A particle of mass $m$ and charge $q$ is fastened to one end $A$ of a massless string having equilibrium length $l$, whose other end is fixed at point $O$. The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field is switched on along the direction shown in the figure, then the speed of the particle when it crosses the $x$-axis is
Answer: (B) $\sqrt{\dfrac{qEl}{m}}$
The force $qE$ acts along $+x$, so the particle swings about $O$ on a circle of radius $l$ from the $60^\circ$ position down to the $x$-axis. The string tension does no work.
Displacement along the field: $l - l\cos60^\circ = \dfrac{l}{2}$.
$$\frac{1}{2}mv^2 = qE\cdot\frac{l}{2} \Rightarrow v = \sqrt{\frac{qEl}{m}}$$
Solution by Sreeraj P, M.Sc Physics