# How can the surface of a conducting shell be equipotential when a charge is introduced inside the cavity?

Here is the confusion.

Take a spherical conducting shell of a certain thickness.

When a postive charge is introduced into the shell the inside surface collects negative charges and the outter surface postive charges. So far so good.

Now gauss' law tells be there is no field in the conducting shell and thus its outter and inner surfaces must be equipotential.

However the inside surface has negative charges and thus is at a negative potential and the outter surface has postive charges thus is at a postive potential. And thus the inner and outter surfaces have opposite signs for potential and thus are different.

This seems to falsely contradict the earlier gauss' law conclusion which I know is correct.

Where did I go wrong?

• Don't forget the potential (or field) due to the charge that you introduced. – garyp Mar 22 at 19:05
• Could you please expand on that statement. I don't seem to understand – VinalV Mar 22 at 19:06
• The inner surface and outer surfaces are equipotential on their own, they obviously dont have the same potential, but the complete outer surface and complete inner surface are equipotentials – Sidharth Giri Mar 22 at 19:19
• Correct but as they have different sign that would imply a potential difference between surfaces and thus an electric field inside the shell? Which of course is false but why – VinalV Mar 22 at 19:21
• @SidharthGiri This is incorrect. The conductor is at equipotential. – my2cts Mar 23 at 10:46