A parallel plate capacitor is fully charged using a battery. The...

When the battery is disconnected, the charge on the capacitor has no path to flow, so the charge remains constant: $Q = Q_0$. (Statement A is correct)

When a dielectric slab of dielectric constant $K > 1$ is inserted, the capacitance increases: $C = K C_0$. (Statement B is incorrect)

The potential difference is given by $V = \dfrac{Q}{C} = \dfrac{Q_0}{K C_0} = \dfrac{V_0}{K}$. Since $K > 1$, the potential difference decreases. (Statement C is correct)

The electrostatic energy stored is $U = \dfrac{Q^2}{2C} = \dfrac{Q_0^2}{2 K C_0} = \dfrac{U_0}{K}$. Since $K > 1$, the energy decreases. (Statement D is incorrect)

The electric field between the plates is $E = \dfrac{V}{d} = \dfrac{V_0}{K d} = \dfrac{E_0}{K}$. Since $K > 1$, the electric field decreases. (Statement E is correct)

Therefore, the correct statements are A, C, and E.

Answer: A, C and E only