Surface charge on a current carrying conductor is impossible?

Question

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!

Lucas Baldo · Answer

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.

Surface charge on a current carrying conductor is impossible?

One Answer

Add your own answers!

Ask a Question