I am studying EE and have (unfortunately) only found unsatisfactory answers to this question. Here is my understanding and confusion thus far.

When a battery is connected to a wire, the electric field of the battery is said to be "confined" or at least somewhat concentrated to/along the shape of the wire, no matter how many "loops" or whatever strange configuration the wire makes up.

I understand that the cause for current flow in a wire is from the impressed EMF of a voltage source (e.g a battery) forcing the electrons to accelerate, ultimately causing a net drift velocity. What I don't understand is how this electric field travels so well inside the conductor, compared to in the surrounding free space.

One of the puzzling things to me can be seen in this moment in a video which simulates a popular transmission line experiment (Ben Watson's YouTube video "Response to Veritasium - In Depth Explanation").

As can be seen, the electric field lines are strongly concentrated and parallel along the outside of the wire. I am presuming that they are even stronger within the wire, but the simulation seems to not show this (I guess this is due to the idealization of the conductors i.e. they are able to immediately re-arrange in such a way as to reach static equilibrium instantly, and therefore the software considers there to be no net electric field within them).

How does this "concentration" take place? How does the field propagate in such a "kind" manner? How would resistors come into play? Do they decrease this effect?

  • $\begingroup$ Related meta post: physics.meta.stackexchange.com/q/13917/2451 $\endgroup$
    – Qmechanic
    Jun 2, 2022 at 10:50
  • $\begingroup$ @Qmechanic I don't believe my question is about Veritasium's initial thought experiment, which was mainly centered around energy transfer. I think it is more about electric field propagation, which just happened to be showcase in a response to his original video $\endgroup$
    – Gary Allen
    Jun 2, 2022 at 11:01
  • $\begingroup$ Electric field within a wire corresponds to resistive loss; unless you are losing significant amounts of power to heat in a wire there isn't a significant electric field in it. Remember that "objects in motion stay in motion" roughly applies to electrons in a wire too; you only need a significant fields near/in voltage sources/loads to get the electrons moving. The software is probably right to show (near-)zero field in the wire, even during the start-up transient. The fields outside the wire should quickly fall off inside the wire. $\endgroup$
    – HTNW
    Jun 2, 2022 at 14:41
  • $\begingroup$ Ahh okay. So the electrons will redistribute to create an electric field which opposes the impressed field, thereby bringing the electric field inside the conductor to 0 and allowing a constant flow of current in the wire due to the near zero resistance of the wire? How is the parallel electric field just outside the wire explained then? @HTNW I wonder if you could help me understand how this "impressed" field from the battery (is that the right terminology) then still "follows" the wire's path? $\endgroup$
    – Gary Allen
    Jun 2, 2022 at 14:44
  • $\begingroup$ @GaryAllen The perpendicular field comes from the radial (re)distribution of charge: charges on the outside of the wire produce a field, which shows up on the outside but is quickly cancelled by charges on the inside. The wire is basically acting like a line charge, so the field naturally has strength about $1/r$ in the distance from the wire. I don't think it's particularly "concentrated" by anything... can you clarify what you mean by that? $\endgroup$
    – HTNW
    Jun 2, 2022 at 14:58

2 Answers 2


This is indeed something that seems rather mysterious at first. It is well known in the literature, but is often not discussed in a typical EE curriculum. It seems like the physics classes don’t think it is an important concept and the circuit theory classes assume you already know it.

The “field concentration” phenomenon that you are interested in is driven by the surface charges on the conductor.

Outside the wire the E field is mostly radial and the B field is circumferential, so the Poynting vector is mostly longitudinal and energy is transported in the longitudinal direction outside the wire. However, inside the wire the E field is mostly longitudinal and the B field remains circumferential, so the Poynting vector is radial and energy flows inward to be dissipated in the wire.

Now, notice that at the surface of the wire the E field changes abruptly from mostly radial to longitudinal. That sharp bend in the E field implies a charge on the surface, called the surface charge. It is this surface charge that is responsible for producing the longitudinal E field inside the conductor regardless of the external field.

A side effect of the surface charge is that it has its own field. It is this field which you noticed that is concentrated around the wire. In the animation you linked to, the source produces a EM field which moves in the space outside the wire. Because the wire is thin, when the field reaches a segment of the wire it takes very little time for some charge to move to the surface so it almost instantly produces the surface charges I described. Thus almost as soon as the EM fields reach the wire the field takes the concentrated configuration you noticed.


An electric field parallel to the wire suggests a voltage change along the wire. These are associated with resistive losses due to the finite conductivity of the wire material. Due to Ohm's law, if the resistive wire carries a current there has to be an electric field along the wire direction as well.

Note that this means the Poynting vector is directed radially inwards towards the wire. This may seem counterintuitive, as you would expect the energy to be flowing along the wire.


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