A small uncharged conducting sphere is placed in contact with an identical sphere but having $$4\times10^{-8} C$$ charge and then removed to a distance such that the force of repulsion between them is $$9\times10^{-3} N$$. The distance between them is $$ \text{ (Take } \frac{1}{4\pi \epsilon_o} \text{ as }9\times10^{9} \text{ in SI units)}$$

We need to find the distance between two identical spheres after charge sharing.

When identical conducting spheres are brought into contact, total charge distributes equally:

Each sphere gets $$\frac{4 \times 10^{-8} + 0}{2} = 2 \times 10^{-8}$$ C.

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$$

$$9 \times 10^{-3} = 9 \times 10^9 \times \frac{(2 \times 10^{-8})^2}{r^2}$$

$$9 \times 10^{-3} = 9 \times 10^9 \times \frac{4 \times 10^{-16}}{r^2}$$

$$r^2 = \frac{9 \times 10^9 \times 4 \times 10^{-16}}{9 \times 10^{-3}} = 4 \times 10^{-4}$$

$$r = 2 \times 10^{-2} \text{ m} = 2 \text{ cm}$$

The correct answer is Option 2: 2 cm.