I would like to start a discussion in which we have an ongoing debate elsewhere with no convincing solution in sight. So I decide to ask here:

The question is mainly academic: Is a current carrying conductor charged? I do not mean any charges imposed by electrostatic boundary conditions, because it is clear that a voltage drop between two conductors in a loop , which are connected with a resistor at their ends are making up a capacitor and therefore carries charge. I talk about a free conductor without any other boundary conditions.

There are many papers on this topic, in principle they state, that the bulk of a current carrying conductor gets slightly negatively/positively charged when a current of negatively/positively charged carriers flows through.

It took me a while to understand this, but in the meantime I think this result is beyond discussion. As an example there is a slab which carries current in z-Direction. The image depicts the cross section of this slab.

enter image description here

A Lorenz force acts on each moving carrier and it would get deflected in y-direction. Because the conductor is infinite in flow-direction, there must be an equilibrium state, where the carrier flows parallel to the conductor. So an additional force is needed. A simple analysis shows, that the Field B is

$$B = \mu_0 J \cdot y$$

and the Lorenz force (ignore sign now) on an electron, moving moving with speed v is

$$F_L = \mu_0 J \cdot y \cdot e \cdot v$$

So there must be a compensating electric field of amount

$$E = \mu_0 J \cdot y \cdot v$$

Because of

$$\vec \nabla \cdot \vec E = \rho/\epsilon_0$$

We must have a bulk charge density

$$\rho = \mu_0 \epsilon_0 J \cdot v = \frac{v^2}{c^2} \cdot e \cdot n_e$$

This is just a basis for further discussion now. One paper describing the scenario is


but only chapters I, II are relevant in my context, because it goes much further and does not address my problem.

The question that arose here is whether the excess charge inside is balanced out by a surface charge density of opposite sign, or the conductor is charged as a whole.

Personally, I prefer the assumption of surface charge density, but the others disagree. However, none of the opinions has convincing evidence.

This is solution one with a charge density on the surface built up:

enter image description here

While the other possibility is, that the conductor carries a total charge:

enter image description here

Personally, I find case-1 more logically, because the moving carriers are first deflected into the center so that, within a short time, they are finally missing at the surface and distributed homogenously within the bulk, thereby leaving a positive surface charge behind. I would consider this analogous to the charge separation in a metal by means of influence, where surface charge is build up in a way to exactly satisfying boundary conditions.

Case-2 would mean, that the source must provide additional charge to load the whole conductor negatively. However, this means, that the source itself must charge-up positively, because where do the additional negative charges come from? Although the effect is actually very very small, I would still find that pretty absurd.

In my opinion, both variants meet the boundary conditions at the interfaces between air and metal. However, due to the external field, variant 2 has a much higher energy as compared to variant 1 and is therefore energetically less favorable. It would be like a capacitor charging itself as a whole instead of charging both plates with opposite charges in order to keep it neutral. Why should a total charge be built up, when an energetically much cheaper scenario can satisfy the imposed boundary condition (have a fixed potential difference between its plates) as good?

I think over this problem already for more than three days and cannot come to a solution...any ideas?


1 Answer 1


If the two ends of a length of uniform wire are connected to the terminals of a battery, the battery will pull electrons from one end of the wire and place them on the other end. The total charge on the wire will be zero, but one end will have a positive charge/unit length and the other end will be negative. The system quickly reaches equilibrium with a continuous flow of current, driven by an electric field which is proportional to the gradient of the charge density. For the potential difference between the two ends of the wire to be independent of the radius, the parallel component of the field in the wire must also be independent of the radius. Any symmetrical radial component would tend to push charges outward. This suggests that the excess charge in any segment of the wire will be found at the surface, but the magnetic field produced by the current may work against this effect. For Gauss's law to be true for any chosen segment of the wire, field lines must be leaving the sides of the positive half of the wire and reentering along the negative half.

  • $\begingroup$ I understand what you mean but I think this is not the effect I wanted to discuss. Let's assume infinite conductivity, then, when both end of a conducting loop are connected by an ideal shortcut, there is no surface charge due to the reason you describe. Notably , the effect I talk about is only of academic interest since it is a question of just about 10 electrons per mm³. Anyway, we had an endless discussion. $\endgroup$
    – MichaelW
    Commented Nov 18, 2021 at 20:53

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