$A$ parallel plate capacitor has a potential of $20\,kV$ and a capacitance of $2 \times 10^{-4}\,\mu F$. If the area of the plate is $0.01\,m^2$ and the distance between the plates is $2\,mm$,find the dielectric constant of the medium.

In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

$A$ 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 $\mu F$.

When a dielectric slab is introduced between the plates of a parallel plate capacitor that is connected to a battery,the new charge on the plates is:

Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.

$A$ dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $x = 0$ to $x = 3d$: