I understand that any charge distribution in a conductor will always manifest itself onto its surface, and this leads to $\vec \nabla\cdot \vec E = 0$. But $\vec \nabla\cdot \vec E = 0$ doesn't mean that $\vec E$ has to be zero at all. In fact, coupled with the $\vec \nabla \times \vec E = 0$, the governing equation becomes Laplace's equation for which $\vec E = 0$ (or $V =$ constant) is just one solution out of a family of solutions. Is it possible, due to different boundary conditions, that the field in a conductor be any of those other solutions than simply just $\vec E = 0$?

The most given explanation is that any non-zero field in a conductor would immediately produce a non-steady current to dissipate that field until electrostatic equilibrium. But that assumes $\vec J$ to be non-steady, which isn't always the case (e.g. a conducting wire with constant current flow. Zero net internal charge but $\vec E$ is non-zero and $\vec J$ is constant). Could there be any other such steady $\vec J$ that arises electrostatically in conductors in real life?

  • $\begingroup$ "Could there be any other such steady J⃗ that arises electrostatically in conductors in real life?" - steady current density is magnetostatics; current density is zero in electrostatics. $\endgroup$ Commented Jun 3, 2017 at 0:19
  • $\begingroup$ Usually in conductors Ohm's law is valid, so electric field inside conductor implies there is electric current as well. Electrical conduction falls outside of scope of electrostatics. In electrostatics electric current is zero and in most cases this implies that electric field inside conductor is zero as well. However, if there are other electromotive forces present (such as due to thermal, chemical or even gravity potential gradient), it is possible electrical current density vanishes while electric field is non-zero. Technically this would belong to electrostatics too. $\endgroup$ Commented Jun 3, 2017 at 6:13
  • $\begingroup$ The way I understand it is that as long as the spatial charge distribution remains constant in time, it is electro-"static". Charges may move around such as in a current, but as long as charges constantly fill in where another has left, the distribution remains constant in time and so would J, E, and B. Maybe this isn't strictly electrostatics but magnetostatics (or something else??) instead. But, if this is electrostatics, then I don't see why a cylindrical conductor couldn't have a clockwise field E = -phi_hat/r. The electrons would just circulate endlessly without changing any field lines. $\endgroup$ Commented Jun 4, 2017 at 0:38

1 Answer 1


In electrostatics, no (electrodynamics, sure). It is the same concept that says the the surface of water will be flat no matter what container it's in. (the only exception is a river - but that's not hydrostatics ;) ).

And it's really the condition that $\vec{\nabla}\Phi = 0$ inside a conductor that matters; $\vec{\nabla}\cdot \vec{E} = 0 $ is laplace's equation, which is true for any charge free region (including inside a hollow conductor).


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