During my circuit design I came across unusual circuit as shown below:
At first glance this looks trivial however if you tinker a bit with it you will find voltmeter readings that are hard to explain. So for the circuit as it stands above all seems to be clear, capacitor slowly charges to 3V through voltmeter huge 10MOhm resistor. After it reaches steady-state it has equal number of charges on each plate but with opposite sign. Now let's unplug this capacitor and plug it back inversely so the previously negatively charged plate is on battery + side. What happens is voltmeter now shows +6V and slowly discharges to 1.5V. Here is the list of my questions:
If you apply KVL to the newly created circuit it make sense. We have: $$ -3 - 3 + Vu1 = 0 $$ $$ Vu1 = 6V $$ But let's think of it in terms of capacitor charges. What we did is we connected negatively charge plate to positive terman of the battery. So I would rather expect it to quickly (because o the absence of any resistance) disolve alll nagative charge and gain positive ones therefore lowering capacitor pottential to 0 - since second plate slowly (due to 10Mohm resistor of voltmeter) discharges its positiive charge. So we have a situation in which right after the flip one side is +Q and other also has +Q so capacitor net potetnial shall be 0 in my opinion. Where is the mistake in my understanding?
Second thing is, why it stops at 1.5V. For this is have no explanation.
PS: All I am describing is real circuit, the schematic may suggest I simulate it but this is not the case.