If two charges $$q_1$$ and $$q_2$$ are separated with distance $$d$$ and placed in a medium of dielectric constant $$k$$. What will be the equivalent distance between charges in air for the same electrostatic force?

The electrostatic force between two charges in a medium with dielectric constant $$k$$ is given by Coulomb's law:

$$F_{\text{medium}} = \frac{1}{4\pi\epsilon_0 k} \frac{q_1 q_2}{d^2}$$

In air (which is approximately vacuum), the force is:

$$F_{\text{air}} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$

where $$r$$ is the equivalent distance in air for the same force.

Since the force is the same in both cases, set $$F_{\text{air}} = F_{\text{medium}}$$:

$$\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0 k} \frac{q_1 q_2}{d^2}$$

Cancel the common terms $$\frac{1}{4\pi\epsilon_0}$$ and $$q_1 q_2$$:

$$\frac{1}{r^2} = \frac{1}{k d^2}$$

Take the reciprocal of both sides:

$$r^2 = k d^2$$

Take the square root of both sides:

$$r = d \sqrt{k}$$

Thus, the equivalent distance in air is $$d\sqrt{k}$$.

Comparing with the options, the correct answer is A. $$d\sqrt{k}$$.