In free space, a particle $A$ of charge $1\,\mu C$ is held fixed at a point $P.$ Another particle $B$ of the same charge and mass $4\,\mu g$ is kept at a distance of $1\,mm$ from $P$. If $B$ is released, then its velocity at a distance of $9\,mm$ from $P$ is [ Take $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N{m^2}{C^{ - 2}}$ ]
JEE MAIN 2019, Diffcult
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${{\text{q}}_{\text{A}}} = 1\,\mu {\text{C}};\,{{\text{q}}_8} = 1\,\mu {\text{C}},$
${{\text{m}}_{\text{B}}} = 4 \times {10^{ - 9}}\,{\text{kg}},\quad {{\text{r}}_{{\text{AB}}}} = {10^{ - 3}}\,{\text{m}}$
$\frac{1}{2} M_{B}\, V^{2}=k q_{A} q_{B}\left\{\frac{1}{10^{-3}}-\frac{1}{9 \times 10^{-3}}\right\}$
$\frac{1}{2} 4 \times 10^{-9}\, \mathrm{V}^{2}=9 \times 10^{8} \times 10^{-6} \times 10^{-6} \times \frac{8}{9} \times 10^{3}$
$V^{2}=\frac{8}{2} \times 10^{9}=4 \times 10^{9}$
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