While considering the Gauss law for a conducting sphere with a uniform charge distribution on its surface, it can be shown that the electric filed inside the conducting sphere is zero.

But, what would happen if charges are asymmetrically distributed on its surface? What would be the electric field inside the conducting sphere in this case?


closed as unclear what you're asking by sammy gerbil, ACuriousMind, user36790, Wolpertinger, honeste_vivere Aug 29 '16 at 12:16

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  • $\begingroup$ The charges would distribute on the surface in such a way that the field vanishes exactly everywhere in the inside $\endgroup$ – Phoenix87 Aug 25 '16 at 20:31
  • $\begingroup$ Suppose you bring a positive charge close to the conductor (say, which had uniformly distributed negative charges on its surface before), but not touching it. So what will happen to the electric filed inside now? $\endgroup$ – Dinuka Aug 25 '16 at 20:37
  • 3
    $\begingroup$ The argument for steady-state fields in a conductor being zero does not depend on symmetry, uniform change distributions or Gauss' law. It really only depends on the definition of a conductor (charge moves easily) and the idea that the pool of available changes is effectively infinite (true until charge densities reach many Coulombs per molar volume). $\endgroup$ – dmckee Aug 25 '16 at 20:41
  • $\begingroup$ @dmckee : This appears to be an answer-in-comment. $\endgroup$ – sammy gerbil Aug 25 '16 at 20:44
  • $\begingroup$ Please can you give an example of what you mean? Why do you think the result will be different if charge is not distributed symmetrically? $\endgroup$ – sammy gerbil Aug 25 '16 at 20:51