Question

Two small spheres each having mass m kg and charge q coulomb are suspended from a point by insulating threads each l meter long but of negligible mass. If θ is the angle, each string makes with the vertical when equilibrium has been attained, show that q 2 = (4 mg l 2 sin2 θ tan θ) 4π∈0

Answer 1

this may help you ..I think so..

Answer 2

$\bold{q^{2}=\left(4 m g l^{2} \sin ^{2} \alpha \tan \alpha\right) 4 \pi \varepsilon_{0}}$ is proven when angle is θ with vertical when equilibrium is obtained.

Solution:

To attain equilibrium, all the forces in each planes has to be equal in magnitude and opposite in direction, for the same, refer the figure as shown above, from where we have the following forces balanced in each plane ,

Vertical plane: $T \cos \alpha=m g$

Horizontal plane: $T \sin \alpha=\frac{K q^{2}}{(2 l \sin \alpha)^{2}}$  

Therefore the equilibrium in all respect attained, will have the ratio of horizontal plane to the vertical plane amounting to be, $\frac{T \sin \alpha}{T \cos \alpha}=\frac{K q^{2}}{(2 t \sin \alpha)^{2}} \times \frac{1}{m g}$  

$\begin{array}{l}{\tan \alpha=\frac{K q^{2}}{m g(2 l \sin \alpha)^{2}}} \\ \\{q^{2}=\frac{4 m g l^{2} \sin ^{2} \alpha \tan \alpha}{K}} \\ \\{q^{2}=\frac{4 m g l^{2} \sin ^{2} \alpha \tan \alpha}{\frac{1}{4 \pi \epsilon 0}}}\end{array}$

$q^{2}=\left(4 m g l^{2} \sin ^{2} \alpha \tan \alpha\right) 4 \pi \varepsilon_{0}$

Hence the expression achieved.