Change in distribution of charge on plates of capacitor on earthing one plate

Question

If we consider the following situation:

Now the capacitors are already charged so the potential difference between the plates is well defined but the absolute potential isn't.

So what will happen if we close the switch $S_1$. I realize that grounding means the potential is zero at the left side of $C_1$ now but does that mean it automatically becomes $-30 V$ on it's right and no charge flows between the capacitor and the ground?

If yes, then in this situation:

Does it make any difference if we close either of the two switches? I thought that the charge distribution on the plates should change on grounding it but I'm quite confused about how that could happen?

Any help would be appreciated.

Answer 1

When solving “floating” circuits you need to remember that every conductor has self capacitance and is therefore connected to ground. Usually, the self capacitance is so small that it can be neglected, but in a question like this it becomes important. So, including the self capacitance we would re-draw the circuit as follows:

Note that this circuit is still not very solvable. To get a good solution you would also need to add in the wire resistances. Otherwise you wind up with time constants of 0. But the circuit drawing should give you the conceptual idea of how to approach this analysis.