A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are ${Q_0},{V_0},{E_0 }and {U_0}$ respectively. A dielectric slab is inserted in order to fill the complete space between plates, while the battery is still connected. If now the corresponding values Q, V, E and U are in relation with the initially values and stated as:-
(a) $Q > {Q_0}$
(b) $V > {V_0}$
(c) $E > {E_0}$
(d) $U > {U_0}$

Hint: We usually place dielectrics between the two plates of parallel plate capacitors. They can fully or partially occupy the region between the plates. When we place the dielectric between the two plates of a parallel plate capacitor, the electric field polarizes it.

Complete step by step answer:
1) When the condenser is connected to the battery total charge is conserved in the condenser. Here the potential difference on the condenser is equal to the EMF of the battery and while it is still connected to the battery the potential difference (V) will be maintained therefore remain the same.
2) When a dielectric is introduced in the capacitor, the capacitance will be increased, but the potential difference will remain the same because the battery is still connected
3) By the equation,
$Q = CV$
So, the charge will increase that means,
$Q > {Q_0}$
4) The energy is given by:-
$U = \dfrac{1}{2}QV$
So, the energy will increase, which means $U > {U_0}$.
5) The electric field between the plates gets polarized and the electric field gets reduced due to the internal field which is in the opposite direction and thus this reduces the electric field of the condenser.
$K = \dfrac{{{E_0}}}{E}$Here K is the dielectric constant and E is the reduced electric field.
Hence from the above given statements: statements (a) and (d) are correct.

Note: The battery maintains the potential between the two plates and thus it remains the same. While the dielectric slab decreases the electric field by polarization and this leads to the increase in the capacitance of the condenser.