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 $q_{0}$ )

  • 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
AIIMS 2019, Medium
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