Two capacitors are shown in the diagram. Each is charged to potential $V_0$. If capacitor $C$ is filled with a dielectric of constant $2$ and capacitor $2C$ is filled with a dielectric of constant $3$,then the key is closed. The final voltage on the capacitor $C$ will be:

  • A
    $\frac{3}{8} V_0$
  • B
    $\frac{3}{5} V_0$
  • C
    $\frac{2}{5} V_0$
  • D
    $V_0$

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