\(25 \%\) charge from \(A\) is transferred to \(B\)
New force \((\mathrm{F})=\frac{\mathrm{K}\left(\frac{3 \mathrm{q}}{4}\right)\left(\frac{-3 \mathrm{q}}{4}\right)}{\mathrm{r}^{2}}=\frac{-9 \mathrm{kq}^{2}}{16 \mathrm{r}^{2}}=\frac{9 \mathrm{F}}{16}\)