My understanding of this circuit is that the capacitor must 'use up' all the voltage across the terminals of the battery. If you double the distance between the plates, I would assume the voltage drop across the two plates would remain the same as it is still connected to the battery. However as $$C=\epsilon A/d$$ (for a parallel plate capacitor) the capacitance must decrease. In addition, as $$Q=C\Delta Vc$$ for C to change, $\Delta Vc$ or $Q$ must also adjust. I would have thought that as the amount of charge is conserved, this would mean that the Voltage must change....
From other answers on this fantastic forum (linked below), I understand that when disconnected the voltage between the plates will intially increase lineary with distance and that (now) makes sense to me, and when disconnected from the battery I understand how the voltage across the plates can change.
But when connected to the battery I am unsure how this works and would appreciate any insight into the matter.
Why does the voltage increase when capacitor plates are separated? How is Capacitance affected by a seperation distance and with or without a battery?