$A$ parallel plate capacitor with air between the plates has a capacitance of $12 \mu F$. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $4$,what will be the new capacitance of the capacitor (in $\mu F$)?

The distance between the plates of a parallel plate capacitor is $d$. $A$ metal plate of thickness $d/2$ is introduced between the plates. The capacitance will then be

Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distances as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k = 2$. The capacitance of the system between $1$ and $3$ and between $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{C_1}{C_2}$ is

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 $\,V$)

$A$ parallel plate capacitor is immersed in an oil of dielectric constant $K = 2$. The electric field between the plates is:

Four metallic plates $A, B, C$ and $D$ of the same size with the same separation between them are arranged as shown in the figure. Dielectric slabs of dielectric constant $K = 2$ are placed between $B, C$ and $C, D$ respectively. Plates $B$ and $D$ are connected together. The effective capacitance between $A$ and $C$ is (Assume capacitance of each pair of plates without dielectric is $C$):