An air capacitor of capacity $C = 10\,\mu F$ is connected to a constant voltage battery of $12\,V$. Now,the space between the plates is filled with a liquid of dielectric constant $K = 5$. The charge that flows from the battery to the capacitor is......$\mu C$.

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

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 $:1$)

$A$ parallel plate capacitor is charged to a potential of $100 \ V$. $A$ dielectric slab of thickness $2 \ mm$ is inserted between the plates. To maintain the same potential difference,the distance between the plates is increased by $1.6 \ mm$. The dielectric constant of the slab is:

$A$ parallel plate capacitor has a potential of $20 \, kV$ and a capacitance of $2 \times 10^{-4} \, \mu F$. If the area of the plates is $0.01 \, m^2$ and the distance between the plates is $2 \, mm$,find the dielectric constant of the medium.

$A$ parallel plate capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to: