$A$ 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 $V$)

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

$A$ 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

$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)

The function of a dielectric in a capacitor is to

$A$ capacitor of capacitance $20 \ \mu F$ has a distance of $2 \ mm$ between its plates. If a dielectric slab of thickness $1 \ mm$ and dielectric constant $K = 2$ is inserted between the plates,the new capacitance will be ..... $\mu F$.