A conceptual question about current

[O.J.]

Hello!

We know that current flows in a circuit due to an electric field set up from the terminals which resembles that of a dipole, right?
Now because it resembles the dipole's field, the electric field is NOT constant everywhere along the wire, and therefore, the electrons experience a different force at every different position along the wires. My question is: based on that, how can we have a CONSTANT current when the force on the electrons & hence there velocity is changing.


NOTE: I understand that for a constant E Field the force on the electrons will be constant and their velocity will be brought to a limit due to the resistance. (Much like a situation of a free falling body that reaches terminal velocity). Is my understanding correct?

Thanks!

 

[chroot]

Current is a macroscopic phenomenon, not a microscopic one. If you look at individual electrons, their behavior is very erratic. On the other hand, there are so many of them that on a large scale, one coulomb of them pass a point in the wire every second for a current of one ampere. The number of electrons in a coulomb is immense; even if five fewer pass this second and five more pass in the next, you'd never notice.

- Warren

 

[O.J.]

Even on a macroscopic level, shouldn't THE OVERALL NET flow of electrons be different at every different location on the wire?

 

[chroot]

That depends on how precise you want to make your measurements. Are you concerned about a few parts per trillion trillion?

- Warren

 

[Doc Al]

We know that current flows in a circuit due to an electric field set up from the terminals which resembles that of a dipole, right?

No. The electric field that drives the current is due to a gradient of surface charge that builds up (very quickly) along the wire.

 

[pam]

What Doc is saying is that E is fairly constant along the length of a conducting wire.

 

[O.J.]

No. The electric field that drives the current is due to a gradient of surface charge that builds up (very quickly) along the wire.

I don't understand at all. Could you elaborate by explaining or providing a link?

 

[O.J.]

And how can charge build up on the conductor surface? This should only happen if its an EXcESS charge, right? Aren't the electrons given off to the conductor from one of the battery terminals taken away from the conductor at the other terminal (therefore the conductor gains no charge in effect)?

 

[Doc Al]

And how can charge build up on the conductor surface? This should only happen if its an EXcESS charge, right? Aren't the electrons given off to the conductor from one of the battery terminals taken away from the conductor at the other terminal (therefore the conductor gains no charge in effect)?

There's no net charge on the wire surface (taken as a whole), but there's a gradient of surface charge that's positive near the positive terminal and becomes negative near the negative terminal.

See my comments in this thread: https://www.physicsforums.com/showthread.php?t=149617&page=2, starting with post #16. Read the attachment in post #18, an excellent article by Sherwood and Chabay on this very issue.

 

[O.J.]

Thanks Doc! -gives him candy- :P

 

[O.J.]

I'm reading the article; it is very interesting and is exactly what I was looking for. However, I'm stuck at page 9, where it says that the electric field due to the capacitor charges will cause accumulation of +ve charge on the right bend and -ve charge on the left bend. This will cause an electric field between these two accumulations that is opposite to that due to the capacitor plates. Question is: why should the accumulation continue till you have a net E field to the left? Shouldn't it stop once the E to the right is equal to the E to the left?

 

[O.J.]

bump :)

 

[kamerling]

I'm reading the article; it is very interesting and is exactly what I was looking for. However, I'm stuck at page 9, where it says that the electric field due to the capacitor charges will cause accumulation of +ve charge on the right bend and -ve charge on the left bend. This will cause an electric field between these two accumulations that is opposite to that due to the capacitor plates. Question is: why should the accumulation continue till you have a net E field to the left? Shouldn't it stop once the E to the right is equal to the E to the left?

But if the net E field is 0, there will be no current in between the bends, and the charges will continue to pile up.

 

[Doc Al]

I'm reading the article; it is very interesting and is exactly what I was looking for. However, I'm stuck at page 9, where it says that the electric field due to the capacitor charges will cause accumulation of +ve charge on the right bend and -ve charge on the left bend. This will cause an electric field between these two accumulations that is opposite to that due to the capacitor plates. Question is: why should the accumulation continue till you have a net E field to the left? Shouldn't it stop once the E to the right is equal to the E to the left?

Oops... Sorry, meant to respond earlier. But kamerling is correct.

I think you're talking about the thought experiment with the capacitor plates discharging through the wire, which is discussed on pages 5-8 (not page 9). Figure 6 shows a temporary configuration of surface charges as the surface charges build up. As kamerling says, the charge will continue to build up and rearrange until a steady state is reached. (Of course, with the capacitor example this state is only quasi-stationary as the capacitor will eventually discharge.) A steady state must have the net electric field equal throughout the wire, else the current would be different in different sections.

 

[O.J.]

what's quasi?

 

[O.J.]

But if the net E field is 0, there will be no current in between the bends, and the charges will continue to pile up.

No, since E is 0, charges will NOT move anywhere. They won't pile up nor will there be a current.

 

[Doc Al]

what's quasi?

Quasi-stationary, meaning almost stationary. The surface charge distribution is actually slowly changing as the capacitor discharges. Once discharged, no more current flows.

 

[Doc Al]

No, since E is 0, charges will NOT move anywhere. They won't pile up nor will there be a current.

If E were 0 everywhere, that would be true. But if E is 0 at only one point, it won't last long as the charges rearrange around that point and change the field again.