In free space, a particle A of charge 1 μC is held fixed at point P. Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is:
[Take $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$]

We have two identical point charges, each of magnitude $$q = 1\;\mu{\rm C} = 1 \times 10^{-6}\;{\rm C}$$. Charge A is fixed at point P, while charge B (mass $$m$$) is released from rest. Because the charges are like-signed, the electrostatic force is repulsive, so B moves away from A. The mechanical energy of the system is conserved because no non-conservative forces act in free space.

First we translate all given quantities into SI units:

$$q = 1 \times 10^{-6}\;{\rm C}$$

$$m = 4\;\mu{\rm g} = 4 \times 10^{-6}\;{\rm kg}$$  (since $$1\;\mu{\rm g} = 10^{-6}\;{\rm kg}$$)

Initial separation $$r_i = 1\;{\rm mm} = 1 \times 10^{-3}\;{\rm m}$$

Final separation $$r_f = 9\;{\rm mm} = 9 \times 10^{-3}\;{\rm m}$$

The Coulomb constant is provided as $$k = \dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\;{\rm N\,m^2\,C^{-2}}.$$

The electrostatic potential energy of two point charges is given by the formula

$$U = k\,\frac{q_1 q_2}{r}.$$

Here $$q_1 = q_2 = q$$, so

$$U = k\,\frac{q^2}{r}.$$

We calculate the initial potential energy at separation $$r_i$$:

$$U_i = k\,\frac{q^2}{r_i} = \left(9 \times 10^{9}\right) \frac{\left(1 \times 10^{-6}\right)^2}{1 \times 10^{-3}} = 9 \times 10^{-3}\times\frac{1}{1 \times 10^{-3}} = 9\;{\rm J}.$$

Next, the potential energy when the separation becomes $$r_f$$ is

$$U_f = k\,\frac{q^2}{r_f} = \left(9 \times 10^{9}\right) \frac{\left(1 \times 10^{-6}\right)^2}{9 \times 10^{-3}} = 9 \times 10^{-3}\times\frac{1}{9 \times 10^{-3}} = 1\;{\rm J}.$$

The change in potential energy is therefore

$$\Delta U = U_i - U_f = 9\;{\rm J} - 1\;{\rm J} = 8\;{\rm J}.$$

Since particle B starts from rest, its initial kinetic energy is zero. By conservation of mechanical energy, the decrease in potential energy equals the increase in kinetic energy of B:

$$\frac{1}{2} m v^2 = \Delta U.$$

Substituting $$m = 4 \times 10^{-6}\;{\rm kg}$$ and $$\Delta U = 8\;{\rm J}$$ gives

$$\frac{1}{2}\,(4 \times 10^{-6})\,v^2 = 8,$$ $$\Rightarrow\; (2 \times 10^{-6})\,v^2 = 8,$$ $$\Rightarrow\; v^2 = \frac{8}{2 \times 10^{-6}} = 4 \times 10^{6},$$ $$\Rightarrow\; v = \sqrt{4 \times 10^{6}} = 2 \times 10^{3}\;{\rm m\,s^{-1}}.$$

Thus the speed of particle B when it is 9 mm away from P is $$2.0 \times 10^{3}\;{\rm m\,s^{-1}}.$$

Hence, the correct answer is Option 3.