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When we have a spherical conducting shell and charge on outer surface of the shell then the potential inside remains constant i.e, kQ/R (R=radius). But say the inner surface of the shell is charged rather than outer then too does the potential remain constant?

Will the electric field still be 0 inside or will charges move towards outer surface , what would be the result ?

A situation similar to the one in the image below:-

enter image description here

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3 Answers 3

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If the grey and blue parts of your diagram are the conductors and the magnitudes of the two sets of charges are the same then you have drawn a correct diagram with an electric field present only in the region between the two sets of charges.
The field in that region being the same as if there was a $-Q$ charge at the centre of the arrangement. In all other regions, including outside the outer sphere, the electric field will be zero and there will be no charge resident on the outside of the outer sphere. Application of Gauss's law will show that is so.

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The potential will remain constant and the the electric field at any point between the shells will be zero.

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  • $\begingroup$ Why do you think so? Please explain the reasons explicitly and don't just leave any statement without showing the reason for its validity. $\endgroup$
    – user36790
    Commented Jul 8, 2016 at 17:05
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If you are talking of a conducting sphere, this situation cannot happen physically. Charge is supposed to stay on the outer surface of the conducting shell. However if you make the shell infinitely thin (which is again not possible physically,) then the potential inside (for radii less than the radius of the shell) would be still a constant.

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  • $\begingroup$ Is it because we can consider the shell to be made of infinitely many infinitesimally small planes facing each other which cancel each others electric field inside the shell? $\endgroup$
    – Shahbaaz1104
    Commented Apr 8, 2016 at 4:48
  • $\begingroup$ I think you make it more complex than it is. $\endgroup$
    – Benjamin
    Commented Apr 8, 2016 at 5:04
  • $\begingroup$ I am trying to visualize everything that is why maybe. $\endgroup$
    – Shahbaaz1104
    Commented Apr 8, 2016 at 5:09

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