The circuit shown in figure is in the steady state with switch $S_1$ close. At $t=0,S_1$ is opened and switch $S_2$ is closed, 

Derive an expression for the charge ( in $\mu C$) on the capacitor $C_2$ as function of time.

In the steady state $C_1$ and $C_2$ are in series arrangement, their equivalent is 

$C_{\text{eq}}=\frac{C_1{C}_2}{C_1+C_2}=1.2\mu \text{F}$

Charge on the capacitor $C_2,Q_0=C_{aq}V=1.2\times 20=24\mu \text{C}$

When $S_1$ is open and $S_2$ is closed. Capacitor $C_2$ starts discharging through the inductor and let at any time $‘t’$ charge on the capacitor be $Q$, then 

Applying Kirchhoff's voltage law,

$\frac{Q}{C_2}-L\frac{dI}{dt}=0$

$\frac{Q}{C_2}+L\frac{d^{2}Q}{d{t}^2}=0\left[I=-\frac{dQ}{dt}\right]$

The solution of this equation is

$Q=Q_0\sin (\omega t+\varphi )$

where $\omega =\frac{1}{\sqrt{L{C}_2}}=5\times {10}^4\text{rad/s}$

At $t=0,Q=Q_0$

Hence, $\varphi =\frac{\pi }{2}$

Thus the charge on the capacitor at any time $t$ is 

$Q=Q_0\cos \omega t$ Ans.