Ohm's law in microscopic level

Question

Ohm's law tells us that current density $\vec{J}$ is proportional to electric field $\vec{E}$. But from my last post, as well as this blog, a comment introduced me to the concept that it is instead the surface charge density that created the electric field and helps in current conduction.

So which electric field is the $\vec{E}$ in $\vec{J}=\sigma \vec{E}$ here? Is it of the surface charge densities of the wire, or of the battery? Or is the electric fields inside the wire due to the surface charges the same as that of the battery? Kindly enlighten me with the proper concept.

Answer 1

You may find this paper by Jackson (the guy who wrote the famous E&M text book) interesting.

The significance of the surface electric charge densities associated with current‐carrying circuits is often not appreciated. In general, the conductors of a current‐carrying circuit must have nonuniform surface charge densities on them (1) to maintain the potential around the circuit, (2) to provide the electric field in the space outside the conductors, and (3) to assure the confined flow of current. The surface charges and associated electric field can vary greatly, depending on the location and orientation of other parts of the circuit.

For DC and low frequency circuits where there is not significant radiation, or redistribution of charge on the time scale we are interested in the system is in steady state. By that there is a current flowing in the wire, but the amount of charge per unit volume everywhere is not changing. Same with the surface charge density, so there is no time dependence when you think about the distribution of charge.

So you divide up the problem to be internal to the wire, external to the wire and have boundary condition at the interface.

Internal to the wire, the physics is wrapped up in the $\sigma$ .This is material dependent. You can think about the mobility of electrons, the electrons as a gas with a a mean free path etc., but all of that is wrapped up in $\sigma $. if you want to make things complicated you can make $\sigma$ a tensor and have it be different in different directions. But, inside the wire $J=\sigma E$ gives the connection between the current and the electric field.

As you go along the length of the wire you also have the voltage drop and if the conductivity and wire cross section is constant the amount of drop is constant per unit length according to the conductivity. Thus you have a the potential changing as you go from one end of the wire to the other. (point 1 of the paper) This is consistent with $E$ being constant and in the same direction of $J$ if the diameter of the wire and $\sigma$ is constant.

If we consider potential along the length of the wire, and that there must be a unique solution by the uniqueness theorem, then if the voltage is changing along the length of the wire there needs to be a charge distribution along the length of the wire, and using a gaussian pill box, there we can look at the boundary conditions and the charge at the interface must be due to the mismatch of the radial components of the electric fields. Thus external to the wire there is a radially pointing electric field even if there is not a radial component of the electric field inside wire. That is o.k. if J is axial to the wire inside the wire there can be an axially pointing E. (point 2 of the paper).

So how does the power get transported? According to point three, we only have electrons moving inside the wire. From this constant current you also have a B field outside the wire. That B field will be in the $\phi$ direction circling the wire by the right hand rule.

For power to be transported, we can then look at the Poynting vector. Often we think of the Poynting vector $S$ when thinking of waves, but it also works for this case with $S=E \times B$. Note it doesn't have any charge density built into the expression. Outside the wire $E\times B$ points in the same direction of the wire and is carrying the power.

Note, that you can also compute the Poynting vector for inside the wire with $E$ pointing in the axial direction and the $B$ field inside the wire in the circumferential direction. This gives the joule heating and is pointing radially into wire and the magnitude would be be the potential difference times the current, or in electrical engineering terms the Power is VI.

Answer 2

The relation $\mathbf{j}(\mathbf{r}) = \sigma(\mathbf{r}) \mathbf{e}(\mathbf{r})$, or its inverse $\mathbf{e}(\mathbf{r}) = \rho_R(\mathbf{r}) \mathbf{j}(\mathbf{r})$, is a local relation between the value of the current vector field $\mathbf{j}$, the resistivity $\rho_R(\mathbf{r})$ or the conductivity $\sigma(\mathbf{r})$ and the electric field $\mathbf{e}$ at the point $\mathbf{r}$ in space, for a Ohmic medium, defined as a medium where the current vector is proportional to the electric field.

If you feel confused by the fact the in the interior of a perfect conductor there is a current, but the electric field is approximately zero, this agrees with the Ohm's law: a perfect conductor has approximately zero resistivity, $\rho_R(\mathbf{r}) = 0$, and thus the electric field is $\mathbf{e}(\mathbf{r}) = 0$, whatever the value of current vector $\mathbf{j}(\mathbf{r})$ is. The surface charge distribution, that can model a very thin layer of net charge distribution (as a first approximation if we're not interested in the charge distribution in the radial dimension), is approximately uniform on a straight perfect conductor wire, since charge gradient would produce non zero electric field and thus a infinite current, since $\sigma = \frac{1}{\rho_R} \rightarrow \infty$. This microscopic description coincides with the macroscopic engineering description of the circuit, with no potential difference between two end of a perfect conductor wire.

In a Ohmic resistor, the resistivity is non-zero, and thus an electric field must exists in every point where the current vector is non-zero. This electric field is generated by non uniform distribution of electric charges. We get its macroscopic constitutive law from the macroscopic description, by means of integration over a volume of the resistor

$\displaystyle \int_{dV} \mathbf{e} \cdot \mathbf{\hat{t}} = \int_{dV} \rho_R \mathbf{j} \cdot \mathbf{\hat{t}}$,

being $\mathbf{\hat{t}}$ the unit vector along the center line of the conductor. Splitting the volume integral on $dV$ as a surface integral on the section of the element $S$ and a line integral along an elementary section of the element $d\ell$, assuming $\rho_R$ uniform, we get

$\displaystyle \int_S \underbrace{\int_{d\ell} \mathbf{e} \cdot \mathbf{\hat{t}}}_{ d V } = \rho_R \int_{d \ell} \underbrace{\int_S \mathbf{j} \cdot \mathbf{\hat{t}}}_{i} $$\qquad \rightarrow \qquad $ $dV = \underbrace{\rho_R \dfrac{d \ell}{S}}_{d R} i = dR \ i$,

we found the elementary potential drop $dV$ in the resistor is proportional to the electric current $i$ by means of the elementary electric resistance $dR$ of the elementary part of the resistor of length $d\ell$. If the electric resistance is constant along the conductor, we get the elementary first and second Ohm's law for a Ohm's resistor

$\Delta V = R i$, with $R = \rho_R \dfrac{\ell}{S}$