completely diluted and dissipated the charge potential with surface area.
To make a potential or voltage, we have to have accumulation....to have accumulation...we must have limited surface area.
Also remember that there is a difference between a dipole source and a monopole source.

Man, this does not seem correct.
If we connect a wire to a battery, the voltage drop at the distant end of the wire depends solely on the resistance of the wire. If we take a wire with a larger surface area, but with a lower resistance, the voltage drop at its end will be lower, despite its surface area being larger.

 

There is a difference: a capacitor when charged is neutral as a whole, a wire, on the other hand, when charged is not neutral.
Do you believe a capacitor can be charged in an open circuit?

And the wire plus the other plate of the capacitor (also known as the positive terminal of the battery) is neutral as a whole.

It's a simple concept.

Take a capacitor -- for our purposes let's consider it a small parallel plate capacitor with an air dielectric and let's further assume that each plate is just the size of the battery terminals.

Now connect one plate of this capacitor to the positive terminal of the battery and run a wire to the other side.

Do you agree that we now have a capacitor that is connected to a battery and that it becomes charged?

Do you agree that there is a net negative charge on one plate of the capacitor and a net positive charge on the other?

Now move the plate of the capacitor away from the other plate by some small distance.

Do you agree that we still have a charged capacitor, albeit the capacitance is less.

Now move the plate further away.

Do you agree that we still have a charged capacitor, albeit the capacitance is less.

Keep doing this until the two plates are separate by whatever distance you want. We still have a capacitor and it is still charged.

Do you understand that we could do this all over again except not have a parallel plate capacitor at all by merely bring a wire from the negative terminal around and close to, but not touching, the positive terminal and that we then have a small capacitor that is charged?

 

Asked and answered.

Ok. Now let's proceed.
Nothing is connected to the battery. The battery is at equilibrium. There are, say, 10 positive charges on its positive terminus, and 10 negative charges on its negative terminus. We connected a wire to the negative terminus. 2 electrons moved from the negative terminus into the wire. The battery pumped additional 2 electrons into the negative terminus.
Now we have a new charge distribution between the battery's termini: there are 12 positive charges on the positive terminus and 10 negative charges on the negative terminus.
Do you believe that now the potential difference between the battery's termini is the same as it was at the equilibrium? Is it higher?

 

There is no charge transfer between negative terminal and wire.

 

Now connect one plate of this capacitor to the positive terminal of the battery and run a wire to the other side.

What do you mean by "run a wire to the other side"? Do you connect the opposite plates of the capacitor to the opposite termini of the battery?

 

There is no charge transfer between negative terminal and wire.

I am inclined to think so too, but I would like to see a proof to be certain.

 

There is no point in discussing your logic.
Batteries and electrons do not work like that. When you connect a wire to a battery terminal, electrons don't move from one place to another the way you have describe it.

 

ok.....measure the voltage on a neutral wire. Record.

Momentarily touch the wire the the negative terminal.

Measure the voltage on the charged wire. Record.

Was there a difference in voltage?

 

There is no point in discussing your logic.
Batteries and electrons do not work like that. When you connect a wire to a battery terminal, electrons don't move from one place to another the way you have describe it.

Can you prove your point? How do you come to know that? Sources, equations ...?

 

ok.....measure the voltage on a neutral wire. Record.
Momentarily touch the wire the the negative terminal.
Measure the voltage on the charged wire. Record.
Was there a difference in voltage?

This is not a proof. If we are talking about small charges, they will dissipate before a voltmeter could measure potential difference. Say, take a 1pF capacitor, charge it, disconnect it, measure voltage between its connectors. My guess is a voltmeter will show 0. Yet we know that there was a charge on the capacitor.