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

Here the work done$\left(W\right)$ by the electric field is the kinetic energy$\left(KE\right)$ of the mass m.
$W=Fdx=qE(OB-OC)=qE(l-l\cos 60)=\frac{1}{2}qEl$
$KE=\frac{1}{2}m{v}^2$
therefore, $\frac{1}{2}qEl=\frac{1}{2}m{v}^2...\left(1\right)$
$\Rightarrow v=\sqrt{\frac{qEl}{m}}$
If T is the tension in the string, $qE=T-\frac{m{v}^2}{l}....\left(2\right)$
using (1), (2) become: $qE=T-qE\Rightarrow T=2qE$
ans: $\left(B\right),\left(C\right)$