If a capacitor with air,a capacitor with a dielectric,and a capacitor with a conducting slab have capacitances $C_1$,$C_2$,and $C_3$ respectively,then:

$A$ parallel plate capacitor has two layers of dielectrics as shown in the figure. This capacitor is connected across a battery. The ratio of the potential difference across the dielectric layers $(V_1/V_2)$ is:

The distance between the plates of a parallel plate capacitor is $0.05\, m$. An electric field of $3 \times 10^4\, V/m$ is established between the plates. The capacitor is disconnected from the battery,and an uncharged metal plate of thickness $0.01\, m$ is inserted. If a dielectric slab of dielectric constant $K = 2$ is inserted instead of the metal plate,what will be the potential difference in $kV$?

Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant $8$,and a copper plate of thickness $d/2$ is introduced between the plates of $N$ ($d$ is the distance between the plates). Then the potential differences across $M$ and $N$ are,respectively,in the ratio:

$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \text{ pF}$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $x/4 \text{ pF}$. The value of $x$ is

As shown in the figure,a dielectric slab of thickness $d/2$ and dielectric constant $K$ is placed between the plates of a parallel plate capacitor of plate separation $d$. The capacitor is charged to a potential $V$ using a battery. If the dielectric slab is pulled out after disconnecting the battery from the capacitor,the final potential difference across the plates of the capacitor is