$A$ parallel plate capacitor having capacitance $C$ farad is connected to a battery of emf $V$ volt. Keeping the capacitor connected to the battery,a dielectric slab of dielectric constant $K$ is inserted between the plates. The dimensions of the slab are such that it fills the whole space between the capacitor plates. Then $:-$

• A
  charge on the capacitor plates remains the same
• B
  charge on the plates becomes $K$ times
• C
  potential difference between the plates decreases to $V/K$
• D
  all of the above