A small metal sphere, carrying a net charge of q_1=−2μC, is held in a stationery...

Let i be the point when the charge q2 is 0.8 m.
Due to the same nature of the two charges, charge q2 will experience an electrostatic force of replusion from charge q1, which will cause charge q2's velocity to drop. Charge Q2 will eventually arrive at point a f, where it is the furthest away from charge Q1 and where its velocity is zero. Let r be the smallest distance between q1 and q2.
By total energy conservation
$$\begin{gathered} ⟹E_i=E_f \\ P{E}_i+K{E}_i=P{E}_i+K{E}_f \end{gathered}$$
Velocity of charge q2 at point i = 20 ms^−1
∴ Kinetic energy of q2 at point i = 12m(20)2 J
Potential energy of charge q2 at point i will be kq1q2/0.8
Potential energy of charge q2 at point f will be kq1q2/r
Kinetic energy of q2 at point f = 0, vf=0
$$\begin{gathered} ⟹\frac{k{q}_1{q}_2}{0.8}+\frac{1}{2}(0.0015)(20{)}^2=\frac{k{q}_1{q}_2}{r}+0 \\ ⟹0.48J=\frac{k{q}_1{q}_2}{r} \\ ⟹r=\frac{9\times {10}^9\times 2\times {10}^{-6}\times 8\times {10}^{-6}}{0.48} \\ \therefore r=0.3m \end{gathered}$$