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 :

 ${\mathrm{F}}^{\mathrm{\prime }}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\left(\frac{\mathrm{d}}{2}\right)}^2}=4\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{d}}^2}$ 

or     F' = 4F