$A$ parallel plate capacitor with air between the plates has a capacitance $C$. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant $6$,then the capacitance will become

$A$ 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 has plate area $40 \ cm^2$ and plate separation $2 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and dielectric constant $5$. The capacitance of the system is ($\varepsilon_0 =$ permittivity of vacuum)

$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

In a parallel plate capacitor,the separation between the plates is $3\,mm$ with air between them. Now,a $1\,mm$ thick layer of a material of dielectric constant $K = 2$ is introduced between the plates,due to which the capacity increases. In order to bring its capacity back to the original value,the separation between the plates must be increased to......$mm$. (in $.5$)

What will be the capacity of a parallel-plate capacitor when the half of the parallel space between the plates is filled by a material of dielectric constant $\varepsilon_r$? Assume that the capacity of the capacitor in air is $C$.