Two identical conducting spheres A and B carry an equal charge. They are separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, force between A and B would be equal to:

Let the identical spheres A and B be separated by a distance $$r$$ (with $$r$$ much larger than their diameters, so we may treat the charges as point charges).

Suppose each of A and B initially carries charge $$q$$.

The electrostatic force between two point charges is given by Coulomb’s law:

$$F \;=\; k\;\dfrac{q_1 q_2}{r^{\,2}}$$

where $$k$$ is the electrostatic constant.

With $$q_1 = q_2 = q$$, the initial force is

$$F \;=\; k\;\dfrac{q^{\,2}}{r^{\,2}} \quad\text{(1)}$$

Now sphere C (identical to A and B) is uncharged at the start.

First contact: C touches A. Being identical conductors, they share their total charge equally.

Total charge on A + C before contact: $$q + 0 = q$$.

After contact, charge on each (A and C) becomes

$$\dfrac{q}{2}$$.

So,

$$q_A = \dfrac{q}{2}, \qquad q_C = \dfrac{q}{2}$$

Second contact: the same sphere C now touches B.

At this moment, charges are

$$q_B = q, \qquad q_C = \dfrac{q}{2}$$

The total charge on B + C before this contact is

$$q + \dfrac{q}{2} = \dfrac{3q}{2}$$.

After they touch, the charge again divides equally (because the spheres are still identical):

Charge on each (B and C) becomes

$$\dfrac{1}{2}\left(\dfrac{3q}{2}\right) = \dfrac{3q}{4}$$.

So, finally,

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

Sphere C is removed, so only A and B remain with the above charges.

New force between A and B. Using Coulomb’s law again:

$$F' = k\;\dfrac{q_A\,q_B}{r^{\,2}} = k\;\dfrac{\left(\dfrac{q}{2}\right)\left(\dfrac{3q}{4}\right)}{r^{\,2}} = k\;\dfrac{3q^{\,2}}{8\,r^{\,2}}$$

Compare this with equation (1): $$F = k\dfrac{q^{\,2}}{r^{\,2}}$$. Hence

$$F' = \dfrac{3}{8}\,F$$

Therefore the force becomes $$\dfrac{3F}{8}$$.

Hence, the correct answer is Option C.