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(C) Since the two spheres are identical,the final charge on each of them after contact is $q = \frac{(q_1 + q_2)}{2}$.The initial electrostatic force

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(C) Since the two spheres are identical,the final charge on each of them after contact is $q = \frac{(q_1 + q_2)}{2}$.The initial electrostatic force between them is $F_1 = \frac{K q_1 q_2}{r^2}$.The final electrostatic force between them after contact is $F_2 = \frac{K q^2}{r^2} = \frac{K}{r^2} \left( \frac{q_1 + q_2}{2} ight)^2$.To compare $F_1$ and $F_2$,we calculate the difference $F_2 - F_1 = \frac{K}{r^2} \left[ \left( \frac{q_1 + q_2}{2} ight)^2 - q_1 q_2 ight]$.Simplifying the expression: $F_2 - F_1 = \frac{K}{4r^2} (q_1^2 + q_2^2 + 2q_1 q_2 - 4q_1 q_2) = \frac{K}{4r^2} (q_1 - q_2)^2$.Since $(q_1 - q_2)^2 > 0$ for $q_1 
eq q_2$,it follows that $F_2 - F_1 > 0$,which means $F_2 > F_1$. Thus,the force increases.

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,the electrostatic force between them in this case will be:

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