The distance between the plates of a parallel plate capacitor is $d$. $A$ metal plate of thickness $d/2$ is introduced between the plates. The capacitance will then be

Which one statement is correct? $A$ parallel plate air condenser is connected to 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 to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as:

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

$A$ capacitor $C$ is charged to a potential difference $V$ and the battery is disconnected. Now,if the capacitor plates are brought closer slowly by some distance,then:

$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced,is

$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. Which graph shows the variation of electric field $(E)$ with distance $(x)$ from the left plate?