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Two positively charged spheres of masses $m_1$ and $m_2$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$. Charges on the two spheres are $q_1$ and $q_2$, respectively. At equilibrium, both strings make the same angle $\theta$ with the vertical. Then
$q_1 m_1=q_2 m_2$
$m_1=m_2$
$m_1=m_2 \sin \theta$
$q_2 m_1=q_1 m_2$
Solution
(b)
In given situation, forces on each of charged sphere are
$(i)$ gravitational pull ( $m g)$
$(ii)$ electrostatic repulsion $\left(\frac{k q_1 q_2}{r^2}\right)$
$(iii)$ tension of string $(T)$
as shown below.
If we resolve tension in horizontal and vertical directions, we have following situation in equilibrium.
$\text { So, } T \sin \theta=\frac{k q_1 q_2}{r^2} \text { and } T \cos \theta=m g$
$\Rightarrow \quad \tan \theta=\frac{k q_1 q_2}{r^2 \cdot m g}$
As angle $\theta$ is same for both spheres, we have
$\tan \theta_1=\tan \theta_2$
or $\frac{k q_1 q_2}{r^2 m_1 g}=\frac{k q_1 q_2}{r^2 m_2 g}$
$\Rightarrow m_1=m_2$
