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(B) From the equilibrium condition of the spheres,the forces acting are tension $T$,weight $mg$,and electrostatic repulsion $F_e = \frac{kq^2}{x^2}$.R

Vedclass · Accepted Answer

$v \propto x^{-1/2}$

Vedclass · Answer

(B) From the equilibrium condition of the spheres,the forces acting are tension $T$,weight $mg$,and electrostatic repulsion $F_e = \frac{kq^2}{x^2}$.Resolving forces: $T \cos 	heta = mg$ and $T \sin 	heta = \frac{kq^2}{x^2}$.Dividing the equations,we get $	an 	heta = \frac{kq^2}{x^2 mg}$.Since $	heta$ is small,$	an 	heta \approx \sin 	heta = \frac{x/2}{l} = \frac{x}{2l}$.Equating the two expressions for $	an 	heta$: $\frac{x}{2l} = \frac{kq^2}{x^2 mg} \implies q^2 = \frac{mg}{2lk} x^3 \implies q \propto x^{3/2}$.Differentiating with respect to time $t$: $\frac{dq}{dt} \propto \frac{3}{2} x^{1/2} \frac{dx}{dt}$.Given that $\frac{dq}{dt}$ is constant,we have $1 \propto x^{1/2} v$,which implies $v \propto x^{-1/2}$.

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