Take the 2-minute tour ×  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 --------------------------------------