The area of the plates of a parallel plate capacitor is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it,one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the capacitor is

In the figure,a capacitor is filled with dielectrics. The resultant capacitance is

Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distances as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k = 2$. The capacitance of the system between $1$ and $3$ and between $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{C_1}{C_2}$ is

$A$ capacitor has a capacitance of $5 \mu F$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material with a dielectric constant of $1.5$,having an area equal to that of the plates but a thickness of $\frac{d}{2}$,is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be $..........\mu F$.

$A$ capacitor of capacitance $15 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.5$,dielectric strength $30 \, MV/m$,and potential difference $V = 30 \, V$. The area of the plate is ....... $\times 10^{-4} \, m^{2}$.

Two identical parallel plate capacitors,of capacitance $C$ each,have plates of area $A$,separated by a distance $d$. The space between the plates of the two capacitors is filled with three dielectrics,of equal thickness and dielectric constants $K_1$,$K_2$,and $K_3$. The first capacitor is filled as shown in fig. $I$,and the second one is filled as shown in fig. $II$. If these two modified capacitors are charged by the same potential $V$,the ratio of the energy stored in the two would be ($E_1$ refers to capacitor $(I)$ and $E_2$ to capacitor $(II)$):