Question

Two identical small charged spheres each having a mass $m$ hang in equilibrium as shown in figure.

The angle made by any string with vertical is $\theta$. Find the magnitude of charge on sphere.

• A. $q=\sqrt{\frac{4l^{2}mg\tan \theta {\sin }^2\theta }{k}}$
• B. $q=\sqrt{4k{l}^{2}mg\cos \theta {\sin }^2\theta }$
• C. $q=\sqrt{\frac{4l^{2}mg{\sin }^2\theta }{k\tan \theta }}$
• D. $q=\sqrt{2l^{2}mg{\sin }^2\theta }$

Answer 1

The correct option is A $q=\sqrt{\frac{4l^{2}mg\tan \theta {\sin }^2\theta }{k}}$

The forces acting on each sphere are tension $\left(T\right)$, weight $\left(mg\right)$, electrostatic force of repulsion $\left(F_e\right)$.


Applying equilibrium condition on sphere.
$T\cos \theta =mg\dots \dots \left(i\right)$
$T\sin \theta =F_e\dots \dots \left(ii\right)$
Thus, $\frac{F_e}{mg}=\tan \theta$
or, $F_e=mg\tan \theta \dots \dots \left(iii\right)$
The repulsion force is, (from Coulomb's law)
$F_e=\frac{k{q}^2}{x^2}=\frac{k{q}^2}{(2l\sin \theta {)}^2}\dots \dots \left(iv\right)$
From (iii) and (iv),
$\frac{k{q}^2}{\left(4l^2{\sin }^2\theta \right)}=mg\tan \theta$
$\Rightarrow q^2=\frac{4l^2{\sin }^2\theta mg\tan \theta }{k}$
$\therefore q=\sqrt{\frac{4l^2{\sin }^2\theta mg\tan \theta }{k}}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Due to identical shape of sphere and equal charges,}\\ \text{the angle}\theta \text{is same for both spheres from}\\ \text{vertical at equilibrium.}\\ \text{Bottom line: Draw FBD of sphere and apply equilibrium condition}\end{array}$