Two particles each of mass $m$ and charge $q$ are separated by distance $r_1$ and the system is left free to move at $t = 0$. At time $t$ both the particles are found to be separated by distance $r_2$. The speed of each particle is
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Due to symmetry, each particle will have same speed,
${\mathrm{E}_{\mathrm{i}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{1}}+0}$
${\mathrm{E}_{\mathrm{f}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{2}}+\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}}$
As the field is conservative, hence applying law of conservation of energy.
${\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{1}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{2}}+\mathrm{mv}^{2}}$
$\therefore $ ${\mathrm{v}=\mathrm{q} \sqrt{\frac{\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)}{4 \pi \varepsilon_{0} \mathrm{mr}_{1} \mathrm{r}_{2}}}}$
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