JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Work, Power & Energy

The field of an infinite line charge falls off as $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$, so moving the point charge from $r_0$ out to $r$ takes it through a potential difference $\Delta V=\frac{\lambda }{2\pi {\epsilon }_0}\ln \frac{r_0}{r}$. Starting from rest, the work-energy theorem gives $\frac{1}{2}m{v}^2=q\Delta V=\frac{q\lambda }{2\pi {\epsilon }_0}\ln \frac{r_0}{r}$. Both charges are positive, so they repel and the charge speeds up as $r$ grows beyond $r_0$; writing this with the positive quantity, $v^2\propto \ln \frac{r}{r_0}$. Taking the square root, $v\propto \sqrt{\ln \frac{r}{r_0}}$.