The capacitance of a parallel plate capacitor is $2\mathrm{\mu }\mathrm{F}$ and the charge on its positive plate is $2\mathrm{\mu }\mathrm{C}$. If the charge on its plates is doubled, the capacitance of the capacitor

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,

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 $Y ?$

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 :

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 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: