1. Initial charges on spheres $A$ and $B$

Since the two identical conducting spheres carry equal charge, let each initially have charge $q$.

So, $q_A = q, \qquad q_B = q$

They are far apart, so we can treat them as point charges. The initial force between them is $F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}$

where $r$ is the separation.

2. Sphere $C$ is initially uncharged and touched to $A$

Sphere $C$ is identical to $A$, so when two identical conducting spheres touch, charge gets shared equally.

Before touching: $q_A = q, \qquad q_C = 0$

Total charge on $A$ and $C$: $q+0=q$

After touching, each gets $\frac{q}{2}$

So now, $q_A = \frac{q}{2}, \qquad q_C = \frac{q}{2}$

3. Now sphere $C$ is touched to $B$

Before touching $B$: $q_B = q, \qquad q_C = \frac{q}{2}$

Total charge on $B$ and $C$: $q + \frac{q}{2} = \frac{3q}{2}$

Since $B$ and $C$ are identical, after touching they share equally: $q_B = q_C = \frac{1}{2}\cdot \frac{3q}{2} = \frac{3q}{4}$

Thus after removal of $C$, $q_A = \frac{q}{2}, \qquad q_B = \frac{3q}{4}$

4. New force between $A$ and $B$

The new electrostatic force is $F' = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{r^2}$

$F' = \frac{1}{4\pi\varepsilon_0}\frac{3q^2}{8r^2}$

But initially, $F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}$

Therefore, $F' = \frac{3}{8}F$

5. Checking options

• A: $F$
• B: $\frac{3F}{4}$
• C: $\frac{3F}{8}$
• D: $\frac{F}{2}$

Hence the correct option is: $\boxed{\frac{3F}{8}}$

So, Option C is correct.