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2. Electric Potential and Capacitance
hard
Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle $\theta $ from the vertical. If the mass of each charge is $m,$ then the electrostatic potential at the centre of line joining them will be $\left( {\frac{1}{{4\pi { \in _0}}} = k} \right).$
A$2\sqrt {k\,\,mg\,\,\tan \theta } $
B$\sqrt {k\,\,mg\,\,\tan \theta } $
C$4\sqrt {k\,\,mg\tan \theta } $
D$6\sqrt {k\,\,mg/\tan \theta } $
(JEE MAIN-2013)
Solution
In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$$\mathrm{mg}=\mathrm{T} \cos \theta$
$\tan \theta=\frac{F_{e}}{m g}=\frac{q^{2}}{4 \pi \epsilon_{0} x^{2} \times m g}$
$\therefore x = \sqrt {\frac{{{q^2}}}{{4\pi {\varepsilon _0}\,\tan \,\theta \,mg}}} $
Electric potential at the centre of the line
$V=\frac{k q}{x / 2}+\frac{k q}{x / 2}=4 \sqrt{k m g \tan \theta}$
Standard 12
Physics
