Charge conservation in ohmic material - Apparent paradox

Question

Consider an ohmic material with constant conductivity, which obeys the relation:

$$\vec J = \sigma \vec E$$

where $\vec J$ is the current density and $\sigma$ is the conductivity. Applying the first of Maxwell's laws, we have:

$$\vec \nabla \cdot \vec E = \frac{\rho}{\varepsilon_0}=\frac{1}{\sigma}\vec \nabla \cdot \vec J$$

Therefore the expression for the divergence of the current density will be:

$$\vec \nabla \cdot \vec J=\frac{\sigma\rho}{\varepsilon_0}$$

Introducing this result in the charge density continuity equation and assuming there are no charge sources or sinks:

$$\frac{\partial \rho}{\partial t}+\vec \nabla \cdot \vec J = 0$$ $$\frac{\partial \rho}{\partial t}+\frac{\sigma \rho}{\varepsilon_0}=0$$

Now, integrating the density over the whole volume of the material should yield the total charge $Q$, which should be constant since we assumed there are no charge sources or sinks present in the material. Therefore, we obtain:

$$\frac{\partial Q}{\partial t}+\frac{\sigma}{\varepsilon_0}Q=0 \Longrightarrow \boxed{Q=0}$$

How can this be explained? The conclusion one can draw from these calculations is an ohmic material will always be neutral if there are no other agents acting on it, modifying its charge. Nonetheless, if this is true, then a charged dielectric should not exist, which is completely false.

Comment: I figured perhaps the step where I get it wrong is when interchanging the volume integral and the partial derivative, but this issue is solved easily by considering just those charge distributions described by a density function that is analytic in all of its variables

Answer 1

Differential problems are defined in a domain and require boundary conditions, and the solution to be "regular enough" for the differential equations to hold.

If you're dealing with a body of finite extension, assuming electric charges can't leave the body, they are accumulating in the (assumed to be, or modeled) infinitesimal-thickness layer at the boundary: charges inside the body goes to zero, as it moves to the this boundary layer, preserving the total charge.

You can easily realize it, once you have evaluated the density $\rho(\mathbf{r},t) = \rho(\mathbf{r},0) e^{-\frac{t}{\tau}}$, you can also evaluate the flux (i.e. the electric current) flowing across the boundary from the inner region of the body to the boundary layer as a function of time,

$$\Phi_{\partial V}(\mathbf{j})(t) = \oint_{\partial V} \mathbf{j} \cdot \mathbf{\hat{n}} = \int_V \nabla \cdot \mathbf{j} = - \int_V \partial_t \rho = - \frac{d}{dt} \int_V \rho(\mathbf{r},t) = -\dfrac{d Q}{dt} = \frac{1}{\tau} Q_0 e^{-\frac{t}{\tau}} \ .$$

Integrating from $t=0$ to $t^*$ you get the current accumulated in the boundary layer,

$$\int_{t=0}^{t^*} \Phi_{\partial V}(\mathbf{j})(t) dt = \int_{t=0}^{t^*} \frac{1}{\tau} Q_0 e^{-\frac{t}{\tau}} dt = Q_0 \left(1 - e^{-t^*/\tau} \right) \ ,$$

that approximates $Q_0$ after a very short time $t^*$ for conductors and media with small time constant $\tau$ in general.

Answer 2

As I understand it, the questioner's claim is that the equation $\frac{\partial Q}{\partial t}+\frac{\sigma}{\epsilon_0}Q=0$ is derived only from the standard Maxwell equations. As a result, $Q$ is always $0$ when $\frac{\partial Q}{\partial t}=0$, for example, even in electrostatic problems.

For simplicity, let us assume that there is a conductor and a vacuum. Also assume that the electric permitivity is also $\epsilon_0$ in the conductor. In the conductor, $\sigma$ is positive and constant, while in the vacuum $\sigma$ is 0. At the boundary of the conductor and space, the function $\sigma(x,y,z)$ is expressed as a Heaviside step function, i.e., $\sigma$ is not constant over the whole domain. And we know that the derivative of Heaviside step function is the Dirac's delta function. Using the math identity, we get, $$ \vec{\nabla}\cdot\vec{J}=\vec{\nabla}\cdot\left(\sigma\vec{E}\right) =\sigma\vec\nabla\cdot\vec{E}+\vec{\nabla}\sigma\cdot\vec{E}. $$ As stated abobe, the $\vec{\nabla}\sigma(=\vec{\nabla}\sigma(x,y,z))$ includes Dirac's delta function. Using this relationship, the following equation is derived. $$ \frac{\partial \rho}{\partial t}+\frac{\sigma\rho}{\epsilon_0}+\vec{\nabla}\sigma\cdot\vec{E}=0. $$ Thus $\frac{\partial Q}{\partial t}+\frac{\sigma}{\epsilon_0}Q=0$ is not derived.