# Why is potential on a conducting shell always constant?

I've read that electric field inside a conductor is zero, and that potential of a two points say A and B on the surface of the conductor is same (constant). The latter puts me in a state of confusion. If $$V$$ is constant then there is no electric field. But there is an accumulation of charges on the surface of a conductor. Then why is $$V$$ constant? Shouldn't it vary since the electric field on the surface of the conductor is NOT zero?

The potential difference between two points in an electric field can be zero in two cases. The first one, as you point out, is when $$\vec E = 0$$.
The second case however, is when $$\vec E.d \vec s$$ is always $$0$$, where $$d\vec s$$ is an infinitesimal displacement used in path integrals. This can happen when vectors $$\vec E$$ and $$d\vec s$$ are perpendicular.
On the surface of a conductor, the electric field $$\vec E$$, due to charges on it's surface are always perpendicular to the surface at any point (Otherwise, there would be a component of electric field along the surface, and the charges would move. The conductor would then no longer be in electrostatic equilibrium.) Hence, for any two points $$A$$ and $$B$$ on the surface of the conductor: $$\Delta V = -\int_{A}^B \vec E.d \vec s = 0 => V_A = V_B$$