Exercise 5.4.1: Charge relaxation in a conductor

With $J=\sigma E$ and $\nabla \cdot D=\epsilon \nabla \cdot E={\rho }_v$,

With ${\epsilon }_0=8.842\times 10^{-12}F/m$: ${\tau }_{\mathrm{Cu}}=8.842\times 10^{-12}/(5.8\times 10^7)=1.5\times 10^{-19}\mathrm{s}$, ${\tau }_{\text{sea}}=81\times 8.842\times 10^{-12}/4=1.8\times 10^{-10}\mathrm{s}$, ${\tau }_{\text{glass}}=5\times 8.842\times 10^{-12}/10^{-12}=44\mathrm{s}$.

Charge cannot vanish: integrating over a volume, the loss of $Q$ inside is the current $\oint J\cdot dS$ leaving through the surface. The charge is driven outward by its own field until it reaches the boundary of the conductor, where it stays as surface charge, which is the electrostatic result of Section 3.2 that a conductor carries its charge on its surface.

Units: $\epsilon /\sigma$ is $(F/m)/(S/m)=\mathrm{F}\cdot \mathrm{\Omega }=\mathrm{s}$. A medium behaves as a conductor for processes slower than $\tau$ and as a dielectric for faster ones: $\omega \tau =\omega \epsilon /\sigma$ is exactly the inverse of the loss tangent of Section 5.2.3. (For copper the formula gives a time shorter than the collision time of the electrons, so there the simple Ohm's law no longer applies, but the conclusion "instantaneous" stands.)