Question

Two small spheres each having mass $mkg$ and charge $q$ coulomb are suspended from a point by insulating threads each I metre long but but of negligible mass. If $\theta$ is the angle, each thread makes with the vertical when equilibrium has been attained, show that
$q^2=\left(4mg{l}^2{\sin }^2\theta \tan \theta \right)4\pi {\epsilon }_0$

Answer 1

$T\cos \theta =mg$

$T\sin \theta =Fe$

$\tan \theta =\frac{Fe}{mg}$

$\because$ $Fe=\frac{{Kq}^2}{r^2}$

$Fe=\frac{{Kq}^2}{{\left(2d\right)}^2}$

$Fe=\frac{{Kq}^2}{4.l^2{\sin }^2\theta }$

$\therefore$ $\tan \theta =\frac{{Kq}^2}{4l^2{\sin }^2\theta \times mg}$

$=\frac{q^2}{\left({4l}^2{\sin }^2\theta mg\right)4\pi {ϵ}_0}$

$q^2=\left(4mg{l}^2{\sin }^2\theta \tan \theta \right)4\pi {ϵ}_0$

proved