We have an arbitrary hollow shaped conductor, with small thickness. Suppose we place a charge $q$ at any arbitrary location within the hollow shaped conductor. By Gauss's law there will be a total charge on the inner surface $-q$. Since the conductor is neutral there will be a charge of $+q$ on the outer surface.

Is it true that considering the contribution of field from the charges induced on the inner surface and the charge kept inside the conductor only, the field is zero everywhere outside the conductor? In different words, is the field at any point outside the conductor due to only the induced charges developed at the outer surface?

My attempt- Consider any external point P, for finding the flux at P, it is due to the electric field due to the two charge distribution,

  1. Charge distribution due to inner induced charges and the charge kept inside.
  2. Charges induced on the outer surface.

Considering body 1 only, the flux comes to be zero, since the net charge enlcosed is zero, but how can proceed to prove that the electric field is zero as well due to body 1.

Note that the total flux at the point P is non zero, due to the flux of body 2 , as since the total induced charge on the outer surface is $q$ , the flux due to this $q$/$\epsilon $



1 Answer 1


In electrostatic situation, yes. Field of charges inside the outer conductor boundary vanishes everywhere outside.

  • $\begingroup$ Thanks for the answer, can you please prove this rigoursly? $\endgroup$
    – green_32
    Commented Jun 8, 2021 at 22:31
  • 1
    $\begingroup$ green, this very much seems your homework? $\endgroup$ Commented Jun 8, 2021 at 23:42
  • $\begingroup$ @MarcusMüller, this is not homework question, i have read this statement many times, but the proof was not given there, i myself tried to prove it but i was able to show that flux to be zero only. $\endgroup$
    – green_32
    Commented Jun 8, 2021 at 23:50
  • 2
    $\begingroup$ If you could edit your question to show your derivation, that would be super useful – an answer might simply pick up where you've left of, maybe explain the boundary conditions you've not seen before, and then resolve this pretty solidly! $\endgroup$ Commented Jun 9, 2021 at 9:33

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