$A$ parallel plate capacitor has a capacitance of $2\ \mu F$ with a separation of $0.4\ cm$ between the plates. If the distance between the plates is halved and the space is filled with a dielectric material of constant $K = 2.8$,the final capacitance of the capacitor will be .....$\mu F$.

$A$ combination of parallel plate capacitors is maintained at a certain potential difference. When a $3 \, mm$ thick slab is introduced between all the plates,in order to maintain the same potential difference,the distance between the plates is increased by $2.4 \, mm$. Find the dielectric constant of the slab.

$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

$A$ parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers,parallel to its plates,each with thickness $\delta = \frac{d}{N}$. The dielectric constant of the $m^{\text{th}}$ layer is $K_m = K(1 + \frac{m}{N})$. For a very large $N (> 10^3)$,the capacitance $C$ is $\alpha \left( \frac{K \varepsilon_0 A}{d \ln 2} \right)$. The value of $\alpha$ will be. . . . . . . .
[$\varepsilon_0$ is the permittivity of free space]

An air-filled parallel plate capacitor has capacity $C$. If the distance between the plates is doubled and it is immersed in a liquid,the capacity becomes twice. The dielectric constant of the liquid is:

Two parallel plates of area $A$ are separated by two different dielectrics as shown in the figure. Find the resultant capacitance.