Capacitor Charge Sharing

Question

Consider the following circuit

Capacitor C2 is charged up to 1V from an external source before-hand. The top plate of C2 is connected to the bottom plate of C1. The top plate of C1 is at a voltage of 0.25V

Now, let's say we increased 0.25V to 0.35V:

Due to the increase in voltage across C1, the charge on it must increase according to Q = CV. This is fine for the top plate of C1 which can get it's +Q charge from the 0.35V source, the bottom plate however must receive the increased (-Q) charge from the C2 capacitor.

Thus, negative charge flows from top plate of C2 to bottom plate of C1, hence conventional current flows to C2, charging it up further and increasing it's voltage to 1.1V (random chosen value to illustrate).

Is this correct? It seems wrong to me. I know I can calculate this with maths, but I really want to understand intuitively how charge is moving. I can't find any resources online that talk about charge moving in detail. I don't care about exact values, I just want to understand roughly what's happening.

Answer 1

No, not correct because the 1V precharge of C1 is able to overcome the 0.25V of the supply. Since C1 acts like a DC blocker, once C1 and C2 equalize in voltage it will just there. It's kind of a weird scenario and I'm having some trouble thinking about it.

But the end result should be as if C1 and C2 are connected in parallel with both bottom plates connected to GND. The top plates will both reach some voltage less than 1V due to the finite charge available from C2's precharge redistributing voltage. However, C1 will hold less charge than it would if its bottom plate were actually connected to GND since its negative terminal has been lifted above ground by 0.25V/0.35V. In effect, C1 would appear as a smaller capacitor whose bottom plate was connected to GND.