$A$ parallel plate capacitor has a parallel sheet of copper inserted between and parallel to the two plates,without touching the plates. The capacity of the capacitor after the introduction of the copper sheet is:

If the equivalent dielectric constant of the given system is $K$,then...

$A$ parallel plate capacitor has a separation between plates of $0.885$ mm. It has a capacitance of $1$ $\mu$$F$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13}$ $\Omega$m and resistance $17.7 \times 10^{14}$ $\Omega$. The relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is . . . . . . . (Take permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ $F$/m)

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:

$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

$A$ capacitor has a capacitance of $5 \mu F$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material with a dielectric constant of $1.5$,having an area equal to that of the plates but a thickness of $\frac{d}{2}$,is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be $..........\mu F$.