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 option is 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 initial electrostatic force of repulsion $F$ between them is given by:

$F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q\cdot q}{r^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{r^2}$

2. Sphere C touches Sphere A:
A third identical uncharged conducting sphere C (initial charge = 0) is brought in contact with sphere A.
Since spheres A and C are identical, the total charge distributes equally between them upon contact:

$q_A=q_C=\frac{q+0}{2}=\frac{q}{2}$

3. Sphere C touches Sphere B:
Now, sphere C (carrying charge $\frac{q}{2}$) is brought in contact with sphere B (carrying charge $q$).
The total charge on B and C is shared equally between them:

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

4. Final Force calculation:
Sphere C is finally removed. The new charges on spheres A and B are $q_A=\frac{q}{2}$ and $q_B=\frac{3q}{4}$ respectively.
The new force of repulsion $F^{\prime }$ between them is:

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_A\cdot q_B}{r^2}$

Substituting the new charges:

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\left(\frac{q}{2}\right)\cdot \left(\frac{3q}{4}\right)}{r^2}$

$F^{\prime }=\frac{3}{8}\cdot \left(\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{r^2}\right)$

$F^{\prime }=\frac{3F}{8}$