$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

• A
  $\frac{\varepsilon_{0} A}{d} V$
• B
  $\frac{k \varepsilon_{0} A}{d} V$
• C
  $\frac{\varepsilon_{0} A}{k d} V$
• D
  Zero