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(B) Initially,the charges are $q_A = Q$ and $q_B = -Q$. The force between them at a distance $r$ is given by Coulomb's Law:$F = \frac{k Q (-Q)}{r^2} =

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$\frac{9 F}{16}$

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(B) Initially,the charges are $q_A = Q$ and $q_B = -Q$. The force between them at a distance $r$ is given by Coulomb's Law:$F = \frac{k Q (-Q)}{r^2} = -\frac{k Q^2}{r^2}$When $25 \%$ of the charge from $A$ is transferred to $B$,the amount transferred is $\Delta q = 0.25 Q = \frac{Q}{4}$.The new charges are:$q_A' = Q - \frac{Q}{4} = \frac{3Q}{4}$$q_B' = -Q + \frac{Q}{4} = -\frac{3Q}{4}$The new force $F'$ between the charges is:$F' = \frac{k q_A' q_B'}{r^2} = \frac{k (\frac{3Q}{4})(-\frac{3Q}{4})}{r^2}$$F' = -\frac{9}{16} \frac{k Q^2}{r^2}$Since $F = -\frac{k Q^2}{r^2}$,we substitute this into the expression for $F'$:$F' = \frac{9}{16} F$

Two point charges $A$ and $B$,having charges $+Q$ and $-Q$ respectively,are placed at a certain distance apart and the force acting between them is $F$. If $25 \%$ of the charge of $A$ is transferred to $B$,then the force between the charges becomes

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