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 $C$.

In the given figure,a capacitor is formed by placing a compound dielectric between the plates of a parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $= A$):

$A$ parallel plate capacitor is charged and then isolated. If the separation between the plates is increased,which one of the following statements is $NOT$ correct?

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

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

Consider a parallel plate capacitor with plates in the shape of a square in the $XY$-plane. The gap between the plates is filled with a dielectric material. The dielectric constant $k$ varies along the $X$-axis as $k(x) = \left[1 + \left(\frac{x}{L}\right)^\alpha\right]$,where $\alpha$ is a constant. Let $C_d$ and $C_a$ be the capacitance in the presence of the dielectric and air,respectively. If the ratio $\frac{C_d}{C_a} = \frac{7}{6}$,then the value of $\alpha$ must be