Question

A Particle $A$ having a charge of $5\cdot 0\times {10}^{-7}C$ is fixed in a vertical wall. $A$ second particle $B$ of mass $100g$ and having equal charge is suspended by a silk thread of length $30cm$ from the wall. The point of suspension is $30cm$ above the particle $A$. Find the angle of the thread with the vertical when it stays in equilibrium.

Answer 1

The force is given as,

$F=\frac{KqQ}{r^2}$

Since $q=Q$

Then the force between the charges are given at distance $r=0.6\cdot \sin \left(\frac{\theta }{2}\right)$

$F=8.99\times {10}^9\times \frac{{\left(5\times {10}^{-7}\right)}^2}{0.6\sin {\left(\frac{\theta }{2}\right)}^2}$

By using Lami’s theorem

$\frac{F}{\sin \theta }=\frac{mg}{\cos \frac{\theta }{2}}$

The force is given as,

$F=\frac{KqQ}{r^2}$

Since $q=Q$

Then the force between the charges are given at distance $r=0.6\cdot \sin \left(\frac{\theta }{2}\right)$

$F=8.99\times {10}^9\times \frac{{\left(5\times {10}^{-7}\right)}^2}{0.6\sin {\left(\frac{\theta }{2}\right)}^2}$

By using Lami’s theorem

$\frac{F}{\sin \theta }=\frac{mg}{\cos \frac{\theta }{2}}$

By using trigonometric equation, We get the angle for the equilibrium is $\theta ={16.9}^{\circ }$