Two identical conducting spheres carrying dif | vedclass

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(C) Let the initial charges on the two spheres be $Q_1$ and $Q_2$. Since they attract each other,they must have opposite signs,so $Q_1 Q_2 < 0$.The in

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the initial charges on the two spheres be $Q_1$ and $Q_2$. Since they attract each other,they must have opposite signs,so $Q_1 Q_2 The initial force is $F = \frac{k |Q_1 Q_2|}{d^2}$. Since they attract,$F = -\frac{k Q_1 Q_2}{d^2}$.When the spheres are brought into contact,the total charge is shared equally because the spheres are identical. The new charge on each sphere is $Q' = \frac{Q_1 + Q_2}{2}$.After returning to their original positions,the new force is $F' = \frac{k (Q')^2}{d^2} = \frac{k (Q_1 + Q_2)^2}{4d^2}$.According to the problem,the magnitude of the new repulsive force is equal to the magnitude of the initial attractive force: $\frac{k (Q_1 + Q_2)^2}{4d^2} = \frac{k |Q_1 Q_2|}{d^2}$.Since $Q_1$ and $Q_2$ have opposite signs,$|Q_1 Q_2| = -Q_1 Q_2$. Thus,$(Q_1 + Q_2)^2 = -4 Q_1 Q_2$.$Q_1^2 + Q_2^2 + 2 Q_1 Q_2 = -4 Q_1 Q_2 \Rightarrow Q_1^2 + Q_2^2 + 6 Q_1 Q_2 = 0$.Dividing by $Q_2^2$,we get $(\frac{Q_1}{Q_2})^2 + 6(\frac{Q_1}{Q_2}) + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $x = \frac{Q_1}{Q_2}$:$\frac{Q_1}{Q_2} = \frac{-6 \pm \sqrt{36 - 4}}{2} = \frac{-6 \pm \sqrt{32}}{2} = -3 \pm \sqrt{8}$.

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