Second Ionisation Energy Confusion It is intuitive that the second ionisation will require more energy than the first. However, I'm having trouble explaining exactly why. I came across this other question on a similar topic: Why second ionisation energy greater than the first?

The answer made sense until the answer said:


  Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus.


Considering, say, chlorine, when a valence electron is removed, it makes sense that the electrostatic repulsion between valence electrons would decrease. However, by removing 1 electron, why does this increase the attractive force? I don't see how would increase the coulombic force between the nucleus and a valence electron, since the magnitude of both charges are constant. 

If it was the radius changing, why should the electrons be closer to the nucleus just because 1 electron is removed?
Beside that the proportional relationship proposed could be discussed further, as r rearrange, all this statement are related to the sums of forces. They stated that there would be a greater attractive force because they discuss ionization and this is the case. But the attractive force is the result of all attractions and repulsions. Any distribution of electric charges would lead to this situation and is generally an exercise treated after Coulomb law. Any reader can do that providing the dynamics is not involved.