Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ has a capacitance of $4 \, \mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K = 3$ (as shown in figure) will be .........$ \mu F$.

$A$ parallel plate capacitor has charge $5 \times 10^{-6} \ C$. $A$ dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \ C$, then the dielectric constant of the slab is . . . . . . .

Between the plates of a parallel plate capacitor, a dielectric plate is introduced to completely fill the space between the plates. The capacitor is charged and then disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference $V$ across the plates versus the length $x$ of the dielectric plate drawn out is:

$A$ parallel plate capacitor is filled with two dielectrics as shown in the figure. The area of each plate is $A \; m^2$ and the separation between the plates is $t \; m$. The dielectric constants are $k_1$ and $k_2$ respectively. Its capacitance in farad will be: