$A$ parallel plate capacitor has plates of area $0.4 \pi \,m^2$ and spacing of $0.5 \,mm$. If a slab of thickness $0.5 \,mm$ and dielectric constant $4.5$ is introduced in between the plates of the capacitor, then the capacitance of the capacitor is

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 $:1$)

Two metallic plates of radius $r$ are placed at a distance $d$ apart,and the capacitance is $C$. If a plate of radius $r/2$ and thickness $d$ with a dielectric constant $6$ is placed between the plates of the capacitor,then its new capacitance will be

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

An air capacitor has a capacitance of $1 \mu F$. Now,the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is ($d=$ distance between two plates,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

Two point charges are placed at a distance $r$ in air,experiencing a force $F$. When they are placed in a medium with a dielectric constant $K$,at what distance will the force between them remain the same?