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Two balls of same mass and carrying equal charge are hung from a fixed support of length l. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, x between the balls is proportional to : 

Option: 1

l
 


Option: 2

l^{2}


Option: 3

l^{2/3}

 


Option: 4

l^{1/3}


Answers (1)

best_answer


Two balls of same mass and carrying equal charge are hung from a fixed support of length l.

According to diagram

\\ T \cos \theta=m g \ldots \ldots(I) \\ \\ T \sin \theta=\frac{k q^{2}}{x^{2}} \ldots \ldots( II )\\ \text{Divided equation} \ \ (II) \ \ \text{by equation} \ \ (I) \\ \tan \theta=\frac{k q^{2}}{m g x^{2}} \ldots . .(III)
\begin{aligned} &\tan \theta=\frac{C B}{O B}\\ &\tan \theta=\frac{\frac{x}{2}}{\left(l^{2}-\frac{x^{2}}{4}\right)^{\frac{1}{2}}} \end{aligned}

Put the value from equation (III)

\frac{\frac{x}{2}}{\left(l^{2}-\frac{x^{2}}{4}\right)^{\frac{1}{2}}}=\frac{k q^{2}}{m g x^{2}}$ \\ $x^{3}=\frac{2 k q^{2}}{m g}\left(l^{2}-\frac{x^{2}}{4}\right)^{\frac{1}{2}}$ \\ If is small then, $\frac{x^{2}}{4}<<l^{2}$ \\ $x^{3}=\frac{2 k q^{2}}{m g}\left(l^{2}\right)^{\frac{1}{2}}$ \\ $x=\frac{2 k q^{2}}{m g}(l)^{\frac{1}{3}}$ \\ \\ $x \propto l^{\frac{1}{3}}

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