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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(6\sqrt {k\,\,mg/\tan \theta } \)
Answer & Solution
Correct Answer
(C) \(4\sqrt {k\,\,mg/\tan \theta } \)
Step-by-step Solution
Detailed explanation
In equilibrium, \(\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta\) \(\mathrm{mg}=\mathrm{T} \cos \theta\) \(\tan \theta=\frac{F_{e}}{m g}=\frac{q^{2}}{4 \pi \epsilon_{0} x^{2} \times m g}\) \(\therefore x = \sqrt {\frac{{{q^2}}}{{4\pi {\varepsilon _0}\,\tan \,\theta \,mg}}} \)…
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