1
$\begingroup$

Is the redistribution of excess charge in a conductor onto its surface, thereby reaching static equilibrium (a steady state), only an empirical observation?

Or, equivalently, is it guaranteed, i.e. it can be theoretically proved, that there exists a static equilibrium which conductors having excess charge can attain by reallocating these charges?

$\endgroup$
  • $\begingroup$ Think about what if this was not true. $\endgroup$ – Aaron Stevens Aug 11 at 21:36
  • $\begingroup$ If these charges do not attain a static state at the surface after a while, then they must be in constant motion, not necessarily on the surface of the conductor. So perhaps one way to prove that there is a such a static state is to show that the charges' total energy (kinetic+potential) is minimized when this static equilibrium is attained. When the charges are static, the kinetic energy vanishes, so we must prove that the potential energy is minimized when all the charges are distributed over the surface, I can't really tell why this latter is the most optimal configuration. $\endgroup$ – Hilbert Aug 11 at 21:48
  • $\begingroup$ ...Suppose we have a solid conducting sphere, if we inject positive charges right in the center, why would they reallocate themselves only onto the surface instead of another configuration, why wouldn't they distribute themselves, for instance, with a charge density $\rho(r) \propto r$? $\endgroup$ – Hilbert Aug 11 at 21:52
  • $\begingroup$ So are you asking how we know an equilibrium will be reached? Or are you asking once it is achieved how we know charge only resides on the outer surface? $\endgroup$ – Aaron Stevens Aug 12 at 0:03
  • $\begingroup$ I think I digressed a little bit from what I was looking for, I am mainly interested in the first question, "How do we know an equilibrium will be attained?". $\endgroup$ – Hilbert Aug 12 at 9:17

Your Answer

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

Browse other questions tagged or ask your own question.