Question

A particle of mass $m$ and charge $q$ is fastened to one end of a string of length $l$. The other end of the string is fixed to the point $O$. The whole system lies on a frictionless horizontal plane. Initially, the mass is at rest at $A$. A uniform electric field in the direction shown is then switched on. Then

• A. the speed of the particle when it reaches $B$ is $\sqrt{\frac{2qEl}{m}}$
• B. the speed of the particle when it reaches $B$ is $\sqrt{\frac{qEl}{2m}}$
• C. the tension in the string when particles reach at $B$ is $qE$.
• D. the tension in the string when the particles reach at $B$ is $2qE$.

Answer 1

The correct option is D the tension in the string when the particles reach at $B$ is $2qE$.

By the work energy theorem,
$\frac{1}{2}m{v}^2=qE[l-l\cos {60}^{\circ }]$
$v=\sqrt{\frac{qEl}{m}}$
At point $B$,
$T-qE=\frac{m{v}^2}{l}$
$T=qE+\frac{m}{l}{\left(\sqrt{\frac{qEl}{m}}\right)}^2=2qE$
$\therefore$ (d) is correct.