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  • Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating massless strings of equal lengths. They come to equilibrium with the each string making a

Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating massless strings of equal lengths. They come to equilibrium with the 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 \epsilon _{0}}=k \right ).

Option: 1

2\sqrt{k\, \, mg\tan \theta }
 


Option: 2

\sqrt{k\, \, mg\tan \theta }


Option: 3

4\sqrt{k\, \, mg/\tan \theta }

 


Option: 4

4\sqrt{k\, \, mg\tan \theta }


Answers (1)

best_answer

\\ \text{In equilibrium, the expressions are given as},\\\ F=T \sin \theta \dots (1) \\ \\ m g=T \cos \theta \dots (2)\\ \\ \text{From above equations, it can be written as},\\ \tan \theta=\frac{F}{m g} \dots (3) \\ \text{The electric force is given as}, F=\frac{q^{2}}{4 \pi \varepsilon_{0} x^{2}}

\\ Substitute \ the \ value \ of \ F$ in equation $(3),$ we get $\tan \theta=\frac{q^{2}}{4 \pi \varepsilon_{0} m g x^{2}}$\\ \\ \\ $x=\sqrt{\frac{q^{2}}{4 \pi \varepsilon_{0} m g \tan \theta}}$ \\ $x=\sqrt{\frac{k q^{2}}{m g \tan \theta}}$ \\ The electric potential is given as, $V=\frac{k q}{\frac{x}{2}}+\frac{k q}{\frac{x}{2}}$ \\ $V=\frac{4 k q}{x}$ \\ $V=4 \sqrt{k m g \tan \theta}$ \\ \\ Thus, the electric potential at the centre of the line is $4 \sqrt{k m g \tan \theta}


 

Posted by

Kuldeep Maurya

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