Two identical small conducting spheres carry charges of $Q_{1}$ and $Q_{2}$ with $Q_{1} >> Q_{2}$ . The spheres are d distance apart. The force they exert on each other is $F_{1}$ . The spheres are made to touch each other and then separated by a distance d. The force they exert on each other now is $F_{2}$ . Then $F_{1} / F_{2}$ is :-

\frac{4 Q_{1}}{Q_{2}}

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\frac{Q_{1}}{4 Q_{2}}

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\frac{4 Q_{2}}{Q_{1}}

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\frac{Q_{2}}{4 Q_{1}}

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Solution

The correct option is C $\frac{4 Q_{2}}{Q_{1}}$
$\begin{matrix} F = \frac{K Q_{1} Q_{2}}{d^{2}} Q_{1}^{^{'}} + Q_{2}^{^{'}} = Q_{1} + Q_{2} . . . . . . . . (1) \frac{K Q_{1}^{^{'}}}{R {1 p t}_{1}} = \frac{K Q_{2}^{^{'}}}{R_{2}} \Rightarrow \frac{Q_{1}^{^{'}}}{R_{1}} = \frac{Q_{2}^{^{'}}}{R_{2}} \Rightarrow Q_{1}^{^{'}} = (Q_{1} + Q_{2}) - \frac{Q_{1}^{^{'}} R_{2}}{R_{1}} \Rightarrow Q_{1}^{^{'}} (\frac{R_{1} + R_{2}}{R_{1}}) = (Q_{1} + Q_{2}) \Rightarrow Q_{1}^{^{'}} = \frac{(Q_{1} + Q_{2}) R_{1}}{(R_{1} + R_{2})} = \frac{(Q_{1} + Q_{2})}{2} \Rightarrow Q_{2}^{^{'}} = \frac{(Q_{1} + Q_{2}) R_{2}}{(R_{1} + R_{2})} = \frac{Q_{1} + Q_{2}}{2} F_{2} = \frac{K Q_{1}^{1} Q_{2}^{2}}{d^{2}} = \frac{K}{d^{2}} \frac{{(Q_{1} + Q_{2})}^{2}}{4} \approx \frac{K Q_{1}^{2}}{4 d^{2}} ∴ \frac{F_{1}}{F_{2}} = \frac{K Q_{1} Q_{2}}{d^{2}} \times \frac{4 d^{2}}{K Q_{1}^{2}} = \frac{4 Q_{2}}{Q_{1}} H e n c e, o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Two identical small conducting spheres carry charges of Q1 and Q2 with Q1>>Q2. The spheres are d distance apart. The force they exert on each other is F1. The spheres are made to touch each other and then separated by a distance d. The force they exert on each other now is F2. Then F1/F2 is :-