The distance between two plates of a capacitor is $d$ and its capacitance is $C_1$,when air is the medium between the plates. If a metal sheet of thickness $\frac{2d}{3}$ and of same area as the plate is introduced between the plates,the capacitance of the capacitor becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is: (in $:1$)

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \ F$ when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \ V$. The dielectric slab is then removed. The charge passing through the wire is:

Two identical capacitors $1$ and $2$ are connected in series. Capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0 \text{ volts}$. The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charges stored in the capacitors before removing the slab,and $Q'_1$ and $Q'_2$ be the values after removing the slab. Then:

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

Half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,then the new capacitance will be:

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: