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. The speed of the particle when it reaches $B$ is -

• A. $\sqrt{\frac{2qEl}{m}}$
• B. $\sqrt{\frac{qEl}{m}}$
• C. $\sqrt{\frac{3qEl}{m}}$
• D. $\sqrt{\frac{5qEl}{m}}$

Answer 1

The correct option is B $\sqrt{\frac{qEl}{m}}$


Let work done by tension on string is $W_{Tension}$
Work done by electrostatic force is $W_{\text{Electric force}}$
On applying work-energy theorem,
$W_{\text{Tension}}+W_{\text{Electric force}}=\text{change in kinetic energy}$
$\Rightarrow 0+qE[l-l\cos {60}^{\circ }]=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{1}{2}m{v}^2=\frac{qEl}{2}$
$\Rightarrow v=\sqrt{\frac{qEl}{m}}$

Why this question ? Concept - At every instant, Tension and displacement will be perpendicular to each other. Thus, $W_T=0$