Suppose I energize one capacitor by connecting it across a battery, allowing it to achieve some potential difference V0 across its plates, then disconnect it and allow both of its leads to float in air. Next I take a second identical capacitor, whose leads have recently been shorted together, and connect one of its leads to one of the leads of the first capacitor. If the charges on the plates of the first capacitor were +Q and -Q before connecting the two capacitors (and of course 0 on both plates of the second capacitor), how do the charges distribute themselves among the four plates of the two connected capacitors?
I have attempted to answer this for myself using the method of contributor Manisheath at the post How does instant charging of one plate affect the potential of the other plate of a floating capacitor?. What I get, with the charged capacitor on the left and the charge of +Q on the leftmost plate of the left capacitor, is
0|+Q -Q|0 ----------------------------------------- 0|0 0|0
where the "--------" line represents the wire connecting the two capacitors.
In other words, no charge will flow. Is this correct? To me, this seems counter-intuitive, since on energetic grounds, I would expect stored energy to 'want' to spread out to a condition of lowest energy density per unit volume, mass, moles, whatever (without violating conservation of energy, of course). Recall, the energy stored in one capacitor is
E = 0.5*C*V^2 = 0.5*Q*V, since Q = C*V.