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Two identical conducting spheres carrying different charges attract each other with a force F when placed in air medium at a distance d apart. The spheres are brought into contact and then taken to their original positions. Now, the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force The ratio between initial charges on the spheres is

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$(3 + \sqrt{8}) only$

$- 3 + \sqrt{8} only$

$(3 + \sqrt{8}) or (3 - \sqrt{8})$

$+ \sqrt{3}$

answer is C.

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Detailed Solution

$F_{1} = \frac{k Q_{1} Q_{2}}{d^{2}} and F_{2} = \frac{k {(\frac{Q_{1} - Q_{2}}{2})}^{2}}{d^{2}}$

According to question, F₁ = F₂

$Q_{1} Q_{2} = \frac{{(Q_{1} - Q_{2})}^{2}}{4}$

$\Rightarrow 4 Q_{1} Q_{2} = Q_{1}^{2} + Q_{2}^{2} - 2 Q_{1} Q_{2}$

$0 = Q_{1}^{2} + Q_{2}^{2} - 6 Q_{1} Q_{2}$

$\Rightarrow \frac{Q_{1}}{Q_{2}} = 3 \pm \sqrt{8}$

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