Two plates of a capacitor of capacitance $C$ are given charges $q_1\text{and}{q}_2$. This capacitor is connected across a resistance $R$ as shown. Key is closed at $t=0$. Find the total charge on each plate after time $t$

The correct option is C $\begin{array}{cc}\left(\frac{q_1+q_2}{2}\right)+\left(\frac{q_1-q_2}{2}\right)e^{\frac{-t}{CR}}& \\ & \left(\frac{q_1+q_2}{2}\right)-\left(\frac{q_1-q_2}{2}\right)e^{\frac{-t}{CR}}\end{array}$
Initially the charges on the plates will be distributed as shown.


Charge on outer surface $\frac{q_1+q_2}{2}$
is constant because $q_1+q_2$ is constant.
$P.D.$ across the plates is given by $\frac{q_1-q_2}{2C}$
and is independent of the charge on the outer surfaces.
Thus, when capacitor is discharged, the charges on the outer surface do not change. Charge on the inner surface decreases according to the equation.

$q=\left(\frac{q_1-q_2}{2}\right)e^{\frac{-t}{RC}}$

$\therefore \text{Total charge on the left plate}$

$q_1=\frac{q_1+q_2}{2}+\left[\frac{q_1-q_2}{2}\right]e^{-\frac{t}{Rc}}$

And total charge on the right plate

$q_2=\frac{q_1+q_2}{2}-\left[\frac{q_1-q_2}{2}\right]e^{-\frac{t}{RC}}$