If a dielectric slab of dielectric constant $3$ is introduced between the plates of a capacitor having electric field $1.5 \times 10^{-9} \pi \ N C^{-1}$,then the electric displacement is

Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be

When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

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 has a capacitance of $10 \ \mu F$ with air between its plates. Now,half of the space between the two plates is filled with a dielectric material of dielectric constant $K = 4$ as shown in the figure. Find the capacitance of the capacitor in $\mu F$.

$A$ parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The initial level of the liquid is $\frac{d}{3}$. Suppose the liquid level decreases at a constant speed $V$. Find the time constant $\tau$ as a function of time $t$.