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(C) Initially,both spheres have a charge $q$. The force between them is given by Coulomb's law:$F = \frac{k q^2}{r^2}$When $10 \%$ of electrons are tr

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$0.99 F$

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(C) Initially,both spheres have a charge $q$. The force between them is given by Coulomb's law:$F = \frac{k q^2}{r^2}$When $10 \%$ of electrons are transferred from one sphere to the other,the charge on the sphere that loses electrons becomes $q + 0.1q = 1.1q$,and the charge on the sphere that gains electrons becomes $q - 0.1q = 0.9q$.Note: Since electrons are negatively charged,losing them makes a sphere positive,and gaining them makes it negative. However,the problem implies a transfer of charge magnitude $0.1q$. Assuming the spheres were initially positive,the new charges are $1.1q$ and $0.9q$.The new force $F^{\prime}$ is:$F^{\prime} = \frac{k(1.1q)(0.9q)}{r^2}$$F^{\prime} = 0.99 \frac{k q^2}{r^2}$$F^{\prime} = 0.99 F$

Two identical charged spheres separated by a distance $r$ repel each other with a force $F$. If $10 \%$ of electrons are transferred from one sphere to the other,then the force between them becomes

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