An air-filled parallel plate capacitor has a capacity of $2 \ pF$. If the separation between the plates is doubled and the interspace between the plates is filled with a dielectric material,the capacity increases to $6 \ pF$. The dielectric constant of the material is:

$A$ capacitor is half-filled with a dielectric $(K=2)$ as shown in Figure $A$. If the same capacitor is to be filled with the same dielectric as shown in Figure $B$, what would be the thickness $t$ of the dielectric so that the capacitor still has the same capacity?

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

$A$ parallel plate capacitor filled with oil of a dielectric constant $3$ between the plates has capacitance $C$. If the oil is removed,then the capacitance of the capacitor will be

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

The plates of a parallel plate capacitor are separated by a distance $d$ with air as the medium between them. $A$ dielectric slab of dielectric constant $K = 3$ is introduced between the plates so as to increase the capacity by $50 \%$. The thickness of the dielectric slab is