The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium,the capacitance becomes $30\, \mu F$. The permittivity of the medium is..........$C^{2} N^{-1} m^{-2}$.
$(\varepsilon_{0} = 8.85 \times 10^{-12} C^{2} N^{-1} m^{-2})$

$A$ parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential difference $V$ and then the battery is disconnected. $A$ slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then:

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

The electrostatic force between two charges kept in air is $F$. If $30 \%$ of the space between the charges is filled with a medium,then the electrostatic force between the charges becomes $\frac{F}{2.56}$. The dielectric constant of the medium is

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?