Force between two point charges $$q_1$$ and $$q_2$$ placed in vacuum at $$r$$ cm apart is $$F$$. Force between them when placed in a medium having dielectric $$K = 5$$ at $$\frac{r}{5}$$ cm apart will be:

We need to find the force between charges in a dielectric medium at a reduced distance.

In vacuum at distance $$r$$ the force is given by $$F = \frac{kq_1q_2}{r^2}$$. Since the dielectric constant is $$K = 5$$ and the separation is reduced to $$r/5$$, Coulomb's law in the medium yields:

$$F' = \frac{kq_1q_2}{K \cdot (r/5)^2} = \frac{kq_1q_2}{5 \cdot r^2/25} = \frac{25kq_1q_2}{5r^2} = \frac{5kq_1q_2}{r^2} = 5F$$

This shows that although the dielectric reduces the force by a factor of $$K = 5$$, the shorter separation increases it by a factor of 25, resulting in a net factor of $$\frac{25}{5} = 5$$ times the original force. Therefore, the correct answer is Option 2: $$5F$$.