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(C) The initial electrostatic force between the two spheres is given by $F_1 = k \frac{q_1 q_2}{r^2}$.When the two identical spheres are touched,the t

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More than before

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(C) The initial electrostatic force between the two spheres is given by $F_1 = k \frac{q_1 q_2}{r^2}$.When the two identical spheres are touched,the total charge is redistributed equally between them because they are identical. The new charge on each sphere is $q' = \frac{q_1 + q_2}{2}$.The new electrostatic force between the spheres at the same distance $r$ is $F_2 = k \frac{q' q'}{r^2} = k \frac{(\frac{q_1 + q_2}{2})^2}{r^2} = k \frac{(q_1 + q_2)^2}{4r^2}$.Comparing the two forces,we look at the product of the charges: $(q_1 + q_2)^2$ vs $4q_1 q_2$.Since $(q_1 + q_2)^2 - 4q_1 q_2 = (q_1 - q_2)^2$,and $(q_1 - q_2)^2 > 0$ for $q_1 
eq q_2$,it follows that $(q_1 + q_2)^2 > 4q_1 q_2$.Therefore,$F_2 > F_1$,meaning the force is more than before.

Two identical conducting spheres having unequal positive charges $q_1$ and $q_2$ are separated by a distance $r$. If they are made to touch each other and then separated again to the same distance $r$,the electrostatic force between the spheres in this case will be (neglect induction of charges):

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