Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

The correct answer is:
$\frac{3F}{8}$

Step-by-Step Explanation:

1. Initial State:
Let two identical conducting spheres A and B have charge $q$ each. Let the distance between their centers be $r$.
According to Coulomb's Law, the electrostatic force of repulsion $F$ between them is given by:
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

2. Contact between Sphere A and a Third Identical Uncharged Sphere (C):
Let C be the third identical uncharged conducting sphere (initial charge $q_C=0$).
When sphere C is brought in contact with sphere A, the total charge shared between them is distributed equally because the spheres are identical.
The new charge on both sphere A and sphere C becomes:
$q_A\text{'}=q_C\text{'}=\frac{q+0}{2}$

3. Contact between Sphere C and Sphere B:
Now, sphere C (carrying charge $\frac{q}{2}$) is brought in contact with sphere B (carrying charge $q$).
Since they are identical, the total charge is again shared equally between B and C:
$q_B\text{'}=q_C\text{'}\text{'}=\frac{q+{\displaystyle \frac{q}{2}}}{2}=\frac{3q}{4}$

4. Final Repulsion Force:
Sphere C is finally removed. The remaining charges on A and B are:
$q_A\text{'}=\frac{q}{2}$
and
$q_B\text{'}=\frac{3q}{4}$

The new force of repulsion $F\text{'}$ between spheres A and B is:
$F\text{'}=\frac{1}{4\pi {\epsilon }_0}\frac{q_A\text{'}·q_B\text{'}}{r^2}$

Substituting the values of the new charges:
$F\text{'}=\frac{1}{4\pi {\epsilon }_0}\frac{\left({\displaystyle \frac{q}{2}}\right)·\left({\displaystyle \frac{3q}{4}}\right)}{r^2}$

Simplifying the fraction:
$F\text{'}=\frac{3}{8}\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}\right)$

Using the definition of the initial force $F$:
$F\text{'}=\frac{3F}{8}$