Two positive point charges $Q$ are fixed at $x=-a$  and $x=a$ on the x-axis. Another positive point charge $q$ is placed at the origin. If $q$ is slightly displaced along the x-axis and released, the velocity acquired by $q$ when it reaches origin is proportional to: 

The potential due to the two-charge system at any point x from the origin is:

$V\left(x\right)=\frac{KQ}{a+x}+\frac{KQ}{a-x}=\frac{2KQa}{a^2-x^2}$

Potential at origin is:

$V\left(o\right)=\frac{2KQ}{a}$

If $v$ is the velocity required by q when it reaches origin, using $\mathrm{\Delta }K+\mathrm{\Delta }U=0$, we have 

$\begin{array}{l}\frac{1}{2}{\mathrm{mv}}^2+\mathrm{q}\left(\frac{2\mathrm{KQ}}{\mathrm{a}}-\frac{2\mathrm{KQa}}{{\mathrm{a}}^2-{\mathrm{x}}^2}\right)=0\\ \Rightarrow \frac{1}{2}{\mathrm{mv}}^2-\frac{2{\mathrm{KQqx}}^2}{\mathrm{a}\left({\mathrm{a}}^2-{\mathrm{x}}^2\right)}=0\\ \Rightarrow \frac{1}{2}{\mathrm{mv}}^2=\frac{2{\mathrm{KQqx}}^2}{{\mathrm{a}}^3}(\text{ for }\mathrm{x}\ll \mathrm{a})\end{array}$

Hence, $\mathrm{v}\propto {\mathrm{a}}^{-3/2}.$