Distance between charge q₁ and q₂ is d, then electric force between them is :
$\mathrm{F}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{d}}^2}$        ........(i)
If distance reduced to half, then new distance, d’ = d/2, so new force :
 $\mathrm{F}\text{'}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\left(\frac{\mathrm{d}}{2}\right)}^2}=4\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{d}}^2}$       
or    F’ = 4F