The space between the plates of a parallel plate capacitor is halved and a dielectric medium of relative permittivity $10$ is introduced between the plates. The ratio of the final and initial capacitances of the capacitor is

$A$ parallel plate air capacitor, with plate separation $d$, has a capacitance of $9 \text{ pF}$. The space between the plates is now filled with two dielectrics, the first having $K_1=3$ and thickness $d_1=d/3$, while the second has $K_2=6$ and thickness $d_2=2d/3$. The capacitance of the new capacitor is: (in $\text{ pF}$)

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

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 is charged to a potential of $100 \ V$. $A$ dielectric slab of thickness $2 \ mm$ is inserted between the plates. To maintain the same potential difference,the distance between the plates is increased by $1.6 \ mm$. The dielectric constant of the slab is:

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in the figure,will be (area of plate $= A$)