# Ohm's Law in a conductor

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?

• 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 $\text{div} \vec{E} = 0.$ – Ján Lalinský Jun 4 '19 at 18:08
• 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. – Ján Lalinský Jun 4 '19 at 18:10