Griffith briefly addresses the uniformity of the electric field in a wire using Laplace's equation by assuming that the electric field is always parallel to the wire (i.e. $\mathbf{E}\cdot\hat{n}=0$).

On the cylindrical surface, $\mathbf{J}\cdot\hat{n}=0$, else charge would be leaking out into the surrounding space (which we assume to be nonconducting). Therefore $\mathbf{E}\cdot\hat{n}=0$, and hence $\partial V/\partial n = 0$. Having specified V or its normal derivative on all surfaces, the potential is uniquely determined.

-Griffith, Introduction to Electrodynamics

I am likely missing something, but this seems like extremely fallacious logic. The current density perpendicular to the wire is obviously zero, but by Ohm's law,


If the current density $\mathbf{J}$ is equal to zero, and, as he says, the surrounding medium is non-conductive, then doesn't an essentially zero conductivity and essentially zero current density allow $\mathbf{E}\cdot\hat{n}$ to be whatever it wants?

And then for my question. If this is true, there must be some mechanism by which $\mathbf{E}\cdot\hat{n}$ is kept equal to zero so the Laplace's solution holds. The horizontal motion of electrons (i.e. electron buildup on the surface of the wire negating any perpendicular component) seems the obvious choice, but given the slow drift velocity of the electrons and the near-c speed of the "current wave-front", can this mechanism act fast enough? The horizontal motion of electrons in a wire might just be fast enough to achieve this. And does this account for the uniformity of the electric field in the wire, as well?



The statement "On the cylindrical surface $\mathbf{J}\cdot\hat{n}=0$..." refers to just inside the wire so $\sigma\neq0$ and $\mathbf{E}$ cannot be whatever it wants.

| cite | improve this answer | |
  • $\begingroup$ Okay. That's helpful. But why is that true? Is it due to the buildup of electrons on the surface of the wire negating any perpendicular field? Because the conductivity outside the wire is zero, electrons moving with the perpendicular field will accumulate on the surface, unable to move further, and neutralize the field? $\endgroup$ – JAustin Jul 24 '16 at 14:44
  • $\begingroup$ That is correct. $\endgroup$ – James Rowland Jul 24 '16 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.