A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be

(If initial charge is $q_{0}$ )

  • [AIIMS 2019]
  • 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

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  • [JEE MAIN 2024]

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