The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:

$K(x) = K_0 + \lambda x$ ( $\lambda  =$ constant)

The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation

A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$, the dielectric constant of the medium is

If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air cored parallel capacitor charged to a potential $V$. The two capacitors share charges and the common potential is $V$. The dielectric constant $K$ is

. Three identical capacitors $C _1, C _2$ and $C _3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C _1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{ r }$. The cell electromotive force (emf) $V_0=8 V$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $C _3$ is found to be $5 \mu C$. The value of $\varepsilon_{ r }=$. . . . . 

Two dielectric slab of dielectric constant $K_1$ and $K_2$ and of same thickness is inserted in parallel plats capacitor and $K_1 = 2K_2$ . Potential difference across slabs are $V_1$ and $V_2$ respectively then