Two equally charged small balls placed at a fixed distance experience a force 'F'. A similar uncharged ball after touching one of them is placed at the middle point between the two balls. the force experienced by this ball is

Let each small sphere initially have charge $q$, with separation $r$, such that the Coulomb force is $F = k\frac{q^2}{r^2}$.

After the uncharged ball touches the first, both share the charge equally, resulting in $q_A' = q_C' = \frac{q}{2}$, while the other ball retains charge $q$.

The third ball is placed midway, at distances $\frac{r}{2}$ from the others. The force due to ball A is $F_{AC} = k\frac{(q/2)(q/2)}{(r/2)^2} = k\frac{q^2}{r^2}$ (toward B), and from ball B is $F_{BC} = k\frac{q(q/2)}{(r/2)^2} = k\frac{2q^2}{r^2}$ (toward A).

The net force on C is the difference: $F_{net} = F_{BC} - F_{AC} = k\frac{q^2}{r^2} = F$.

Final answer: $\boxed{B}$