The function of a dielectric in a capacitor is to

The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)

The distance between the plates of a parallel plate capacitor is $0.05\, m$. An electric field of $3 \times 10^4\, V/m$ is established between the plates. The capacitor is disconnected from the battery,and an uncharged metal plate of thickness $0.01\, m$ is inserted. If a dielectric slab of dielectric constant $K = 2$ is inserted instead of the metal plate,what will be the potential difference in $kV$?

$A$ parallel plate capacitor has an area of $6 \, cm^2$ and a separation of $3 \, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$,and $K_3 = 14$. The dielectric constant of a material which,when fully inserted in the above capacitor,gives the same capacitance would be:

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

$A$ dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $x = 0$ to $x = 3d$: