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Two similar small balls having mass m and charge q are suspended by silk strings having length l , according to the figure. If in the figure q is an acute angle then for equilibrium what will the distance between the centre of the two balls.

The force and acting on the system are as follows.
T is tension in string, F is coulomb force and mg is weight.
For equilibrium,
$T\; \cos\; q = mg \;and\; T \sin q = F$
$\therefore \tan \theta= \large\frac{F}{mg}=\frac{1}{4 \pi \in_0 X^2} \frac{q^2}{mg}$
If $\theta$ is small.
$\therefore \tan \theta \sin \theta \approx \large\frac{x}{2l}$
$\therefore \large\frac{x}{2l} = \frac{1}{4 \pi \in_0 x^2} \frac{q^2}{mg}$
$x= \bigg\{ \large\frac{q^2 l}{2 \pi \in _0 mg}\bigg\}^{1/3}$
Hence A is the correct answer.