I was wondering what would happen in the following case:

enter image description here

There's an electric field and points A and B nearby the field. As you can see, the's no electric field at those points. Will there be a current flow from point A to B once I connect them with a wire?

According to the definition of voltage, it's a line integral of the electric field over the path from A to B, so in this case the voltage across A and B won't be zero. But I think that only electrons in the area of non-zero electric field will move and because A is not in a non-zero electric field, no electrons will flow from that point towards B.

If I'm wrong, then is it because the electrons located in the middle part of the wire will move towards B, and that will create a shortage of electrons "on the left edge of the non-zero electric field" (looking at the image) and the electrons will be pulled from the area near A?

  • $\begingroup$ You cannot have electric field lines starting and finishing in space as you must have charges present. So how is it that the electric field at $A$ and $B$ is zero? $\endgroup$ – Farcher Mar 2 '17 at 21:16

Yes, the "shortage" thinking is the thing. If you cram electrons together on a wire, they will spread out and place themselves as far from each other as they can - that will end in an even spread throughout. They all set up an electric field (all charges do that), but that cancels out when they have found their equilibrium positions; everything is then pushed equally in all directions.

If you now remove one electron, the others will rearrange. The neighbours to this "hole" will now be repelled more from the other side than from this side - that is a potential difference, and so they will move towards lowest potential as always.

You simply remove a source of electric field, and the others' fields will not balance out, so they have to rearrange until they do again.

As @garyp mentions in comments, the current that is introduced when an external electric field is set up on a isolated piece of wire of course lasts very, very short. This is an open circuit, and very soon the moving charges accumulate at one end and prevent any further charge from arriving. No current then flows. But we could imagine the wire closing as a ring, and then current could flow.

  • $\begingroup$ @user5539357 You mean "positive charge carriers" rather than electrons, right? Note that the current that is produced during the rearrangement lasts for a very very short interval of time before a new equilibrium is reached and current stops. $\endgroup$ – garyp Mar 2 '17 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.