Two point charges $q_1=q_2=2\mu C$ are fixed at $x_1=+3m$ and $x_2=-3m$ as shown in figure. A third particle of mass $1g$ and charge $q_3=-4\mu C$ are released from rest at $y=4.0m$. Find the speed of the particle as it reaches the origin.

 Let $v$ be the speed of particle at origin. From conservation of mechanical energy

$U_i+K_i=U_f+K_f$

or $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{q_3{q}_2}{{\left(r_{32}\right)}_i}+\frac{q_3{q}_1}{{\left(r_{31}\right)}_i}+\frac{q_2{q}_1}{{\left(r_{21}\right)}_i}\right]+0=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{q_3{q}_2}{{\left(r_{32}\right)}_f}+\frac{q_3{q}_1}{{\left(r_{31}\right)}_f}+\frac{q_2{q}_1}{{\left(r_{21}\right)}_f}\right]+\frac{1}{2}m{v}^2$

Here, ${\left(r_{21}\right)}_i={\left(r_{21}\right)}_f$

Substituting the proper values, we have

$\left(9.0\times {10}^9\right)\left[\frac{\left(-4\right)\left(2\right)}{\left(5.0\right)}+\frac{\left(-4\right)\left(2\right)}{\left(5.0\right)}\right]\times {10}^{-12}=\left(9.0\times {10}^9\right)\left[\frac{\left(-4\right)\left(2\right)}{\left(3.0\right)}+\frac{\left(-4\right)\left(2\right)}{\left(3.0\right)}\right]\times {10}^{-12}+\frac{1}{2}\times {10}^{-3}\times v^2$

$\therefore \left(9\times {10}^{-3}\right)\left(-\frac{16}{5}\right)=\left(9\times {10}^{-3}\right)\left(-\frac{16}{3}\right)+\frac{1}{2}\times {10}^{-3}\times v^2$

$\left(9\times {10}^{-3}\right)\left(16\right)\left(\frac{2}{15}\right)=\frac{1}{2}\times {10}^{-3}\times v^2$

$\therefore v=6.2m/s$