Transient Charge Density in Conductor

Question

Veritasium's (Derek Muller's) video How Electricity Actually Works makes the claim at 4:44 that:

The truth is, if you average over a few atoms, you find the charge density everywhere inside a conductor is 0.

I agree that in a steady state, this is true. It follows from the continuity equation:

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

together with the microscopic version of Ohm's Law,

$$\vec{J} = \sigma\vec{E}$$

and Gauss's Law:

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$

Since we are dealing with a steady state regime, all derivatives with respect to time are identically 0, so

$$\nabla \cdot \vec{J} = 0$$

So,

$$\nabla \cdot \sigma \vec{E} = 0$$

Assuming \$\sigma \ne 0\$

$$\nabla \cdot \vec{E} = 0$$

And since \$\epsilon_0 \ne 0\$ we have

$$\rho = 0$$

This proof that the charge density in a conductor is 0 relies upon the conductor being in a steady state, but says nothing about a system that is not steady state. Indeed, we often assume that a system is in some arbitrary state and predict how it will evolve over time. Such starting assumptions might include a non-zero charge density within a conductor.

Now, let's consider Kirchhoff's Current Law (KCL).

KCL applies universally if we understand by current the sum of both conduction current and (Maxwell's) displacement current. That is,

$$\nabla \cdot \left(\vec{J} + \epsilon_0\frac{\partial\vec{E}}{\partial t}\right) = 0$$

But when an E field is not in a steady state, KCL applied to conduction current alone, does not hold.

That is, with a time varying E field

$$\nabla \cdot \vec{J} \ne 0$$

Kirchhoff's Current Law (when applied only to conduction current) does not apply to a circuit immediately upon an EMF being applied, but must be established. Current within a wire is not immediately equal in all sections of the wire, although such a state is quickly established once the E field becomes constant.

Since \$\nabla \cdot \vec{J} \ne 0\$, it must be that \$\nabla \cdot \vec{E} \ne 0\$ and consequently

$$\rho \ne 0$$

Showing that the original assertion applies only in the case of time-invariant E fields.

My question is whether my reasoning is correct (I believe it is) and that Veritasium's/Derek Muller's assertion, that charge density within a conductor is 0, only applies to steady state, and therefore does not apply to a thought experiment that involves transients in the electric field due to throwing a switch. Am I correct or not?

Note: I am well aware that there are other questions on this site that deal with other aspects of Veritasium's/Derek Muller's thought experiment. This question is not about the thought experiment as a whole, but is specifically about his claim regarding charge density within the interior of a conductor.

Answer 1

I haven't watched the video, but what you are saying is essentially correct, inside perfect conductors E=0 and no fields penetrate (e.g. magnets 'float' on superconductors), all excess charge and charge movement occurs on the surface.

This is not the case for finite conductivity. Take a copper mains cable, at 60Hz, the skin depth in copper is about 1cm, so the current density will be pretty uniform throughout the cable. Due to the finite conductivity, ohms law says there must be a small internal E field for there to be internal current. The current direction reverses at 60Hz, so these fields are not static.

In a good conductor though, the energy stored in the electric field is very small compared to that in the magnetic field. For a perfect conductor the Poynting vector is parallel to the surface (as \$E_t=0\$), but for imperfect conductors \$E_t\ne0\$ and you should see the Poynting vector tilt in towards the conductor showing that fraction of the power flow that feeds the conductor losses.

Some EM texts cover EM waves in good conductors.

To answer your question on how quickly any localised charge density dissipates, in regions of constant conductivity and permittivity:

Combining:

\$\vec{J} = \sigma\vec{E}\$

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

\$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon}\$

you can get:

\$\frac{\partial \rho}{\partial t} + \frac{\sigma}{\epsilon}\rho = 0\$

which has the simple solution:

\$\rho(t)=\rho_0 e^{-t/\tau}\$

with \$\tau\$. the charge relaxation time, given by:

\$\tau = \frac{\epsilon}{\sigma}\$

For copper, the relaxation time is around \$10^{-19}\$ sec. So for good conductors, at the sorts of frequencies we use, there is no internal localised charge.