201903310557 Solution to 2004-AL-PHY-II-16

The original electric potential $V_B$ at point $B$ is due to charge $+Q$ at point $A$ :

$V_B =V_{BA} = k \displaystyle{\frac{(+Q)}{(2r)}}=k \displaystyle{\frac{Q}{2r}}$ .

The final electric potential $V_B$ is due to both charges $+Q$ and $-2Q$ at points $A$ and $C$ :

$V_B' = V_{BA} + V_{BC} = k \displaystyle{\frac{Q}{2r}} + k \displaystyle{\frac{(-2Q)}{(r)}}=-3 \bigg( k \displaystyle{\frac{Q}{2r}} \bigg) =-3V_B$ .

And the answer is B.