Two small equal point charges of magnitude q | vedclass

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Question

(C) Let $L$ be the length of each string. The distance between the two charges is $x = 2L \sin 	heta$.In equilibrium,the forces acting on each charge

Vedclass · Accepted Answer

$4\sqrt{k\,mg	an 	heta}$

Vedclass · Answer

(C) Let $L$ be the length of each string. The distance between the two charges is $x = 2L \sin 	heta$.In equilibrium,the forces acting on each charge are tension $T$,weight $mg$,and electrostatic force $F_e = \frac{kq^2}{x^2}$.Resolving forces: $T \sin 	heta = F_e$ and $T \cos 	heta = mg$.Dividing these gives $	an 	heta = \frac{F_e}{mg} = \frac{kq^2}{x^2 mg}$.Thus,$x^2 = \frac{kq^2}{mg 	an 	heta}$,which implies $x = q \sqrt{\frac{k}{mg 	an 	heta}}$.The electrostatic potential $V$ at the centre of the line joining the charges (at distance $x/2$ from each charge) is:$V = \frac{kq}{x/2} + \frac{kq}{x/2} = \frac{4kq}{x}$.Substituting $x$:$V = \frac{4kq}{q \sqrt{\frac{k}{mg 	an 	heta}}} = 4 \sqrt{k^2 \cdot \frac{mg 	an 	heta}{k}} = 4 \sqrt{k \, mg 	an 	heta}$.

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