The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.

$A$ 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 a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

$A$ parallel plate air capacitor has capacitance $C_p$. It is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$. Now its capacity becomes $C_K$. The ratio $C_P$ to $C_K$ is

$A$ parallel plate capacitor having plates of area $S$ and plate separation $d$,has capacitance $C_1$ in air. When two dielectrics of different relative permittivities $(\varepsilon_1=2$ and $\varepsilon_2=4)$ are introduced between the two plates as shown in the figure,the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ has a capacitance of $4 \, \mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K = 3$ (as shown in figure) will be .........$ \mu F$.