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

The correct option is B $6.2\text{m/s}$
Here, the charge $q_3$ is negative and it will get attracted towards $q_1$ and $q_2$ both. So, the net force on $q_3$ is towards origin as shown.


Due to this force, the charge will be accelerated towards the origin, but this acceleration is may or may not be constant.

So, to obtain the speed of particle at origin using our concepts of kinematics, we will first find the acceleration at some intermediate position and then integrate it with proper limits.

Alternatively, it is easy to use energy conservation principle, as the forces are conservative.

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

$U_i+K_i=U_f+K_f$

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

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

Substituting the proper values, we have

$(9\times {10}^9)[\frac{(-4)\left(2\right)}{\left(5.0\right)}+\frac{(-4)\left(2\right)}{\left(5.0\right)}]\times {10}^{-12}=(9.0\times {10}^9)[\frac{(-4)\left(2\right)}{\left(3.0\right)}+\frac{(-4)\left(2\right)}{\left(3.0\right)}]\times {10}^{-12}+\frac{1}{2}\times {10}^{-3}\times v^2$
$\therefore$ $(9\times {10}^{-3})(-\frac{16}{5})=(9\times {10}^{-3})(-\frac{16}{3})+\frac{1}{2}\times {10}^{-3}\times v^2$

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

$\therefore$ $v=6.2\text{m/s}$

Hence, option (b) is the correct answer.