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Consider a conductor containing a charge $q$ inside its cavity. Clearly, a negative charge $-q$ will be induced on the surface of the cavity. Let S denote the system of $q$ and $-q$.

In a discussion, my high school teacher told me that S will not have any contribution to potential outside the conductor but will have inside it. I couldn't get a satisfactory explanation, I thought that the electric field outside is only contributed by the outer surface charge so it must be true for potential as well. However, if we use the relationship between electric field and potential, it's clear that the system S contributes to potential everywhere as electric field of the system is zero outside the cavity.

$$V{_o} = V{_i} - \int_i^o \vec E{_s}. d\vec r $$ Here, $E_s$ is the electric field due to the system S, and $i$ and $o$ are any two points inside and outside the cavity respectively.

Is $V_i$ not non-zero or is there any logical mistake on my part?

Also, the potential due to outer surface charge is equal everywhere in the conducting region, easily noticeable in spherical conductors as the distribution of outer charge comes out to be uniform, how to see this in cases where distribution is non-uniform mathematically? (An intuition on the mathematical break up would be really appreciated.)

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Think simpler, use Gauss' Law. What is the electric field at any point outside the conductor? Now, if you use that to find the electric potential, you see that it's basically the same as if there was a point charge q at the center, or if there was a shell of charge q there. Your teacher may have been referring to the charge q on the shell('s outer surface). To put it simply, S is not the entire system and thus does not give the entire picture.

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