Question

Two identical small balls, each of mass m and charge q is hung with silk thread (the length of each thread is l) as shown is given fig. The distance between them is x and the angle between the threads ($2\theta ={10}^{\circ }$). Then calculate the distance x in the equilibrium state.

Answer 1

Following forces are acting on ball $A$ and $B$

(i) Weight of ball = mb

(ii) Tension in string $T$

(iii) Repulsive force between both balls

$F=\frac{1}{4\pi {ϵ}_0}\times \frac{g^2}{x^2}$...........(i)

Forces are in equilibrium

Refer image,

Therefore at $A$

$\frac{F}{AC}=\frac{mg}{OC}=\frac{T}{AO}$

or $F=mg\times \frac{AC}{OC}$............(ii)

From equation (i) and (ii)

$\frac{1}{4\pi {ϵ}_0}\times \frac{q^2}{x^2}=mg\times \frac{AC}{OC}$

but $AC=l\sin \theta ,OC=l\cos \theta$

$\frac{1}{4\pi {ϵ}_0}\frac{q^2}{x^2}=mg\times \frac{l\sin \theta }{l\cos \theta }$

or $\frac{1}{4\pi {ϵ}_0}\frac{q^2}{x^2}=mg\tan \theta$

$\therefore x^2=\frac{q^2}{4\pi {ϵ}_0\times mg\tan \theta }=\frac{9\times {10}^9\times q^2}{mg\tan 10}$

$=\frac{9\times {10}^9\times q^2}{m\times 9.8\times 0.648}=1.42\frac{q^2}{m}$

$\Rightarrow$ $x=1.19\frac{q}{\sqrt{m}}$