Answer
$V=-1.8\times 10^2 V$
Work Step by Step
The total potential at the point halfway between the centers is
$V=V_1+V_2$
$V=K\frac{q_1}{r}+K\frac{q_2}{r}$
We plug in the known values to obtain:
$V=9\times 10^9(\frac{1\times 10^{-8}}{1})+9\times 10^9(\frac{-3\times10^{-8}}{1})$
$V=-1.8\times 10^2 V$
