Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be

$A$ parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $U$ as $\varepsilon = \alpha U$,where $\alpha = 2 \ V^{-1}$. $A$ similar capacitor with no dielectric is charged to $U_0 = 78 \ V$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors. (in $V$)

$A$ parallel plate capacitor having capacitance $C$ farad is connected to a battery of emf $V$ volt. Keeping the capacitor connected to the battery,a dielectric slab of dielectric constant $K$ is inserted between the plates. The dimensions of the slab are such that it fills the whole space between the capacitor plates. Then $:-$

$A$ parallel plate capacitor of area $A$,plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants $k_1, k_2$ and $k_3$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor,then its dielectric constant $k$ is given by:

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 two plates of a capacitor is $d$ and its capacitance is $C_1$,when air is the medium between the plates. If a metal sheet of thickness $\frac{2d}{3}$ and of same area as the plate is introduced between the plates,the capacitance of the capacitor becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is: (in $:1$)