Griffiths, Example 8.3. Long coaxial cable is connected to a battery at one end and a resistor at the other. The inner conductor carries a uniform charge per unit length $\lambda$ and a steady current $I$.
If a current is steady then the charge density must be zero because $\nabla E = \frac {1}{\sigma} \nabla J = 0$. Why is it not the case in Griffiths's example 8.3?
Best Answer
In the setup of Griffiths' example, there's no resistance opposing the motion of the charge along the central electrode: There will be no voltage there due to Ohm's law. To put it another way, there's no $\nabla E$ because $\sigma$ is infinite.
The charge on the inner conductor is there to provide the potential difference between the conductors; Griffiths could have just told you the potential difference and had you work out the field (via the charge or not), but that would have been a longer example. He tends to create examples that focus on just one thing, in this case the Poynting vector from $\vec{E}$ and $\vec{B}$.