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

The distance between two plates of a capacitor is $d$ and its capacitance is $C_1$,when air is the medium between the plates. If a metal sheet of thickness $\frac{2d}{3}$ and of same area as the plate is introduced between the plates,the capacitance of the capacitor becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is: (in $:1$)

The plates of a parallel plate capacitor are separated by a distance $d$ with air as the medium between them. $A$ dielectric slab of dielectric constant $K = 3$ is introduced between the plates so as to increase the capacity by $50 \%$. The thickness of the dielectric slab is

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

Two identical capacitors $A$ and $B$ are connected as shown in the circuit. Initially,the switch $S$ is closed. Now,the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $K = 3$. The ratio of total electrostatic energy stored in the capacitors before and after the introduction of the dielectric is

Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.