When switch S is open, the equivalent capacitance is
C =
=
=
µF
Charge on the R.H.S. plate of
C1 and upper plate of
C2 is
Q = CE =
µF × 3V = 4µC
When switch S is closed, the potential difference across
C1 is zero, because the batteries inopposition. Therefore, charge on the R.H.S. plate of
C1 = 0 . If the charge flowing through Bis q, then
Q + q = 0 ⇒ q = -Q = -4µC
Now the charge on the upper plate of
C2 =
C2E = 4µF × 3V = 12µC
which is the chargeflowing through A.