# Charge per unit length in coaxial cable with steady current

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?

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 $\vec{B}$.
• Take a close look at Example 8.3. There is a resistance $R$. Coaxial cable has capacitance $C$ that is why it has a uniformly distributed charge on the inner and outer surfaces. – alch Apr 29 '18 at 12:03