How Does Potential Difference Cause Electron Movement in Electric Circuits?

[Tosh5457]

Hi, I'm studying this electric circuits in electromagnetism now, and I never studied this before, so my questions probably have very simple answers :P.

So, if there is a battery generating an electric field between its plates, there is a potential difference between the plates. If the electric field is uniform, \DeltaV = Ed, d is the distance between the plates.
I understand this, what I don't understand is how does this potential difference make charges move on the wire? Due to the electric force exerted by the protons on the electrons, electrons should stay on the plates, right? How can there be an electric field on the wire that makes electrons move?

 

[Drakkith]

A potential difference means that there is an unequal charge between two points. There are MORE protons than there are electrons on the positive side, and LESS protons than electrons on the negative side. As the electrons cannot go through the battery itself to reach the positive charges, they exert a force on the wire through the battery terminal, pushing the wire electrons away while the positive terminal attracts them. If the battery were "dead", that means that there is no more potential difference. The battery and its terminals are all neutral, so no force is attracting or repulsing the electrons in the wire anymore.

 

[Tosh5457]

Oh so what makes electrons move is the force exerted by other electrons, and as they move, more electrons will move, I see :) But can we assume the electric field is uniform on the wire though?

If for example I put a battery, that generates a potential difference, in a closed circuit, would it go dead?
And if I put 2 batteries in a closed circuit, with the same potential difference between its plates, in this disposition: -+ +-, there would be no current? And if one of the batteries had a bigger potential difference than the other, in that case there could be a current right?

Thanks!

 

[Drakkith]

Oh so what makes electrons move is the force exerted by other electrons, and as they move, more electrons will move, I see :) But can we assume the electric field is uniform on the wire though?

The excess of negative charge on one end and an excess of positive charge on the other end is what causes them to move, yes.

If for example I put a battery, that generates a potential difference, in a closed circuit, would it go dead?
And if I put 2 batteries in a closed circuit, with the same potential difference between its plates, in this disposition: -+ +-, there would be no current? And if one of the batteries had a bigger potential difference than the other, in that case there could be a current right?
Thanks!

If you attached a copper wire between the two terminals on a battery it would most definitely go dead. And probably in a sparking smoking way, so I don't suggest trying it.

If you attached the positive to the positive and the negative to the negative between 2 batteries, if they were identical batteries then nothing would happen. If one had more or less voltage than the other then yes you would get some current flow. Again, I don't suggest this as you can easily destroy a battery and hurt yourself.

 

[RedX]

If you have two infinite plates with opposite charges on them, and stick a wire between them, then the two plates will discharge because of the electric field between the plates (there is no way to have a wire connecting two infinite plates that is not between the two plates).

If you have two finite plates, you can connect them with a wire without having the wire directly between the two plates. But the fringe field lines still go from positive to negative, so there will still be a electric field in the wires.

Anyways, that's my interpretation of how a battery works. As for the electric field, it will not be uniform unless the wire is uniform. What happens is that if part of the wire is more resistive, there will be a buildup of charge at the leads until the current is the same everywhere, no matter the particular value of the resistance at a certain location on the wire. The buildup of charges creates a stronger electric field, which will cause a greater potential drop across the part of the wire that has a lot of resistance. And as you know, the greater the resistance, the greater the potential drop.

 

[Tosh5457]

Ok, I see, thanks for the answers, I wish my professors explained the way you 2 did :)