Let the initial charge on spheres A and B be $q$. The force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.

1. Contact with A: When uncharged sphere C touches A, the charge $q$ distributes equally. Charge on A becomes $q_A = q/2$, and charge on C becomes $q_C = q/2$.
2. Contact with B: Sphere C (now with charge $q/2$) touches sphere B (charge $q$). Total charge is $q + q/2 = 3q/2$. This splits equally, so $q_B = 3q/4$ and $q_C = 3q/4$.
3. Placed at Midpoint: C is placed at distance $r/2$ from both A and B. Force from A on C ($F_{CA}$): Repulsive. $F_{CA} = \frac{1}{4\pi\epsilon_0} \frac{(q/2)(3q/4)}{(r/2)^2} = \frac{1}{4\pi\epsilon_0} \frac{3q^2/8}{r^2/4} = 1.5 F$. Force from B on C ($F_{CB}$): Repulsive. $F_{CB} = \frac{1}{4\pi\epsilon_0} \frac{(3q/4)(3q/4)}{(r/2)^2} = \frac{1}{4\pi\epsilon_0} \frac{9q^2/16}{r^2/4} = 2.25 F$.
4. Net Force: Since forces are opposite, $F_{net} = F_{CB} - F_{CA} = 2.25F - 1.5F = 0.75F = \frac{3F}{4}$.