$A$ parallel plate capacitor with air as the dielectric has capacitance $C$. $A$ slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

$A$ 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 ....... $\times 10^{-4} \, m^{2}$.

An air capacitor has a capacitance of $1 \mu F$. Now,the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is ($d=$ distance between two plates,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

$A$ parallel plate capacitor has a capacitance $C$. The distance between the two plates is halved and the space between the plates is filled with a dielectric medium. If the new capacitance is $3C$,what is the dielectric constant of the medium?

$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential difference $V$ and then the battery is disconnected. $A$ slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then:

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.