The capacitance of a capacitor becomes $\frac{7}{6}$ times the original value if a dielectric slab of thickness $t = \frac{2d}{3}$ is introduced between the plates, where $d$ is the distance of separation between the plates. The dielectric constant of the slab is

$A$ parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu F$. The area of the capacitor is divided into two equal halves and filled with two media (as shown in the figure) having dielectric constants $K_1=2$ and $K_2=4$. The capacitance of the system will be (in $\mu F$)

$A$ parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

Two point charges are placed at a distance $r$ in air,experiencing a force $F$. When they are placed in a medium with a dielectric constant $K$,at what distance will the force between them remain the same?

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant $k_1 = 3$ and thickness $d/3$,while the other one has a dielectric constant $k_2 = 6$ and thickness $2d/3$. The capacitance of the capacitor is now . . . . . . $pF$.

The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. To maintain the same potential difference,the distance between the capacitor plates is increased by $1.6\,mm$. The dielectric constant of the plate is: