Two identical balls having like charges Q_1 a | vedclass

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Question

(A) Initial force: $F_1 = \frac{k Q_1 Q_2}{r^2}$When the two identical balls are brought into contact,the total charge is shared equally between them.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Initial force: $F_1 = \frac{k Q_1 Q_2}{r^2}$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 is $r' = r/2$. The new force $F_2$ is:$F_2 = \frac{k Q' Q'}{(r')^2} = \frac{k (\frac{Q_1 + Q_2}{2})^2}{(r/2)^2} = \frac{k (Q_1 + Q_2)^2 / 4}{r^2 / 4} = \frac{k (Q_1 + Q_2)^2}{r^2}$Given that $F_2 = 4.5 F_1$:$\frac{k (Q_1 + Q_2)^2}{r^2} = 4.5 \frac{k Q_1 Q_2}{r^2}$$(Q_1 + Q_2)^2 = 4.5 Q_1 Q_2$$Q_1^2 + Q_2^2 + 2 Q_1 Q_2 = 4.5 Q_1 Q_2$$Q_1^2 + Q_2^2 = 2.5 Q_1 Q_2$Dividing by $Q_2^2$:$(\frac{Q_1}{Q_2})^2 - 2.5 (\frac{Q_1}{Q_2}) + 1 = 0$Let $x = \frac{Q_1}{Q_2}$,then $x^2 - 2.5x + 1 = 0$$x^2 - 2x - 0.5x + 1 = 0$$x(x - 2) - 0.5(x - 2) = 0$$(x - 2)(x - 0.5) = 0$$x = 2$ or $x = 0.5 = 1/2$.Thus,the ratio is $2$.

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