Two point charges of equal mass m and charge Q are suspended at a common point by two threads of negligible mass and length \ell. Show that at equilibrium the inclination angle \alpha of each thread to the vertical is given by
Q^{2}=16\pi\varepsilon_{o}mg\ell^{2}\sin^{2}\alpha\tan\alpha
If \alpha is very small, show that
\alpha=\sqrt[3]{\frac{Q^{2}}{16\pi\varepsilon_{o}mg\ell^{2}}}
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}}}
