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Assume steady-state conditions and a homogeneous conductor. Then dp/dt = 0 where p is volume the charge density. If ohm's law apply in the conductor, then

$${\rm div}\ \bar{J} = {\rm div}\ \bar{E} = 0$$

But if there is an electric field, we know that in the conductor surface there is a constant current; this implies that in the surface:

${\rm div}\ \bar{J} = 0 $ , ${\rm div}\ \bar{E} \neq 0 $

So the ohm's law seems not apply in the surface of a conductor. Where am i wrong?

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  • $\begingroup$ You've got it wrong, Ohm's law does not by itself imply $\text{div} \vec{J} = 0$. It is the steady-state condition that implies this. Only then Ohm's law and uniformity of conductivity $\sigma$ implies $\text{div} \vec{E} = 0.$ $\endgroup$ – Ján Lalinský Jun 4 at 18:08
  • $\begingroup$ Also you've got it wrong about current in the surface. Electric field inside conductor only requires that there is electric charge on the conductor surface, but there is no need for electric current on the surface. All current flows inside, through the cross-section, with approximately constant current density across the cross-section. $\endgroup$ – Ján Lalinský Jun 4 at 18:10
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Ohm's law indeed isn't valid at the conductor surface. This is because electric field has non-zero normal component on the surface, but electric current density's normal component there is zero, as charges won't jump out of the conductor into vacuum (unless the electric field is extreme). Ohm's law is usually valid inside the metallic conductor, where the current flows, not on its surface, where it (in case of steady-state current) does not flow.

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A constant current does not necessarily need an E field. Moving electrons in a vacuum somewhere in the universe or in a superconducting ring do not need a force once they are accelerated. If there was an E field in a superconducting ring the energy transfered to an electron would be zero for a complete loop. The E field - if any- just outside of the surface of a (perfect ) conductor is always perpendicular to the surface, under electrostatic conditions. This is an important constraint of the Maxwell equations.

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  • $\begingroup$ Even if "The E field - if any- just outside of the surface of a (perfect ) conductor is always perpendicular to the surface" , div(E) will still be not 0 $\endgroup$ – JDOE Nov 12 '17 at 10:48
  • $\begingroup$ At the surface, divE is not defined, since E is not continuously differentiable if there is an E field component perpendicular to the surface (inside/ at the surface the E field is zero, just outside it is non-zero = step function). If all components are zero, divE is also. $\endgroup$ – xeeka Nov 12 '17 at 11:08
  • $\begingroup$ So the fact that the current is only in the surface of a perfect conductor is experimental ? $\endgroup$ – JDOE Nov 12 '17 at 17:26

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