If the switch is flipped from position A to position B, what's the charge on capacitor C1 and C2?
Please see the attachment below
The Attempt at a Solution
I know that the capacitor C1 gains potential difference of 200V (= potential difference of a battery) and has charge Q = C1*200, but I'm not sure what happens once the flip is switched. My assumption is that it must discharge because it is no longer connected to the battery, but does it discharges until Q = 0 or until the charge on C2 = charge on C1. Could somebody please explain to me what exactly happens once the switch is in position B