Answer
$V_S-V_B=3600V$
Work Step by Step
Potential difference on the surface is:
$V_S=K\frac{q}{r}$
We plug in the known values to obtain:
$V_S=9\times 10^9\times\frac{30\times10^{-9}}{0.03}$
$V_S=9000V$
The electric potential on the surface is equal to the electric potential inside the sphere, so
$V_A=V_S=9000V$
Now electric potential outside the sphere is:
$V_B=K\frac{q}{r}$
We plug in the known values to obtain:
$V_B=9\times10^9\times\frac{30\times 10^{-9}}{0.05}$
Therefore, $V_S-V_B=9000-5400=3600V$
