Write the working principle of a parallel plate capacitor. On what factors, the capacitance of a parallel plate capacitor depends?

A parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of the same area but the thickness, $\frac{d}{2}$, is inserted between the plates as shown in figure having dielectric constant $K(=4).$ The ratio of new capacitance to its original capacitance will be,

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3\times 10^4\text{ V/m}$, the charge density of the positive plate will be close to :

In the figure is shown a system of four capacitors connected across a $10 \mathrm{V}$ battery. The charge that will flow from switch S when it is closed is:

A parallel plate capacitor of two plates, each with area $90\pi {\mathrm{cm}}^2$, is separated by $5 \mathrm{mm}$ air-gap. The capacitor is initially connected to $100 \mathrm{V}$ source. When an unknown liquid is filled between the plates so as to fill the air-gap, an additional charge of $95 \mathrm{nC}$ is found to be flowing onto the capacitor from the $100 \mathrm{V}$ source. The dielectric constant of the liquid is
(Assume $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9 \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-2}$)

In a parallel plate capacitor set up, the plate area of capacitor is $2{ \mathrm{m}}^2$ and the plates are separated by $1 \mathrm{m}$. If the space between the plates are filled with a dielectric material of thickness $0.5 \mathrm{m}$ and are $2{ \mathrm{m}}^2$ (see figure) the capacitance of the set-up will be ${\epsilon }_0$ ________.

(Dielectric constant of the material $=3.2$)

(Round off to the Nearest Integer)

Four metallic plates each of area $A$ and separated from one another by a distance $d$ are arranged as shown in the figure. What is the capacitance between $X$ and $Y ?$