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 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 switched on. Then,

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

Answer 1

Answer: (A), (D)

Work done by tension = 0
Displacement$=l(1-cos{60}^0)$
Work done by Electric field
$=qE\times l(1-cos{60}^0)$
(B) $\sqrt{qEl}2=\frac{1}{2}m{V}^2\Rightarrow V=\sqrt{\frac{qEl}{m}}$
(C) $T-qE=\frac{m{V}^2}{l}\Rightarrow T=2qE$