We know the force between two point charges is given by:

$$ F=k\frac{q_1q_2}{r^2}$$

Here \(k=9\times10^9 Nm^2C^{-2}\), \(q_1\) and \(q_2\) are magnitudes of the charge whereas \(r\) is the distance between them.

If we double the magnitude of both charges but at the same time also double the distance, the force becomes:

$$F_n=k\cdot\frac{(2q_1)(2q_2)}{(2r)^2}=k\cdot\frac{4q_1q_2}{4r^2} = \frac{k\cdot q_1q_2}{r^2}\cdot\frac{4}{4}$$

$$F_n=\frac{F}{4}$$