I specifically used a scenario in which the charge did NOT increase on the plates. In fact, you could increase the charge on the plates and still see the voltage go down. And you can have a material in between the plates that starts with a low dielectric constant and then, through a chemical reaction, ends up with a high dielectric constant such that, without doing anything to the system, the charge on the plates remains the same but the voltage goes down.
If we connect a long wire to a battery, will battery produce more electrons?
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WBahn.........
The setup that I am referring to is a 12 VDC car battery in deep space.....with the negative terminal at the West coordinate and the positive terminal at the East coordinate.
The wire is connected and extends Westward from the negative terminal. The wire never gets closer to the positive terminal....than the negative terminal does.
You had to cheat to get a voltage by cutting the wire.
But that still does not prove your point........The charge you would measure......still does not come from negative terminal. It comes from the other piece of the wire that you cut.
The setup that I am referring to is a 12 VDC car battery in deep space.....with the negative terminal at the West coordinate and the positive terminal at the East coordinate.
The wire is connected and extends Westward from the negative terminal. The wire never gets closer to the positive terminal....than the negative terminal does.
You had to cheat to get a voltage by cutting the wire.
But that still does not prove your point........The charge you would measure......still does not come from negative terminal. It comes from the other piece of the wire that you cut.
Take your battery and plot the electric field lines in space. Now draw the line where you plan to put your wire. Unless it is everywhere perpendicular to the field lines (which means that it never leaves the surface of the negative terminal) then there is a voltage gradient along that path. Now plop down your wire. Gee, you have a conductor in the presence of a voltage gradient and, guess what, current flows! Not much and not for long. The wire takes on a charge distribution such that the fields in space that used to be tangential to the surface of the wire are now normal to it and that requires that there be excess electrons on the wire. Where did they come from? Why, they came from the negative terminal of the battery!
And if you don't touch, you get a different distribution than if you do.
But the charge on the + pole of the battery will change, too.
Look, we are going nowhere, so let's just agree to disagree and call it a day.
I shouldn't do this, but I'll throw something out. This whole discussion is confused by using a battery. Replace it with a charged balloon. Then the scenario works - touching a (neutral) wire to the balloon will draw off some of the charge. The energy required to drive charge onto the wire is provided by the charge reduction experienced by the balloon. You could continue to touch pieces of neutral wire to the balloon until it had lost almost all its charge to the pieces of wire.
This is very different than touching a neutral wire to a neutral battery, where there is no energy with which to force a charge onto the wire. The chemical energy stored inside the battery cannot be accessed for that purpose, without completing a circuit to interact with that chemistry.
This is very different than touching a neutral wire to a neutral battery, where there is no energy with which to force a charge onto the wire. The chemical energy stored inside the battery cannot be accessed for that purpose, without completing a circuit to interact with that chemistry.
So what you're saying is that there's a difference between transfer of charge and the transfer of potential energy?
In your balloon example, the wires would be acting as capacitors, while this is not possible in a battery (an ideal battery, that is) whose circuit is open... is that it?
A much better analogy would be to use two balloons, one positively charged and one negatively charged, that are close together so that you have zero net charge. Now when you touch a wire to one that is negatively charged, you will still transfer some charge from that balloon to the wire. But after you take that away and bring another wire in, you are bringing it in to a system that has a net positive charge which will try to attract some of the electrons off the wire. But in close proximity to the balloon, the nearness of the excess negative charge will still try to push some electrons onto the wire. At some point you will reach a situation in which the overall positive charge near the pair of balloons creates a strong enough field, even near the surface of the negatively charged balloon, so that charge is neither transferred to or from the wire when it is touched to the balloon.
The only thing that is left to make the analogy complete is to add a source between the balloons that transfers electrons from the positively charged balloon to the negatively charged balloon in order to maintain the same voltage between the two. The voltage between the two now has two components. An overall net positive charge for the entire system riding on the positive balloon superimposed with a net charge separation between the two. The result is that if you pull X electrons off of the negative balloon, you need to transfer fewer than X electrons from the positive balloon to the negative balloon in order to establish the same voltage.
The wires are still a capacitor. That is completely separate from whether the battery is ideal or not.
What I meant about ideal, is that the battery would have no internal resistance. But I understand if you're saying that this doesn't affect the answer. So then, are the poles of an ordinary battery charged?
Yes, but not by very much at all. You might read through that blog entry I linked earlier in the thread:
http://forum.allaboutcircuits.com/blog/a-battery-isnt-a-capacitor.588/
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Excellent... thanks for the tip
I'm already done with it... in case you wanna know...
Why the need to close it? Even if the TS is a troll, things have settled down into a rational discussion that is directly related to the original question which, in and of itself, was reasonable and worth discussing.
Having said that and while I'm more than willing to remain involved, I'm finished unless others still want to continue.
Not sure now if the OP was a troll now... maybe he was just someone (most likely young) genuinely struggling to grasp an abstract concept...
I don't believe that electrons will leave the negative terminal of a battery until new electrons enter the positive terminal of the battery. Positive charges within the battery hold the negative charges inside the battery.
Connecting a long wire to one end of the battery does not, in any real sense, make a capacitor. No current will flow into the wire. Positive charges within the battery would prevent that from happening. The wire becomes a capacitor only when it is held close to the positive terminal. With this additional provision, there would be a small amount of current flowing for a short period of time.
I suggest that the original poster see A Kitchen Course in Electricity and Magnetism by David Nightingale and Christopher Spencer.
Connecting a long wire to one end of the battery does not, in any real sense, make a capacitor. No current will flow into the wire. Positive charges within the battery would prevent that from happening. The wire becomes a capacitor only when it is held close to the positive terminal. With this additional provision, there would be a small amount of current flowing for a short period of time.
I suggest that the original poster see A Kitchen Course in Electricity and Magnetism by David Nightingale and Christopher Spencer.
That's the premise I started with, but the question was phrased as an absolute, one single electron, so I had to consider picofarads diminishing at a distance. Then the question was changed and everything went off the tracks.
