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When we write $j=\sigma E$ in a conductor, is $E$ here the net electric field produced by the electrons and the source that drives the current? For example, inside an electrolytic cell (let us assume it is ohmic), the electric field produced in the electrolyte must have 2 components; (i) the electric field produced by the voltage difference induced by the external battery (ii) the electric field produced by moving ions in the solution.

When we write $j=\sigma E$ for this equation, will both of these components be taken into account? I think they should be. This doubt arose while trying to model a leaky capacitor;

An infinite line charge of uniform electric charge density $\lambda_0$ lies along the axis of an electrically conducting infinite cylindrical shell of radius $R$. At time $t = 0$, the space inside the cylinder is filled with a dielectric material of permittivity $\epsilon$ and electrical conductivity $\sigma$. Find the current in the system as a function of time, assuming Ohm's law holds true.

The reported answer is $i(t)=\dfrac{\lambda_0 \sigma}{\epsilon}e^{-\sigma t/\epsilon}$.

My doubt:

While going through the solution for this, it has been written that $E(r)=\dfrac {\lambda}{2\pi \epsilon r}$; thus we only take into account the electric field produced by the line charge at a distance $r$ from the axis. However, we do not take into account the electric field of the charges of the form $dq$ being released in the form of cylindrical shells as the current flows radially, which must be in the vicinity of the wire (but not beyond distance $r$ from it, otherwise the contribution to the electric field at $r$ of that cylindrical shell of $dq$ charge will be $0$). I don't know if I'm able to phrase my query well enough, but all I'm saying is that since some current will flow in this material, there will be charges in that dielectric material as well whose field is not being taken into account when we use Ohm's law. Is my doubt valid? If not, could someone explain why the microscopic form of Ohm's law should not take into account the electric field due to the mobile charge-carrying species?

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When we write j=σE in a conductor, is E here the net electric field produced by the electrons and the source that drives the current?

If you're going to consider the effect of the mobile electrons, you should also consider the effect of the fixed electrons and fixed nuclear protons. If the material is overall neutrally charged, then these three effects will cancel each other out, at least at macroscopic scales (scale larger than a few tens of atomic spacing).

the electric field produced in the electrolyte must have 2 components; (i) the electric field produced by the voltage difference induced by the external battery (ii) the electric field produced by moving ions in the solution.

Positive and negative ions are produced in equal abundance, so that the net charge remains neutral, and the electric field produced by the ions is 0.

the space inside the cylinder is filled with a dielectric material of permittivity ϵ ...

When we describe a material with a permittivity, we are using the macroscopic forms of Maxwell's equations. We are accounting for the effect of polarized molecules (including their electric fields) in the polarization (often denoted $\chi$).

we do not take into account the electric field of the charges of the form dq being released in the form of cylindrical shells as the current flows radially,

At steady state, an equal charge flows out of the outer surface of the cylinder, so that the material remains electrically neutural, and so there is no net field produced by these charges and the fixed charges in the material.

could someone explain why the microscopic form of Ohm's law should not take into account the electric field due to the mobile charge-carrying species?

Because these charges are balanced by fixed charges that produce fields, but are immobile so they don't move and contribute to current density.


In comments you asked,

How do we know that at every point, ... just because of the net charge being zero ... (...the charge distribution might not even be symmetric so we cannot deduce this conclusion from applying Gauss law), the fields of the fixed charges and mobile electrons exactly cancel out?

The fixed nuclear charges and bound electrons might form dipoles, which I mentioned earlier is modeled by the permittivity of the material.

The mobile charges are distributed randomly. They don't have any fixed positions relative to the fixed nuclear charges. And we normally consider volumes containing at least 1000's of charges (otherwise we wouldn't be talking about permittivity of the material). Averaged over that large number of charges, there's no significant net dipole moment (or it is vanishingly unlikely for there to be any dipole moment). So then we can use Gauss's law after all to determine that there's no significant fields sourced from the totality of the charges (mobile and fixed charges considered together).

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  • $\begingroup$ How do we know that at every point, even on a moreover macroscopic level, just because of the net charge being zero and no idea of the distribution of charges (in case of a battery, the charge distribution might not even be symmetric so we cannot deduce this conclusion from applying Gauss law), the fields of the fixed charges and mobile electrons exactly cancel out? Sorry if I'm being a little paranoid and dumb about this, I just really want to understand this perfectly. $\endgroup$ Feb 12 at 3:57
  • $\begingroup$ I understand, that in the case of the infinite line charge and cylindrical shell, we accounted for the effect of the polarized/bound charge in Gauss law by using ε but what about the effect of the charge the current carries? Aren't these 2 different? $\endgroup$ Feb 12 at 4:04

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