What is the ideal way of connecting three capacitors each of capacitance $4\mathrm{ }\mathrm{\mu }\mathrm{F}$ to achieve an effective capacitance of $6\mathrm{ }\mathrm{\mu }\mathrm{F}$ ?

A capacitor of $\mathrm{2 }\mathrm{\mu F}$ is charged as shown in the diagram. When the switch $\mathit{\text{S}}$ is turned to position $2$, the percentage of its stored energy dissipated is

A parallel plate  capacitor has uniform electric field  $E$ in the space between the plates.If the distance between the plates is $d$ and area of each plate is  $A$, the energy stored in the capacitor 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 $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9 \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-2}$)

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 :