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

$A$ dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

$A$ parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now,an insulating slab of same area but thickness $t = d/2$ is inserted between the plates as shown in the figure,having dielectric constant $K = 4$. The ratio of the new capacitance to its original capacitance will be:

The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).

$A$ parallel plate capacitor is immersed in an oil of dielectric constant $K = 2$. The electric field between the plates is:

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The distance between the plates is $d$. Now,the space between the plates is filled with two dielectrics. One dielectric has a dielectric constant $K_1 = 3$ and thickness $d/3$,while the other has a dielectric constant $K_2 = 6$ and thickness $2d/3$. Find the new capacitance of the capacitor in $pF$.