The capacity of an air-filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric of constant $K$ as shown in the figure. The new capacity becomes $C_n$. The ratio of $C_n$ to $C_0$ is:

Two point charges are kept in air with a separation $r$ between them. The force between them is $F_1$. If half of the space between the charges is filled with a dielectric of dielectric constant $K=4$,the force between them becomes $F_2$. If $1/3$ rd of the space between the charges is filled with a dielectric of dielectric constant $K=9$,then the ratio $F_1/F_2$ is:

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is introduced between the plates which results in

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance:

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is:

$A$ capacitor stores $60 \ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \ \mu C$ flows through the battery. If the initial capacitance of the capacitor was $2 \ \mu F$,the amount of heat produced when the dielectric is inserted is ... $\mu J$.