Coulomb potential in nuclear fission?

In the case of symmetrical fission we must consider the Coulomb potential between the two nuclei :

$E=k\frac{{q}^{2}}{r}=k\frac{(\frac{Ze}{2}{)}^{2}}{2{r}_{0}(\frac{a}{2}{)}^{1/3}}=k\frac{{2}^{1/3}}{8}\frac{{Z}^{2}{e}^{2}}{{r}_{0}{A}^{1/3}}\frac{1}{1.6\times {10}^{-13}}=0.157\frac{{Z}^{2}}{A}$

The process is spontaneous only if $Q>E$ (where $Q$ is the difference between final and initial binding energy).

Why is the process allowed only if $Q>E$ ? Don't the nuclei repel each other rendering the process "easier"?

In the case of symmetrical fission we must consider the Coulomb potential between the two nuclei :

$E=k\frac{{q}^{2}}{r}=k\frac{(\frac{Ze}{2}{)}^{2}}{2{r}_{0}(\frac{a}{2}{)}^{1/3}}=k\frac{{2}^{1/3}}{8}\frac{{Z}^{2}{e}^{2}}{{r}_{0}{A}^{1/3}}\frac{1}{1.6\times {10}^{-13}}=0.157\frac{{Z}^{2}}{A}$

The process is spontaneous only if $Q>E$ (where $Q$ is the difference between final and initial binding energy).

Why is the process allowed only if $Q>E$ ? Don't the nuclei repel each other rendering the process "easier"?