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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 the initial attractive force. The ratio between initial charges on the spheres is
$ - \left( {3 + \sqrt 8 } \right)$ only
$ - 3 + \sqrt 8 $
$ - \left( {3 + \sqrt 8 \,} \right)$ or $\left( { - 3 + \sqrt 8 } \right)$
$+\sqrt 3$
Solution
$\mathrm{F}_{1}=\frac{\mathrm{k} \mathrm{Q}_{1} \mathrm{Q}_{2}}{\mathrm{d}^{2}}$ and $\mathrm{F}_{2}=\frac{\mathrm{k}\left(\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{2}\right)^{2}}{\mathrm{d}^{2}}$
According to question. $\quad \mathrm{F}_{1}=\mathrm{F}_{2}$
$\mathrm{Q}_{1} \mathrm{Q}_{2}=\frac{\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right)^{2}}{4} \Rightarrow 4 \mathrm{Q}_{1} \mathrm{Q}_{1}=\mathrm{Q}_{1}^{2}+\mathrm{Q}_{2}^{2}-2 \mathrm{Q}_{1} \mathrm{Q}_{2}$
$0=Q_{1}^{2}+Q_{2}^{2}-6 Q_{1} Q_{2} \Rightarrow \frac{Q_{1}}{Q_{2}}=-3 \pm \sqrt{8}$
