$A$ 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 $q_{0}$)

$A$ parallel plate capacitor has plate area $40 \ cm^2$ and plate separation $2 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and dielectric constant $5$. The capacitance of the system is ($\varepsilon_0 =$ permittivity of vacuum)

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

The parallel combination of two air-filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage $V$. When the capacitors are fully charged,the battery is removed,and after that,a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is

$A$ parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu F$. The area of the capacitor is divided into two equal halves and filled with two media (as shown in the figure) having dielectric constants $K_1=2$ and $K_2=4$. The capacitance of the system will be (in $\mu F$)

Two identical capacitors $A$ and $B$ are connected as shown in the circuit. Initially,the switch $S$ is closed. Now,the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $K = 3$. The ratio of total electrostatic energy stored in the capacitors before and after the introduction of the dielectric is