$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

Two parallel plates with dielectric slabs placed between them are as shown in the figure. The resultant capacity of the capacitor will be: [$A$ = area of plate,$t_1, t_2, t_3$ are thicknesses of dielectric slabs,$k_1, k_2, k_3$ are dielectric constants.]

$A$ parallel plate capacitor is formed by two plates each of area $30 \pi \, cm^{2}$ separated by $1 \, mm$. $A$ material of dielectric strength $3.6 \times 10^{7} \, Vm^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6} \, C$,the value of the dielectric constant of the material is. $\{ \text{Use} : \frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, Nm^{2}C^{-2} \}$

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

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: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]