# what is the electric field intensity inside a charged metallic shell if there is a point charge inside the shell?

What I know so far: - Charges (electrons) inside a conductor will repel (Coulomb's law). - The charges will experience repulsion which results in maximum separation distances between the charges. - The charges will then redistribute along the surface of the conductor in order to achieve electrostatic equilibrium (ie. net force of zero on each charge) - Thus, no electric field exists inside the conductor.

Please feel free to correct me if I am wrong in any of the above statements.

Appreciate the help. Thanks.

• You are right but it's the charges that appear at the surface are not the same charges you put in. If you put in a bunch of a + charge calcium ions, electrons will leave the surface to give it a positive charge. No calcium ions need to make it to the surface. Jan 8 '15 at 23:43

@vamsi, you are almost entirely correct. The charges on a conductor align in such a way that it cancels all external electric field inside the conductor, more specifically inside the BODY of the conductor.

In case of the shell, it will try to cancel out the field inside the thickness of the shell. So what will happen is that the point charge inside will be considered as an external field for the thickness of the shell and the charges on the INSIDE surface of the shell will align in a way to cancel this field.

More specifically, only the charges on the inside surface of the shell will be responsible for cancelling out any internal field(point charges, etc.), and external field will be cancelled by charges on the external surface only. This can be proved using Gauss's theorem.