$A$ parallel plate capacitor has a plate separation of $d$. $A$ metal plate of thickness $d/2$ and the same area as the capacitor plates is inserted between them without touching the plates. The capacitance of the resulting capacitor becomes:

$A$ 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 $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

The outer sphere of a spherical air capacitor is earthed. For increasing its capacitance:

Two dielectric slabs having dielectric constants $K_1$ and $K_2$ of thicknesses $\frac{d}{4}$ and $\frac{3d}{4}$ are inserted between the plates of a capacitor as shown in the figure. The net capacitance between $A$ and $B$ is [where $\varepsilon_0$ is the permittivity of free space].

$A$ parallel plate capacitor with square plates of side $L$ is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be

$A$ parallel plate capacitor has an area of $6 \, cm^2$ and a separation of $3 \, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$,and $K_3 = 14$. The dielectric constant of a material which,when fully inserted in the above capacitor,gives the same capacitance would be: