Electrostatics Part 2

Let two identical spheres have charge q. And distance between them $=r$$\therefore$ Force between the spheres $(F)=\frac{k q^2}{r^2}$Now when an identical uncharged sphere $C$ comes in contact with $A$, charge $q$ on sphere A get's divided equally to both sphere.So, both sphere $A$ and $C$ have charge $=\frac{q}{2}$Now, C get's in contact with B. So their total charge $\left(q+\frac{q}{2}\right)$ gets divided equally.So, charge on both $B$ and $C$ is $=\frac{q+\frac{q}{2}}{2}=\frac{3q}{4}$Now, $C$ is place midpoint between $A$ and $B$.Repulsion force between A and C,$F_{AC}=\frac{k \left(\frac{q}{2}\right) \left(\frac{3q}{4}\right)}{\left(\frac{r}{2}\right)^2}=\frac{4k \times 3 q^2}{8 r^2}$Repulsion force between B and C,$F_{BC}=\frac{k \left(\frac{3q}{4}\right) \left(\frac{3q}{4}\right)}{\left(\frac{r}{2}\right)^2}=\frac{4k \times 9 q^2}{16 r^2}$$\therefore$ Net force on C,$F_{net}=F_{BC}-F_{AC}$$=\frac{4k \times 9 q^2}{16 r^2}-\frac{4k \times 3 q^2}{8 r^2}$$=\frac{k q^2}{r^2} \left(\frac{9}{4}-\frac{3}{2}\right)$$=\frac{k q^2}{r^2} \times \frac{3}{4}$ $=\frac{3}{4} F .$ as $F=\frac{k q^2}{r^2} ]$