Two identical balls having like charges Q_1 a | vedclass

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(A) Initial force between the two charges $Q_1$ and $Q_2$ at distance $r$ is given by Coulomb's law:$F = \frac{k Q_1 Q_2}{r^2}$When the two identical

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$2$

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(A) Initial force between the two charges $Q_1$ and $Q_2$ at distance $r$ is given by Coulomb's law:$F = \frac{k Q_1 Q_2}{r^2}$When the two identical balls are brought into contact, the total charge is shared equally between them. Each ball now has a charge of $Q' = \frac{Q_1 + Q_2}{2}$.The new distance between them is $r' = \frac{r}{2}$.The new force $F'$ is given as $4.5 F$. Using Coulomb's law for the new configuration:$F' = \frac{k Q' Q'}{(r')^2} = \frac{k \left( \frac{Q_1 + Q_2}{2} \right)^2}{\left( \frac{r}{2} \right)^2} = \frac{k (Q_1 + Q_2)^2}{r^2}$Given $F' = 4.5 F$, we substitute the expressions:$\frac{k (Q_1 + Q_2)^2}{r^2} = 4.5 \left( \frac{k Q_1 Q_2}{r^2} \right)$$(Q_1 + Q_2)^2 = 4.5 Q_1 Q_2$$Q_1^2 + Q_2^2 + 2 Q_1 Q_2 = 4.5 Q_1 Q_2$$Q_1^2 + Q_2^2 - 2.5 Q_1 Q_2 = 0$Dividing by $Q_2^2$ and letting $x = \frac{Q_1}{Q_2}$:$x^2 - 2.5 x + 1 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{2.5 \pm \sqrt{6.25 - 4}}{2} = \frac{2.5 \pm \sqrt{2.25}}{2} = \frac{2.5 \pm 1.5}{2}$$x = \frac{4}{2} = 2$ or $x = \frac{1}{2} = 0.5$.Thus, the ratio $\frac{Q_1}{Q_2}$ can be $2$.

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