# steady state voltage in capacitor network

simulate this circuit – Schematic created using CircuitLab

I have this capacitance network in my circuit and I can't figure out the math behind why the final steady-state value of Vx=0.542v when SW2 closes to the other position. Coincidentally I found out this is Vx - 1.(c2 /(c1+c2+c3)) but I would like an intuitive approach. Please ignore the absence of resistance in the circuit to make it practical.

C1 and C2 are initially in parallel for an equivalent capacitance of 31fF. So, the voltage across the parallel combination of C1 and C2 will be V = Q/31fF. The voltage across C3 will be Vx = Q/31fF. From these 2 equations you can see that V = Vx in this state, which leads to Vx = 0.5V. You can reach this same conclusion by applying Laplace transform circuit theory where you would end up with $$Vx = \frac{1}{s}\frac{C3}{Ce+C3}$$ where Ce is the parallel combination of C1 and C2. This is clearly = 0.5 for initial case. The charge on C2 will be Q2 = C2V2 = 8f Coulombs. The charge on C3 will be Q3 = C3Vx = 15.5f Coulombs.