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(A) At any instant,the forces acting on one sphere are tension $T$,gravity $mg$,and electrostatic force $F_e$.Resolving forces:$T \cos 	heta = mg$ $(

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

Vedclass · Answer

(A) At any instant,the forces acting on one sphere are tension $T$,gravity $mg$,and electrostatic force $F_e$.Resolving forces:$T \cos 	heta = mg$ $(i)$$T \sin 	heta = F_e$ (ii)Dividing (ii) by $(i)$:$	an 	heta = \frac{F_e}{mg} = \frac{kq^2}{x^2 mg}$For small $	heta$,$	an 	heta \approx \sin 	heta = \frac{x}{2l}$.So,$\frac{x}{2l} = \frac{kq^2}{x^2 mg} \Rightarrow q^2 \propto x^3 \Rightarrow q \propto x^{3/2}$.Differentiating with respect to time $t$:$\frac{dq}{dt} \propto \frac{d}{dt}(x^{3/2}) = \frac{3}{2} x^{1/2} \frac{dx}{dt}$.Since the charge leaks at a constant rate,$\frac{dq}{dt} = 	ext{constant}$.Therefore,$1 \propto x^{1/2} v$,where $v = \frac{dx}{dt}$ is the velocity of approach.$v \propto x^{-1/2}$.

Column-$I$	Column-$II$
$(A)$ Ring along its axis	$(P)$ Graph with a peak at a distance $r > 0$
$(B)$ Uniformly charged solid sphere	$(Q)$ Graph increasing linearly for $r < R$ and decreasing as $1/r^2$ for $r > R$
$(C)$ Uniformly charged spherical shell	$(R)$ Graph with zero field for $r < R$ and decreasing as $1/r^2$ for $r > R$
$(D)$ Combination of charge $+Q$ and $-Q$ at the perpendicular bisector	$(S)$ Graph with a maximum at the center and decreasing as $r$ increases

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