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

In the given figure,a capacitor is formed by placing a compound dielectric between the plates of a parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $= A$):

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ are filled between the plates of a parallel plate capacitor as shown in the figure. If the area of each plate is $A$ and the separation between the plates is $d$,what is the equivalent capacitance of the capacitor?

$A$ parallel plate air capacitor has capacitance $C_p$. It is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$. Now its capacity becomes $C_K$. The ratio $C_P$ to $C_K$ is

$A$ 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 has a plate area of $40\,cm^2$ and a plate separation of $2\,mm$. The space between the plates is filled with a dielectric medium of thickness $1\,mm$ and dielectric constant $5$. The capacitance of the system is: