JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Electrostatics

At point $O$, the charges form three dipoles: $+q$ at $A$ with $-q$ at $D$, $+q$ at $B$ with $-q$ at $E$, and $+q$ at $C$ with $-q$ at $F$. The potential at $O$ is zero since $\sum \frac{q_i}{r_i}=0$ (equal positive and negative charges at equal distances). The electric field contributions from the three dipoles all point along $OD$ (from positive to negative charges), giving a net field of $6K$ along $OD$. Line $PR$ is the perpendicular bisector of all three dipoles, so the potential is constant along it. Line $ST$ is not a perpendicular bisector, so the potential varies along it.