Two point charges having charge +Q, –q and mass M, m respectively are separated by a distance L. They are released from rest in a uniform electric field E. The electric field is parallel to line joining both the charges and is directed from negative to positive charge. For the separation between particles to remain constant, the value of L is $\left(\mathrm{K}=\frac{1}{4\mathrm{\pi }{\in }_0}\right)$

In order to maintain constant separation, the particles must have the same acceleration.
Assuming the system of both charges to accelerate towards left. Applying Newton's second law.

$\mathrm{QE}-\frac{\mathrm{KQq}}{{\mathrm{L}}^2}=\mathrm{Ma}.....\left(1\right)$
Under given condition the acceleration of both charges should be same and should also be equal to acceleration of centre of mass of both the charges.
$\mathrm{a}=\frac{{\mathrm{F}}_{\text{net }}}{\text{ total mass }}=\frac{(\mathrm{Q}-\mathrm{q})\mathrm{E}}{\mathrm{m}+\mathrm{M}}.....\left(2\right)$
Hence from equation (1) and (2) we get $\mathrm{L}=\sqrt{\frac{(\mathrm{M}+\mathrm{m})\mathrm{KQq}}{\mathrm{E}(\mathrm{qM}+\mathrm{Qm})}}$