Answer :
To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,
[tex]F_{initial} = \frac{kq_1q_2}{r^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge at each object
r = Distance between them
As the distance is doubled so,
[tex]F_{final} = \frac{kq_1q_2}{( 2r )^2}[/tex]
[tex]F_{final} = \frac{ kq_1q_2}{ 4r^2}[/tex]
[tex]F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}[/tex]
[tex]F_{final} = \frac{1}{4} F_{initial}[/tex]
[tex]\frac{F_{final}}{ F_{initial}} = \frac{1}{4}[/tex]
Therefore the factor is 1/4