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(A) Let the initial charges on the two identical balls be $Q_1$ and $Q_2$ separated by a distance $r$. The initial force of repulsion is given by Coul

Vedclass · Accepted Answer

$2$

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(A) Let the initial charges on the two identical balls be $Q_1$ and $Q_2$ separated by a distance $r$. The initial force of repulsion is given by Coulomb's law:$F = \frac{k Q_1 Q_2}{r^2}$When the balls are brought into contact,the total charge is shared equally between them because they are identical. Each ball now has a charge of $\frac{Q_1 + Q_2}{2}$.They are then moved to a new distance $r' = \frac{r}{2}$. The new force of repulsion $F'$ is:$F' = \frac{k (\frac{Q_1 + Q_2}{2})^2}{(\frac{r}{2})^2} = \frac{k (Q_1 + Q_2)^2}{r^2}$Given that $F' = 4.5 F$,we have:$\frac{k (Q_1 + Q_2)^2}{r^2} = 4.5 	imes \frac{k Q_1 Q_2}{r^2}$$(Q_1 + Q_2)^2 = 4.5 Q_1 Q_2$$Q_1^2 + 2 Q_1 Q_2 + Q_2^2 = 4.5 Q_1 Q_2$$Q_1^2 - 2.5 Q_1 Q_2 + Q_2^2 = 0$Dividing by $Q_2^2$ and letting $x = \frac{Q_1}{Q_2}$:$x^2 - 2.5 x + 1 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{2.5 \pm \sqrt{6.25 - 4}}{2} = \frac{2.5 \pm 1.5}{2}$$x = \frac{4}{2} = 2$ or $x = \frac{1}{2} = 0.5$Thus,the ratio of the initial charges is $2$.

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