See attachment. Assume that the two capacitor's plates were originally connected together as one capacitor, and thus their plate separations were equal, along with their charge densities. In the photo below then, the capacitor has been split down the middle and with one halve the distance between the plates has been decreased. Wires are now connecting the plates. Considering the plates were once connected, they should have equal charge densities right after the cut. This should mean that the electric field magnitude between each plate set is approximately equal - even though the plate separation differs. Taking the potential over the two capacitors will give that the one of greater separation has greater potential. Assuming that charge density has not yet changed on any plates, the decrease in the other two plates results in a lesser potential. My question is then, why does taking potential differences imply that charge will move from the capacitor of higher potential to the one of lower potential when there seems to be no apparent forces on the charges? That is, through all this, it seems as if the charges could still be in equilibrium if the electric field between each set of plates is equal at first. I am assuming that the capacitor of greater potential difference will charge the one of less until both have equal potential differences. I just don't see where the charge dis-equilibrium comes from. It actually seems to be out of equilibrium once they are both at equal potential differences as the electric fields will now be unequal.