A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$

A parallel plate capacitor is to be designed, using a dielectric of dielectric constant $5$, so as to have a dielectric strength of $10^9\;Vm^{-1}$ . If the voltage rating of the capacitor is $12\;kV$, the minimum area of each plate required to have a capacitance of $80\;pF$ is

The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then

A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ 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 air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then :

If ${q}_{{f}}$ is the free charge on the capacitor plates and ${q}_{{b}}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates, then bound charge $q_{b}$ can be expressed as