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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?

Griffiths, Example 8.3

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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 $\vec{B}$.

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  • $\begingroup$ 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. $\endgroup$ – alch Apr 29 '18 at 12:03
  • $\begingroup$ @alch Yes, there's a resistor at the end of the cable. But there is no resistance along it, which is what's relevant to the equation you cited in your question asking about the charge density along the cable. $\endgroup$ – Bob Jacobsen Apr 29 '18 at 18:55
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Obviously, the E field of interest here is between the inner and outer conductors of the coaxial cable, not along the inner our outer conductor, which, for ideal conductors, would be zero.

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