I am working on a homework problem where a coaxial cable has an inner charge/length $\lambda$ and outer charge/length $-\lambda$ and a current $I$.

The inner (conducting) cylinder is solid while the outer cylinder is just a shell.

The solution asserts the $\vec{E}$ field inside the inner cylinder is 0, but I was under the impression that that assumption is only valid in electrostatics.

Does a wire that is itself charged keep all the "net" charges on the surface even as a current is running through it?


Your question assumes the wire to be a perfect electrical conductor (PEC). This basically means that its conductivity goes to infinity $\sigma \rightarrow \infty$, or equivalently its resistivity goes to zero. Recall from Ohm's law that: $$\mathbf J (\mathbf x ,t) = \sigma \mathbf E (\mathbf x ,t)$$ Where $\mathbf J$ is the volume charge density. Now for a PEC, as $\sigma \rightarrow \infty$ , you can see that if we had a non-zero electric field inside the conductor, the current density would also go to infinity, which is clearly nonphysical. Thus, the only way to satisfy the above equation is for $\mathbf E (\mathbf x ,t)$ to be zero everywhere inside the conductor, at all times .

Thus, for a perfect electrical conductor, the electric field is zero inside the conductor even in time-varying cases.

Note that this is of course an idealization, and any wire has some finite conductivity, and thus the electric field inside is not exactly zero, although it is very small (I am disregarding things like superconductors). For a quantitative measure of how good the conductor rejects the internal electric field at some frequency , one can use the penetration depth, which is the the length you need to go inside the conductor for the field to become $1/e$ times its surface value. For further information about penetration depth check this answer.

Also note that the fact that your wires carry a charge doesn't affect the above reasoning, which means that the electric field is also zero in your specific problem.


Could the solution you read have meant that the radial component $E_r$ of the electric field is zero inside the inner cylinder? That is certainly always true due to the cylindrical symmetry of the problem. However, the axial component $E_z$ is only zero if the conductor is perfect (has no resistance). Otherwise Ohm's law $ \vec J = \sigma \vec E$ would require an electric field in the same direction of the current.

In general,

The electric field of a conductor can sustain both a surface charge distribution and a constant current with no interaction. Those charges pinned to the surface don't care there is a current generating electric field underneath them; they still only care about satisfying the E-field boundary conditions ($\Delta E_\bot = \frac{\sigma}{\epsilon_0}$ and $\Delta E_{||} = 0$). This condition goes away if the currents time vary.

bonus points: which way is the poynting vector?

Did you know that the electric field from a current carrying wire extends outside the wire? The tangential boundary conditions ( $\Delta E_{||} = 0$) on electric fields require no discontinuities at the surface of the wire. Thus, the axial component $E_z$ is in general not zero in the space between the inner cylinder and the outer shell. This means the Poynting vector $\vec S = \vec E \times \vec B / \mu_0$ has a slight angle away from the inner conductor. Just food for thought since you mentioned Poynting vectors in the header.

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    $\begingroup$ Actually, the Poynting vector between the coaxial conductor gives you the energy flux P=IV. Easy to calculate for DC when the space is narrow. $\endgroup$ – Pieter Jan 9 '18 at 21:49

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