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Two similar tiny balls of mass $m$, each carrying charge $q$ are hung from silk thread of length $l$ as shown in Fig. These are separated by a distance $x$ and angle $2 \theta \sim 10$. Then for equilibrium :-
$x = 2l$
$x = \frac{{l{q^2}}}{{4\pi {\varepsilon _0}mg}}$
$x = {\left( {\frac{{{q^2}mg}}{{4\pi {\varepsilon _0}}}} \right)^{\frac{1}{2}}}$
$x = {\left( {\frac{{{q^2}l}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
Solution
In equilibrium, the forces on one charge is as shown in fig. Balancing the $x$ – and $y$ – components of various forces.
$\mathrm{T} \cos \theta=\mathrm{mg}$
$\mathrm{T} \sin \theta=\mathrm{F}$
$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{x^{2}}$
or $\tan \theta=\frac{1}{4 \pi \varepsilon_{0}}, \frac{q^{2}}{x^{2} m g}$
From figure for $\theta=\frac{\mathrm{x}}{2 l}$
Thus $\frac{x}{2 l}=\frac{q^{2}}{4 \pi \varepsilon_{0} m g x^{2}}$
or $\quad x=\left(\frac{q^{2} l}{2 \pi \varepsilon_{0} m g}\right)^{1 / 3}$
The correct answer is $(4)$
