I have heard from numerous sources (most notably "Surface charges on circuit wires and resistors play three roles" by J. D. Jackson) that there is surface charge on any steady current carrying conductor of uniform conductivity (i.e the conducting wires for a simple battery resistor circuit). This surface charge is required in order to ensure that the electric field within the conducting wires is such that we have a steady uniform current throughout the circuit. However, I do not understand how Ohms law can permit this build up of surface charge on a steady current carrying wire.

For suppose we have such a wire of uniform conductivity $\sigma$. Within the wire, the charge density must be zero since if we take the divergence of ohms law we get $$\vec{E}=1/\sigma \vec{J} \,\,\,\Rightarrow \,\,\,\nabla \cdot \vec{E}=1/\sigma\, \nabla \cdot \vec{J}=0$$ since $\nabla \cdot \vec{J}=0$ for a steady current. Then from Gauss's law, we know that the charge density must be zero everywhere within the wire. The above argument rests on the fact that $\sigma$ is uniform within the wire though. As we approach the surface of the wire though, the conductivity must either abruptly or continuously drop until is reaches the value of the conductivity of the surrounding air (virtually zero). Thus, if we apply ohms law near or at the surface of the wire, we must take into account that $\sigma$ is no longer a constant and so taking the divergence we get $$\vec{E}=1/\sigma \vec{J}\,\,\, \Rightarrow \,\,\, \nabla \cdot \vec{E}=1/\sigma\, \nabla \cdot \vec{J}+\vec{J}\cdot (\nabla \frac{1}{\sigma})=0$$ but now since $\nabla \cdot \vec{J}=0$ we get that $$\Rightarrow \nabla \cdot \vec{E}=\vec{J}\cdot \nabla \rho$$ where $\nabla \rho $ is the gradient of the resistivity ($\rho$ is simply a function that gives the value of resistivity at all points in space and is a constant function within the wire but rises continuously or abruptly near the surface of the wire until it reaches the value of the resistivity of air). My problem is that the current density $\vec{J}$ is always in the axial direction(even near or at the surface) whilst $\nabla \rho$ must always point radially outward near or at the surface (since this gradient by definition points in the direction of increasing $\rho$ and this direction is outwards towards the highly resistive surrounding air). So that means the dot product $\nabla \cdot \vec{E}=\vec{J}\cdot \nabla \rho$ should always equal zero at or near the surface and so by gauss's law, the surface charge density at the surface must also always equal zero. But if this is the case, then how can a surface charge density ever build up on the surface of a conducting wire?

Any help on this issue would be most appreciated because it has been driving me mad recently!

  • $\begingroup$ Have you considered bends in the wire? For an ideal straight wire with a constant current, there is indeed no net charge. $\endgroup$
    – Gilbert
    Mar 26, 2021 at 19:29

2 Answers 2


You are missing an important part of the puzzle. For simplicity, let us consider a wire of uniform resistivity. For a steady current, the charges must be moving at constant speed and along the wire (macroscopically at least). This means that the net forces acting on them must be zero.

The first force to consider is the Lorentz force due to the field generated by the current. Assuming a uniform current density in the wire, we find that the magnetic field inside it is given by \begin{align} \mathbf{B}(r) &= \frac{\mu_0 I}{2 \pi r} \hat{\mathbf{\phi}} \\ &= \frac{\mu_0 r}{2} j \hat{\mathbf{\phi}}. \end{align} This generates a Lorentz force on the charges: \begin{align} \mathbf{F}_B &= \rho ~ \mathbf{v} \times \mathbf{B} \\ &= -\mathbf{j} \times \mathbf{B} \end{align} If there are no other forces acting on the charges, they will move outwards, accumulating on the surface of the wire. After a while, this charge accumulation will generate an electric field in the radial direction which will counteract the Lorentz force. At steady state the two forces cancel out, which means the electric field must be \begin{align} \mathbf{E} &= -\mathbf{F}_B \\ &= -\frac{\mu_0 r j^2}{2} \hat{\mathbf{r}}. \end{align}

From the divergence of this electric field you find that it induces a charge density inside the wire of \begin{align} \rho = -\frac{\mu_0 j^2}{\epsilon_0}. \end{align}

Since the wire must be charge neutral, we must have an opposite charge at the surface, so that for a given cross section the total charge is zero: \begin{align} (\pi R^2) \rho + (2 \pi R) \sigma &= 0. \end{align}

Finally, we find the charge density at the surface of the wire: \begin{align} \sigma = \frac{\mu_0 R j^2}{2}. \end{align}

Notice that this surface charge is generated by the relocation of the charges due to magnetic field induced by the current. There is no need for a varying resistivity. Moreover, you might feel conflicted to see a finite charge density inside a conductor. This is a bound charge, that appears because we are not dealing with electrostatics, since there is a finite current. This is essentially the same mechanism of the Hall Effect, with the peculiarity that here the field generating it is caused by the current in the same wire.

  • $\begingroup$ A current carrying wire is not charge neutral. In order to maintain a uniform longitudinal field inside the wire there must be a gradient in the charge/unit length. One half of the wire will have a net positive charge and the other half will be negative. I would expect that the radial distribution of this excess charge would depend on magnetic effects, but any small selected volume should conform to Gauss's law. $\endgroup$
    – R.W. Bird
    Dec 21, 2021 at 15:46
  • $\begingroup$ @R.W.Bird what do you mean half of the wire has a positive charge? Which half? $\endgroup$ Dec 22, 2021 at 9:31
  • $\begingroup$ The half which is connected to the positive terminal of the power supply $\endgroup$
    – R.W. Bird
    Dec 22, 2021 at 14:52
  • $\begingroup$ I think I don't know to which phenomenon you're referring to. Could you give me some references? $\endgroup$ Dec 23, 2021 at 14:19
  • $\begingroup$ Continuity of flow in a uniform wire requires that the current and longitudinal electric field must both be uniform. A uniform field requires a uniform gradient in the distribution of charge along the wire. That means there will be extra electrons in the half of the wire which is connected to the negative terminal of the power supply and a deficiency of electrons in the other half. Gauss's law requires that there be electric field lines leaving the outer surface of the wire in the positive half and reentering the negative half. $\endgroup$
    – R.W. Bird
    Dec 23, 2021 at 16:51

I think that unlike the answers stated above, there should be a different explanation. Consider a steadily flowing cathode ray. In a region around it, it will definitely create an electric field but unlike that in electrostatics, this is not due to static but dynamic charges. Similarly what I think is that in a particular cross section of the thick wire,there will be definitely a non uniform charge distribution in wire but not a static charge, but that due to flowing electrons. Thus there will be non uniform current density in the wire which shows that the calculations are wrong. However I think that the surface charge density is static one and one must take its contribution to electric field also.

As pointed by RW bird, the surface charge density and the current density distribution will also vary along the length without violating Kirchhoff 1st law

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