Solution :
Given $q_A= 2 \mu C= 2 \times 10^{-6} \;C$
$q_B= -2 \mu C= -2 \times 10^{-6} \;C$
and $r=5\;cm$
potential $V= \large\frac{2 \times 10^{-6}}{4 \pi \in_ 0 x \times 10^{-2}}$
$\quad= \large\frac{-2 \times 10^{-6}}{4 \pi \in_0 (5-x) \times 10^{-2}}$
$x= 5-x$
$x= 2.5 $