Q.
Two small spheres of masses $M_{1}$ and $M_{2}$ are suspended by weightless insulating threads of lengths $L_{1}$ and $L_{2}$. The spheres carry charges $Q_{1}$ and $Q_{2}$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $\theta_{1}$ and $\theta_{2}$ as shown. Which one of the following conditions is essential, if $\theta_{1} = \theta_{2}$
A) $M_{1} \neq M_{2}$ but $Q_{1} = Q_{2}$
B) $M_{1} = M_{2}$
C) $Q_{1} = Q_{2}$
D) $L_{1} = L_{2}$
Explanation
(b) The three forces acting on each sphere are :
(i) Tension
(ii) Weight
(iii) Electrostatic force of repulsion
[image:1_image257.emf]
For sphere 1
In equilibrium, from figure
$T_{1}{\cos\theta}_{1} = M_{1}g;T_{1}{\sin\theta}_{1} = F_{1}$
$\therefore\tan\theta_{1} = \frac{F_{1}}{M_{1}g}$
For sphere 2
In equilibrium, from figure
$T_{2}{\cos\theta}_{2} = M_{2}g;T_{2}{\sin\theta}_{2} = F_{2}$
$\therefore\tan\theta_{2} = \frac{F_{2}}{M_{2}g}$
Force of repulsion between two charges are same.
$\therefore F_{1} = F_{2}$
$\theta_{1} = \theta_{2}$ only if $\frac{F_{1}}{M_{1}g} = \frac{F_{2}}{M_{2}g}$
But $F_{1} = F_{2},$ then $M_{1} = M_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.