Answer
$V_B=-7.8\times 10^5V$
Work Step by Step
We can find electric potential at corner $B$ as
$V_B=\frac{1}{4\pi \epsilon_{\circ}}(\frac{q_1}{a}+\frac{q_2}{b})$
We plug in the known values to obtain:
$V_B=\frac{1}{4\times 3.1416\times 8.85\times 10^{-12}}(\frac{-5\times 10^{-6}}{0.05}+\frac{2\times 10^{-6}}{0.15})$
$V_B=-7.8\times 10^5V$
