$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. The graph between electric field $(E)$ and distance $(x)$ from the left plate will be:

$A$ $60 \ \mu F$ parallel plate capacitor whose plates are separated by $6 \ mm$ is charged to $250 \ V$,and then the charging source is removed. When a slab of dielectric constant $5$ and thickness $3 \ mm$ is placed between the plates,find the change in the potential difference across the capacitor (in $V$)?

The capacity of an air condenser is $2.0 \, \mu F$. If a medium is placed between its plates,the capacity becomes $12 \, \mu F$. The dielectric constant of the medium will be:

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:

$A$ capacitor is connected to a $10\,V$ battery. The charge on the plates is $10\,\mu C$ when the medium between the plates is air. The charge on the plates becomes $100\,\mu C$ when the space between the plates is filled with oil. The dielectric constant of the oil is:

$A$ parallel plate capacitor has a capacitance of $C_0$. If a dielectric slab of relative permittivity $\varepsilon_r$ and thickness equal to one-fourth of the distance between the plates is inserted,the new capacitance is $C$. Then,the ratio $\frac{C}{C_0}$ is: