$A$ capacitor of capacitance $C_{0}$ is charged to a potential $V_{0}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S,$ the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by

• A
  $\frac{C_{0}(V_{0}-V)}{V_{0}}$
• B
  $\frac{C_{0}(V-V_{0})}{V_{0}}$
• C
  $\frac{C_{0}(V+V_{0})}{V}$
• D
  $\frac{C_{0}(V_{0}-V)}{V}$