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 uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x -axis is

Electric field acts along +x, so force on charge is

F=qE

Initially particle is at angle

$$60^{\circ}$$

with x-axis and string length l.

Initial coordinates:

$$x_i=l\cos60^{\circ}=\frac{l}{2}$$

When it crosses x-axis, particle is at

$$x_f=l$$

(since still constrained on circle radius lll).

Displacement along field:

$$\Delta x=l-\frac{l}{2}=\frac{l}{2}$$

Work done by electric field:

$$W=qE\Delta x$$

$$=\frac{qEl}{2}$$

Tension does no work.

Using work-energy theorem:

$$\frac{1}{2}mv^2=\frac{qEl}{2}$$

So

$$mv^2=qEl$$

so

$$\sqrt{\frac{qEl}{m}}$$