Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
I have done that many times.... but one more time...
Can we agree that the total charge on the top half of the system is:
Q
total = C
c1 V
c1 + C
c2 V
c2So here is the initial state of the system:
Q
total = 1F * 3V + 1F * 0V = 3 Coulomb.
I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.
The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:
Q
total = C
c1 V
c1 + C
c2 V
c2 Q
total = 1F * 1.5V + 1F * 1.5 = 3 Coulomb.
This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:
Q
total = 2F * 1.5V = 3 Coulomb.
We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down, as half the energy is missing, but ideal capacitors and wires have no resistive element to dissipate it.
A simpler example that shows how flawed your position isWhat about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.