In the given figure,two tiny conducting balls | vedclass

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(A) In equilibrium,the forces acting on one ball are the electrostatic force $F_e$,the tension $T$ in the thread,and the gravitational force $mg$.Reso

Vedclass · Accepted Answer

${\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$

Vedclass · Answer

(A) In equilibrium,the forces acting on one ball are the electrostatic force $F_e$,the tension $T$ in the thread,and the gravitational force $mg$.Resolving the tension $T$ into components,we have:$T \sin 	heta = F_e = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{x^2}$ ....... $(i)$$T \cos 	heta = mg$ ....... $(ii)$Dividing $(i)$ by $(ii)$,we get $	an 	heta = \frac{F_e}{mg} = \frac{q^2}{4\pi \varepsilon_0 x^2 mg}$.Given that $	heta$ is very small,$	an 	heta \approx \sin 	heta = \frac{x/2}{L} = \frac{x}{2L}$.Equating the two expressions for $	an 	heta$:$\frac{x}{2L} = \frac{q^2}{4\pi \varepsilon_0 x^2 mg}$$x^3 = \frac{2q^2 L}{4\pi \varepsilon_0 mg} = \frac{q^2 L}{2\pi \varepsilon_0 mg}$Therefore,$x = {\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} ight)^{\frac{1}{3}}}$.

In the given figure,two tiny conducting balls of identical mass $m$ and identical charge $q$ hang from non-conducting threads of equal length $L$. Assume that $\theta$ is so small that $\tan \theta \approx \sin \theta$,then for equilibrium,$x$ is equal to:

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