Question

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

Answer 1

Initially charges are shown as in fig (1) the left plate is $q_1$ and on right plate $q_2$ but when the switch is closed on outer surface of plates charge will be $\frac{q_1+q_2}{2}$ and will remain intact charge will occur only in inner surface charges.

Charge will remain conserved, on each plate so inner surface charge will be $\frac{q_1+q_2}{2}$ on left plate $\frac{q_1+q_2}{2}$ on right plate

We know that during discharging charge at any time on inner surface of left plate:-

$q=\left(\frac{Q_1-Q_2}{2}\right)e^{\frac{-t}{RC}}$

and net charge on left plate $\frac{Q{1}_1+Q_2}{2}+\left(\frac{Q_1-Q_2}{2}\right)e^{\frac{-t}{RC}}$

Similarly on right plate :-

$\left(\frac{Q_1+Q_2}{2}\right)+\left(\frac{Q_2-Q_1}{2}\right)e^{\frac{-t}{RC}}$