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(D) In equilibrium,the forces acting on one charge are as shown in the figure. Balancing the horizontal and vertical components of the forces:$T \cos

Vedclass · Accepted Answer

$x = \left(\frac{q^2 l}{2\pi\varepsilon_0 mg}\right)^{1/3}$

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(D) In equilibrium,the forces acting on one charge are as shown in the figure. Balancing the horizontal and vertical components of the forces:$T \cos 	heta = mg$  (Vertical equilibrium)$T \sin 	heta = F_e = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{x^2}$ (Horizontal equilibrium)Dividing the two equations:$	an 	heta = \frac{F_e}{mg} = \frac{q^2}{4\pi\varepsilon_0 x^2 mg}$From the geometry of the figure,for small $	heta$,$\sin 	heta \approx 	an 	heta \approx \frac{x/2}{l} = \frac{x}{2l}$.Substituting this into the equilibrium equation:$\frac{x}{2l} = \frac{q^2}{4\pi\varepsilon_0 x^2 mg}$$x^3 = \frac{2l q^2}{4\pi\varepsilon_0 mg} = \frac{l q^2}{2\pi\varepsilon_0 mg}$$x = \left(\frac{q^2 l}{2\pi\varepsilon_0 mg}ight)^{1/3}$The correct answer is $D$.

Two similar tiny balls of mass $m$,each carrying charge $q$,are hung from silk threads of length $l$ as shown in the figure. These are separated by a distance $x$ and the angle $2\theta$ is small. Then,for equilibrium:

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