$A$ parallel plate air capacitor has capacity $C$ farad, potential $V$ volt, and energy $E$ joule. When the gap between the plates is completely filled with a dielectric of dielectric constant $K$, what happens to the potential $V$ and energy $E$?

$A$ parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ and $\varepsilon_2$,as shown in the figures. The distance between the plates is $d$ and the area of each plate is $A$. If the capacitance in the first configuration and second configuration are $C_1$ and $C_2$ respectively,then $\frac{C_1}{C_2}$ is

$A$ 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.

$A$ capacitor has capacity $C$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material of dielectric constant $K$ having an area equal to that of the plates but thickness $\frac{d}{2}$ is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be:

$A$ parallel plate capacitor is connected to a battery. If a dielectric slab is inserted between the plates,which of the following quantities increases?

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery,then the charge on the plates relative to the earlier charge: