1. Initial charges on spheres A and B

Since the two identical metallic spheres repel each other with force $F$, they must carry like charges. Let their initial charges be equal:

$q_A=q_B=q$

The distance between them is $r$. Hence initially,

$F = \frac{kq^2}{r^2}$

2. Sphere C touches A

Sphere $C$ is identical and initially uncharged.

Before contact:

• Charge on $A = q$
• Charge on $C = 0$

After contact, since the spheres are identical, charge gets shared equally:

$q_A = q_C = \frac{q}{2}$

3. Sphere C now touches B

Before contact:

• Charge on $B = q$
• Charge on $C = \frac{q}{2}$

Total charge on $B$ and $C$:

$q + \frac{q}{2} = \frac{3q}{2}$

Since they are identical, this total is shared equally:

$q_B = q_C = \frac{1}{2}\cdot \frac{3q}{2} = \frac{3q}{4}$

So finally:

$q_A = \frac{q}{2}, \qquad q_B = \frac{3q}{4}, \qquad q_C = \frac{3q}{4}$

4. Sphere C is placed at the midpoint of A and B

Now $C$ is at the midpoint between $A$ and $B$, so its distance from each is:

$\frac{r}{2}$

We find the forces on $C$ due to $A$ and $B$.

Force on C due to A

$F_{AC} = \frac{k\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{\left(\frac{r}{2}\right)^2}$

Since

$\left(\frac{r}{2}\right)^2 = \frac{r^2}{4}$

we get

$F_{AC} = k\cdot \frac{3q^2}{8}\cdot \frac{4}{r^2} = \frac{3}{2}\frac{kq^2}{r^2} = \frac{3F}{2}$

Force on C due to B

$F_{BC} = \frac{k\left(\frac{3q}{4}\right)\left(\frac{3q}{4}\right)}{\left(\frac{r}{2}\right)^2}$

$F_{BC} = k\cdot \frac{9q^2}{16}\cdot \frac{4}{r^2} = \frac{9}{4}\frac{kq^2}{r^2} = \frac{9F}{4}$

5. Net force on C

Both forces are along the line joining the spheres, but in opposite directions because $C$ is between $A$ and $B$ and all charges are positive.

So the net force on $C$ is the difference:

$F_{\text{net}} = F_{BC} - F_{AC}$

$F_{\text{net}} = \frac{9F}{4} - \frac{3F}{2} = \frac{9F}{4} - \frac{6F}{4} = \frac{3F}{4}$

6. Final answer

$\boxed{\frac{3F}{4}}$

So the correct option is B.