Question

A particle of mass $m$ and charge $(+q)$ is fastened to one end of a light string of length $l$. The other end of string is fixed to the point $O$. The whole system lies on a frictionless horizontal plane. Initially, the mass is at rest at $\text{A}$. A uniform electric field in the direction shown is 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 $\sqrt{\frac{qEl}{m}}$
• C. The tension in the string when particles reaches at B is $2qE$
• D. The tension in the string when the particle reaches at B is zero.

Answer 1

Answer: (B), (C)

The tension in the string when particles reaches at B is $2qE$


From diagram,
$\text{OC}=l\cos {60}^{\circ }=\frac{l}{2}$

$\therefore \text{CB}=\text{OB}-\text{OC}=l-\frac{l}{2}=\frac{l}{2}$

Work done by electric force from A to B,

$W=F\times d=qE\times \text{CB}=\frac{qEl}{2}$

Work done by tension is zero because it is perpendicular to the path taken.

So, using work energy theorem,
$\text{Net work done}=\text{change in kinetic energy}$

$\frac{qEl}{2}=\frac{1}{2}m{v}^2-\frac{1}{2}m(0{)}^2$

$\Rightarrow v=\sqrt{\frac{qEl}{m}}$

Now, at $\text{B}$ let tension be $T$,

So, $T-qE=\frac{m{v}^2}{R}$

$\Rightarrow T-qE=\frac{m\left(\frac{qEl}{m}\right)}{l}$

$\therefore T=2qE$

Hence, options (b) and (c) are the correct answers.