Two charged particles $(M,+Q)$ and $(m, -q)$ are placed in a uniform electric field $E$. After the particles are released, they stay at a  constant distance from each other. What is this distance$\left(L\right)$?

In order to maintain a constant separation, the particles must have the same acceleration. To find that acceleration, let us analyze the forces acting on the particles. Consider the forces acting on particle $Q;$ the leftward force due to$E$-field acting on $Q$ equals$QE$, the rightward force due to particle $q$ acting on $Q$ equals$kQ\left|q\right|L^2$

Assume that$Q> \left|q\right|;$: then the system of both charges will accelerate to the left. Newton's second-law equation of motion for particle $Q$ is then $QE-kQ\left|q\right|/L^2=Ma\Rightarrow \Rightarrow \Rightarrow \left(1\right)$

Considering the system of both charges immersed in an E-field and accelerating together once they are the equilibrium distance L apart, we find their common acceleration to be

$a=\frac{F_{net}}{total mass}=\frac{\left(Q-q\right)E}{\left(M+m\right)}\Rightarrow \Rightarrow \Rightarrow \left(2\right)$

Substituting eqn (2) into eqn (1)

$QE-\frac{KQ\left|q\right|}{L^2}=M\left\{\frac{\left(Q-q\right)E}{\left(M+m\right)}\right\}$

Solving Eq. (3) for L gives the desired result

$L=\sqrt{\frac{\left(M+m\right)kQq}{E\left(qM+Qm\right)}}$

The same result can be obtained when charge (-q, m) is analyzed and, of course, it makes no difference which charge is assumed to have the greater absolute value.