well I know what electric potential is and it's zero outside C1 and the two fields add inside, you should ask what happens during charging and this I'm not sure about because you give energy to charge the plates of C2 to reach the Q of C1 and in this period you form two other capacitors that lower the voltage of C1 temporarily , displacement current depends on change in electric flux, if C1 E field decreases then there's an opposite displ. current in C1 because the change is a negative term and the displacement current has the same direction in the two other capacitors let's call them C3 C4 so in total there's displ. current going in the opposite direction of current flow , when Q1=Q2 there's positive displacement current going through all plates (this should be a larger one).. .. does this look logical to you?? \