When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes:

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

An air capacitor has capacitance $C_1$. The space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The new capacitance of the capacitor is $C_2$. The ratio $\frac{C_1}{C_2}$ is ($d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the two dielectrics respectively).

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

$A$ parallel plate capacitor has plates of length $l$,width $w$,and separation $d$. It is connected to a battery of emf $V$. $A$ dielectric slab of the same thickness $d$ and dielectric constant $k = 4$ is being inserted between the plates. At what length $x$ of the slab inside the plates will the energy stored in the capacitor be two times the initial energy stored?