The plates of a capacitor of capacitance C are given the charges `Q_1 and Q_2` as shwon in figure. Now the switch is closed at t =0. Find the charges on plates after time t.

Explanation

Initially charges on the various surfaces
fo lthe capacitor paltes are as shown in the figure. When the switch is
closed, the outer surface charges will
remain intact. Change will occur
only in inner surface charges. Charge
at any time on the inner surface of left plate is given by
`q = ((Q_(1) -Q_(2))/(2))e^(-t//RC)` and the charge on the inner surface of the
right will be -q. So the net charge at any time on left plate is
`q_(1) =((Q_(1)+ Q_(2))/(2)) +((Q_(1) - Q_(2))/(2))e^(-t//RC)` and the net charge at any time
on right plate is
`q_(2) = ((Q_(1)+Q_(2))/(2)) - ((Q_(1) - Q_(2))/(2))e^(-t//RC)`