A thin spherical shell is charged by some source. The potential difference between the two points $C$ (center) and $P$ (on the surface) shown in the figure is (Take $\frac{1}{4\pi {ϵ}_0}=9\times 10^9\text{SI units}$, $q=1\mu \text{C}$, $R=3\text{cm}$):

For a charged conducting spherical shell:The potential at any point on the surface is $V_s=\frac{1}{4\pi {ϵ}_0}\frac{q}{R}$.The potential at any point inside the shell (including the center $C$) is constant and equal to the potential on the surface.
$$V_C=V_P=\frac{kq}{R}$$
The potential difference $\Delta V=V_P-V_C=0$.