In the figure shown the capacitance of capacitor is C. The left plate of the capacitor is given charge Q and right plate is uncharged. Select the correct statement/s

 

The charge appearing on the inner face of the left plate at steady state after closing key k is CV

The charge appearing on the inner face of the right plate at steady state after closing key k is CV

The amount of charge that will flow through the battery till the steady state is reached when key k is closed will be  $CV+\frac{Q}{2}$

If $V=\frac{2Q}{C}$ , then the work done by cell after closing the key k is  $\frac{3Q^2}{C}$