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Consider a spatially finite region of conductor, with conductivity $\sigma$. Now imagine a net charge density $\rho$ spatially distributed throughout the volume of the conductor at time $t=0$. Usually, it is then said that the charge "quickly" redistributes to the surface of the conductor establishing an equipotential condition. I'm interested in the transient behavior. A quick attempt to get at an equation for charge density provides a weird answer. Starting with charge continuity and Ohm's law: $$-\frac{\partial\rho}{\partial t}=\nabla\cdot\bf{J}=\nabla\cdot(\sigma\bf{E})=\sigma\nabla\cdot\bf{E}+\bf{E}\cdot\nabla\sigma$$ Then using Poisson's Equation and Gauss's Law: $$\frac{\partial\rho}{\partial t}=-\frac{\sigma\rho}{\epsilon}+\nabla\phi\cdot\nabla\sigma$$ If we assume that the conducting region is uniform, then the last term is dropped and we have a solution which makes no pretense of conserving charge and has no spatial dependence. I suspect one of two things may have happened:

  1. Substituting $\sigma\bf E$ for $\bf J$ may be doing things the wrong way around such that the information about where the charge was going contained in $\bf J$ gets lost
  2. Using a local formulation may just be plain wrong-headed due to e.g. Debye shielding type effects (is treating a coupled system of PDEs, one for $\rho$ and another for $\phi$ a solution?)

I would greatly appreciate any insight into this problem.

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  • $\begingroup$ There is no obvious problem. The solution you mention (spatially uniform charge density in some region, decreasing in time) can obey local charge conservation, with proper $\mathbf J$. The lost charge from the region accumulates on the boundary. $\endgroup$ Commented Oct 23 at 22:14
  • $\begingroup$ It is clear that in the long-time limit that the charge accrues on the boundary, but is it implicit in assuming that the material is Ohmic that the conduction currents are neutralized by opposite sign carriers everywhere except the surface, at all times? $\endgroup$ Commented Oct 24 at 15:15
  • $\begingroup$ Said another way: are electrically charge neutral currents implicit in requiring the material to be ohmic, or may they be charged, and if so, how does one account for that? $\endgroup$ Commented Oct 24 at 15:24
  • $\begingroup$ I have no idea what you're talking about. Why would conduction currents be "neutralized by opposite sign carriers"? When there is non-zero conduction current density, there is motion of mobile charge, there is no neutralization of current. Neutralization happens, if left alone, to charge density inside. In time, it decreases to zero. $\endgroup$ Commented Oct 24 at 19:05
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    $\begingroup$ What is $dJ$ and $d\rho$, differentials in time, or do you actually mean $JdV$ and $\rho dV$? $\endgroup$ Commented Oct 25 at 7:03

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In usual conductors like metals you have to take into account that in equilibrium the negative mobile charge (electrons) is neutralized by the charge of the positive ions of the constituent atoms. Only a deviation $n_1$ of the electron concentration $n$ from this equilibrium value $n_0$ produces the charge density $$\rho= -q(n-n_0)=-qn_1 \tag{1}$$ and an electric field according to Gausses law $$\nabla\cdot\mathbf{E_1}=\frac{\rho}{\epsilon}=-\frac{qn_1}{\epsilon} \tag{2}$$ Such electron concentration deviations $n_1$ are usually extremely small compared to the equilibrium concentration $$n_1<<n_0 \tag{3}$$ On the other hand, the drift current density is determined by the total mobile electron concentration $n=n_0+n_1$ and the electron mobility $\mu$: $$\mathbf{J=J_0+J_1}=qn\mu\mathbf{E}=q(n_0+n_1)\mu\mathbf{(E_0+E_1)} \tag{4}$$ If we assume that there is no external electric field, then $\mathbf{E_0=J_0}=0$, and considering the condition (3) we can linearize the expression for the current density $$\mathbf{J\approx J_1}=qn_0\mu\mathbf{E_1}=\sigma_0 \mathbf{E_1}\tag{5}$$ where $\sigma_0$ is the constant bulk conductivity of the conductor. Thus the current continuity equation becomes $$\nabla\cdot\mathbf{J_1}=\frac{\partial qn_1}{\partial t} \tag{7}$$ Inserting expression (5) for the current density into equ. (7) and using Gauss' law (2) we obtain the simple differential equation for the electron density deviation $$\frac{\partial n_1}{\partial t}=-n_1\frac{\sigma_0}{\epsilon}=-\frac{n_1}{\tau} \tag{8}$$ with the exponentially decaying solution $$n_1(t)=n_1(t=0)\exp (-t/\tau) \tag{9}$$ Thus any initial electron concentration distribution $n_1(x)$ at time $t=0$ (and thus charge distribution $\rho(x)$) decays exponentially to zero with a speed determined by the time constant $$\tau=\epsilon/\sigma_0 \tag{10}$$ which is called the dielectric relaxation time. In normal metals this dielectric relaxation time is extremely small. It can be easily estimated from the known conductivities and dielectric constants.

Later Note: In your last equation for $\rho$ you discuss the (very sensible) neglect of the gradient of $\sigma$, but then state that, "we have a solution which makes no pretense of conserving charge and has no spatial dependence". This is incorrect! This equation, as my above equation (8) with the exponentially decaying solution (9), has been derived from the charge conserving continuity equation and is connected with an out- or inflow of conduction current. The spatial dependence of the electron charge deviation is arbitary and doesn't change shape during the decay.

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