I am plagued by this simple question that what force exactly causes the motion of an electron in the middle of a lattice of electrons of a wire?

a. Is it an electric field created inside a wire due to battery?

b. Or is it that the initial electrons at the ends of the wire are pushed/sucked by battery which in turn push/suck more electrons right to the middle of the wire.

My bet is that second option is the major player. All help is appreciated.

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    $\begingroup$ It's the electric force. Since there is a voltage drop on the wire, there is an electric field. Stay away from electron pushing/sucking, that's not a good concept. $\endgroup$ – CuriousOne Apr 30 '16 at 18:31
  • $\begingroup$ A drop can also be caused by the lack of electrons or their excess. No? $\endgroup$ – Kraken Apr 30 '16 at 18:32
  • $\begingroup$ It's an equilibrium between charge and field and we can describe it with Maxwell's equations and a model for the interaction between electrons and the metal lattice, but electrons still don't "push" each other. They are moving in the field. $\endgroup$ – CuriousOne Apr 30 '16 at 18:37
  • $\begingroup$ I did not mean pushing in the physical sense. I was like: don't the electrons have field of their own. If one is pushed forward by the battery, wouldn't it push the next one in turn due to its own field interacting with the other electron. $\endgroup$ – Kraken Apr 30 '16 at 18:39
  • $\begingroup$ Also, the field of battery is bound to get weaker as it moves away from the battery, isn't it? $\endgroup$ – Kraken Apr 30 '16 at 18:41

I would say it is both! Because of the abundance of electrons, the electric field at the battery pole/boundary, at the instant of turning on the switch (t=t0), is quickly (within a few Debye lengths) screened and cannot possibly reach the electrons further down the wire. However, the electrons at the vicinity of the pole that do feel the effect of electric field reposition themselves (pushed/pulled by the force), creating electric dipoles (by partnering up with the positive nuclei), so that the original field can further penetrate into the wire. This reconfiguration of the charges mediates the electric field through the metallic wire and happens very very quickly when the circuit is formed (switch is turned on). So you see, the electrons do push each other, but only through the field they themselves create. It only takes a tiny fraction of the electrons (small polarization density) to establish the field. This is no other than the subject of electric field in a material medium by the way. The only difference being usually they do the time-independent treatment whereas I believe you are asking about the time-dependent aspect of it. The timescale all this happens is $\epsilon/\sigma$ where $\sigma$ is conductivity and $\epsilon$ the permittivity of the medium.

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  • $\begingroup$ So, going down the wire, its majorly electrons pushing each other. Copy? $\endgroup$ – Kraken May 1 '16 at 8:10
  • $\begingroup$ If by "pushing" you mean creating the electric field that does the pushing, then yes. Also see this great answer. $\endgroup$ – Pooya May 1 '16 at 16:43
  • $\begingroup$ Hey I know its not the right place to ask but I was wondering if you could answer my other question: physics.stackexchange.com/questions/254516/…" .Thanks. $\endgroup$ – Kraken May 8 '16 at 12:29

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