In the above image, one capacitor is uncharged while the other is at a potential Vi. If we close the switch then potential difference across both capacitor would become equal(because the upper and lower plates of both capacitors respectively would gain same potential)
My question is, why does the switch need to be closed? Sure, the circuit would remain open. But if we consider, say, 2 separately charged conductors and connect them with a copper wire, charge flows till both are at the same potential. Why cannot this happen with the capacitors in the picture? The bottom plate of both the capacitors are still connected, irrespective of the switch being closed or opened. Still, charge doesn't flow between the plates. Why is this?
My thought process was that in the charged capacitor, initially the net potential on the positive plate is actually the potential difference between both plates. And this plate is connected with the bottom plate of the uncharged capacitor. So, since there is a difference between both plates(Vi and zero), charge should flow.