A particle of mass m and charge q is fastened | vedclass

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(C) The particle moves in a circular path of radius $\ell$ centered at $O$ due to the constraint of the string.Initially,the particle is at an angle o

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$\sqrt{\frac{q E \ell}{m}}$

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(C) The particle moves in a circular path of radius $\ell$ centered at $O$ due to the constraint of the string.Initially,the particle is at an angle of $60^{\circ}$ with the $x$-axis. The initial $x$-coordinate is $x_i = \ell \cos(60^{\circ}) = \frac{\ell}{2}$.When the particle crosses the $x$-axis,its final $x$-coordinate is $x_f = \ell$.The displacement of the particle in the direction of the electric field is $\Delta x = x_f - x_i = \ell - \frac{\ell}{2} = \frac{\ell}{2}$.Using the work-energy theorem,$W_{	ext{electric}} = \Delta K$.The work done by the electric field is $W = qE \Delta x = qE \left(\frac{\ell}{2}ight)$.Since the system starts from rest,$K_i = 0$ and $K_f = \frac{1}{2}mv^2$.Thus,$qE \frac{\ell}{2} = \frac{1}{2}mv^2$.Solving for $v$,we get $v^2 = \frac{qE\ell}{m}$,which implies $v = \sqrt{\frac{qE\ell}{m}}$.

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