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(D) Initially,both spheres have charge $q$ and are separated by a distance $r$. According to Coulomb's law,the force between them is $F = \frac{Kq^2}{

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$\frac{7F}{16}$

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(D) Initially,both spheres have charge $q$ and are separated by a distance $r$. According to Coulomb's law,the force between them is $F = \frac{Kq^2}{r^2}$.When $75\%$ of the charge from one sphere is transferred to the other,the amount of charge transferred is $0.75q = \frac{3q}{4}$.The charge on the first sphere becomes $q - \frac{3q}{4} = \frac{q}{4}$.The charge on the second sphere becomes $q + \frac{3q}{4} = \frac{7q}{4}$.The new force $F'$ between the spheres is given by $F' = \frac{K(\frac{q}{4})(\frac{7q}{4})}{r^2} = \frac{7}{16} \cdot \frac{Kq^2}{r^2}$.Since $F = \frac{Kq^2}{r^2}$,we substitute this into the equation to get $F' = \frac{7}{16}F$.

The force between two identical spheres charged with the same charge is $F$. If $75\%$ of the charge from one sphere is transferred to the other sphere,then the new force will be:

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