Question 4.2: Two point charges of equal mass m and charge Q are suspended...

Consider the system of charges as shown in Figure 4.3, where $F_{e}$ is the electric or Coulomb force, T is the tension in each thread, and mg is the weight of each charge. At A or B

$T\sin\alpha=F_{e}$

$T\cos\alpha=mg$

Hence

$\frac{\sin\alpha}{\cos\alpha}=\frac{F_{e}}{mg}=\frac{1}{mg}\cdot \frac{Q^{2}}{4\pi\varepsilon_{o}r^{2}}$

But $r=AB$ is given by

$r=2\ell\sin\alpha$

Hence

$Q^{2}\cos\alpha=16\pi\varepsilon_{o}mg\ell^{2}\sin^{3}\alpha$

or

$Q^{2}=16\pi\varepsilon_{o}mg\ell^{2}\sin^{2}\alpha\tan\alpha$

as required. When $\alpha$ is very small

$\tan\alpha \simeq \alpha \simeq \sin\alpha$

and so

$Q^{2}=16\pi\varepsilon_{o}mg\ell^{2}\alpha^{3}$

or

$\alpha=\sqrt[3]{\frac{Q^{2}}{16\pi\varepsilon_{o}mg\ell^{2}}}$