When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

$A$ parallel plate capacitor has a separation between plates of $0.885$ mm. It has a capacitance of $1$ $\mu$$F$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13}$ $\Omega$m and resistance $17.7 \times 10^{14}$ $\Omega$. The relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is . . . . . . . (Take permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ $F$/m)

$A$ parallel plate capacitor has a capacitance $C$. The distance between the two plates is halved and the space between the plates is filled with a dielectric medium. If the new capacitance is $3C$,what is the dielectric constant of the medium?

Two point charges are kept in air with a separation $r$ between them. The force between them is $F_1$. If half of the space between the charges is filled with a dielectric of dielectric constant $K=4$,the force between them becomes $F_2$. If $1/3$ rd of the space between the charges is filled with a dielectric of dielectric constant $K=9$,then the ratio $F_1/F_2$ is:

The outer sphere of a spherical air capacitor is earthed. For increasing its capacitance:

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).