When a fully charged capacitor is used to charge another capacitor, how does voltage disappear? I understand that the voltages of both capacitors must be equal, but how does some of the voltage end up disappearing after calculating the charge of each capacitor here:
How does charge redistribute when a charged capacitor is connected to an uncharged capacitor?
The total voltage between the two capacitors has been reduced. Why is this?
 A: Calculations, based on conservation of charge, show that, for a charged capacitor, A, connected to an identical, initially uncharged, capacitor, B, the new voltage between the plates of A or B is half the original voltage between the plates of A.
I assume that you can do calculations of this type, but are seeking a more intuitive reason for the voltage falling. You need to start with a good concept of voltage, or, in this case, potential difference (pd). The pd between the plates of a capacitor is the work you would need to do to take a small 'test' charge, $q$, from one plate to the other, divided by $q$. In other words
$$\text{pd between plates}=\frac{\text{work needed}}{q}.$$
That's what potential difference or voltage means!
When you have connected the capacitors together, A has lost (to B) half the charge on each of its plates. This halves the electric field strength in the gap between the plates. It therefore halves the force on $q$. So the work needed to take $q$ from one plate to the other is halved – so the potential difference is halved!
Charge and energy are conserved quantities (though energy can be transferred from one 'form' to another). Voltage is not a conserved quantity, so there's nothing odd about what happens to the voltage when you connect capacitors together. What is, at first sight, odd, is that the total energy stored in A and B together is half what you started with, in A. Half the energy has gone missing! But energy is never lost or gained, only transferred! Can you figure out what has happened to 'the missing half'?
