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 \(\left( {\frac{1}{{4\pi { \in _0}}} = k} \right).\)

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Answer & Solution

Correct Answer

(C) \(4\sqrt {k\,\,mg/\tan \theta } \)

Detailed explanation

In equilibrium, \(\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta\)

\(\mathrm{mg}=\mathrm{T} \cos \theta\)

\(\tan \theta=\frac{F_{e}}{m g}=\frac{q^{2}}{4 \pi \epsilon_{0} x^{2} \times m g}\)

\(\therefore x = \sqrt {\frac{{{q^2}}}{{4\pi {\varepsilon _0}\,\tan \,\theta \,mg}}} \)

Electric potential at the centre of the line

\(V=\frac{k q}{x / 2}+\frac{k q}{x / 2}=4 \sqrt{k m g / \tan \theta}\)

Same subject

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