${\mathrm{E}_{\mathrm{i}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{1}}+0}$

${\mathrm{E}_{\mathrm{f}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{2}}+\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}}$

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 $ ${\mathrm{v}=\mathrm{q} \sqrt{\frac{\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)}{4 \pi \varepsilon_{0} \mathrm{mr}_{1} \mathrm{r}_{2}}}}$