I actually have three related questions:
An open circuit chemical cell separates charges creating a surplus of electrons on its negative terminus and a shortage of electrons on its positive terminus. The separation of charges is conducted by chemical reactions. As electrons accumulate on the negative electrode, they repulse each other, so it takes more work to put any additional electrons in the negative electrode, so eventually, after a certain potential difference is achieved, the chemical reactions are inhibited and stopped.

1) If we connect a long wire to the negative terminus, the accumulated electrons will evenly distribute themselves along the wire, because of their mutual repulsion. Thus, the charge density at the negative electrode will decrease. Will this cause the battery to relaunch the chemical reactions and put more electrons in the negative electrode?

Let's say we connected two long wires to a chemical cell, one to it's negative electrode, another to it's positive electrode (the wires do not connect to each other). The negative wire will thus acquire some extra electrons, the positive wire will have a shortage of electrons.

2) If now we disconnect the wires from the battery, will they remain charged? And how is this charge mathematically related to the emf of the battery?

If we connect the wires to the battery as described above, the potential difference between their termini will be about the same as the emf of the battery.

3) If we disconnect the wires from the battery and they remain charged, what will be the potential difference between the wires?

Thank you.

 

Think of the wires as the plates of a tiny capacitor, or think of a tiny capacitor as two wires that are glued together and almost touching.

Yes, a connected wire will have the voltage it's connected to. Yes, the battery has to supply that charge movement. Yes, a capacitor stays charged when you disconnect it. That capacitance will store the voltage that was applied to it.

Did I get them all?

 

PLEASE STOP POSTING THE SAME QUESTION OVER AND OVER!

 

now there are 4 copies of this post . wont one do?

 

Sorry for posting it multiple times!
My bad. I did not see it was posted. The form froze and did not switch as they usually do.
Is there any way to combine the answers into one post and remove the others?

 

A moderator will log in soon and fix this.

 

Think of the wires as the plates of a tiny capacitor, or think of a tiny capacitor as two wires that are glued together and almost touching.

Yes, a connected wire will have the voltage it's connected to. Yes, the battery has to supply that charge movement. Yes, a capacitor stays charged when you disconnect it. That capacitance will store the voltage that was applied to it.

Did I get them all?

Thank you for the answer, #12.
Do I understand you correctly, that the voltage between wires in question # 3 will be equal to the emf of the battery, after the wires are disconnected from the battery?
And the longer a wire we connect to the negative electrode of the battery, the more surplus electrons this wire will get? Thus, connecting a battery to different open circuit wires might eventually deplete the battery?

 

yes, yes, yes
but this is only in theory. YOU are never going to connect enough different wires to a battery to deplete it.

 

yes, yes, yes
but this is only in theory. YOU are never going to connect enough different wires to a battery to deplete it.

Thank you for dissipating my bewilderment.
Somehow I just can not go on without understanding things.

 

yes, yes, yes
but this is only in theory. YOU are never going to connect enough different wires to a battery to deplete it.

Hi, #12. Now I am in doubt again.
Let's say we have a battery. We connected an open circuit wire to it's negative electrode. The wire took some electrons. We disconnect the wire, and connect another one. And so on, many times. Each time a wire takes the same amount of electrons from the battery.
Somehow this does not seem possible.
Since the battery as a whole becomes positively charged, and it won't be giving up it's electrons, unless there is a closed circuit.
Which brings us back to my original questions.

 

This goes back to the tiny capacitor example. The positive button on the battery is the other end of the tiny capacitor. It may be less than a picofarad of capacitance compared to the wire you attached, still, it's the other end of the circuit. If you connected a billion, billion different stray wires to one end of the battery, you might have some practical effect.

or...you could just attach one massive capacitor and get back to getting some work done today.

 

This goes back to the tiny capacitor example. The positive button on the battery is the other end of the tiny capacitor. It may be less than a picofarad of capacitance compared to the wire you attached, still, it's the other end of the circuit. If you connected a billion, billion different stray wires to one end of the battery, you might have some practical effect.

or...you could just attach one massive capacitor and get back to getting some work done today.

I still do not understand. The battery as a whole is neutral, it has the same number of negative and positive charges. We connect a wire to the negative electrode of the battery, and then disconnect it. Now we have a wire that is negatively charged, and a battery, which as a whole has a positive charge. If we connect another wire to the negative end of the battery, the electrons will not flow into the wire, because the positive charge of the whole battery will pull them back. Is it not so?

 

It is not. No electrons flow into the wire that are not replaced from the air at the other pole of the other battery. You are correct that the battery maintains a neutral overall charge.

 

It is not. No electrons flow into the wire that are not replaced from the air at the other pole of the other battery. You are correct that the battery maintains a neutral overall charge.

So, in the vacuum, if we connect a wire to the negative electrode of a battery, will this wire receive a surplus of electrons from the battery, or not?

 

Nope. What force would drive them there?

 

Nope. What force would drive them there?

But there is a surplus of electrons on the negative terminus of the battery. They repel each other. When we attach a conductor to the negative terminus, the surplus electrons spread out along the conductor because of this mutual repulsion, so that the system attains the minimum possible energy state. Where I am wrong?

 

So your position is that when I move this switch, the second, third, and 4th capacitor will not charge.

 

If you are going to ask about pico-amp, absolute limits, you have to be willing to think in pico amp terms.

 

If you are going to ask about pico-amp, absolute limits, you have to be willing to think in pico amp terms.

If we consider it all in the vacuum, is it correct to say that conductance of the vacuum can be approximated to that of a 0.001 pF capacitor?

 

The last time I measured a vacuum with my ohm meter, it read infinite ohms, so that makes the conductance of a vacuum zero. The tiny capacitor is caused by the distance between the battery terminal and the spare piece of wire you connected. It is true that the dielectric constant of a vacuum is different from that of air, but that's not the point here. You are asking about very, very, very, very small quantities and they will still be very, very, very, very small in a vacuum.

We can make this as complicated as you want. You show me your calculation for this stray capacitance between a wire and a nearby piece of battery terminal in air and I will show you the conversion factor for that capacitance in a vacuum.