$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

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

$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \ F$ when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \ V$. The dielectric slab is then removed. The charge passing through the wire is:

$A$ 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:

Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$,then the amount of charge which will flow through the battery is (neglect internal resistance of the battery).