The potential gradient at which the dielectric of a condenser just gets punctured is called

A parallel plate capacitor having plates of area $S$ and plate separation $d$, has capacitance $C _1$ in air. When two dielectrics of different relative permittivities $\left(\varepsilon_1=2\right.$ and $\left.\varepsilon_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C _2$. The ratio $\frac{ C _2}{ C _1}$ is

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is

After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :- 

A parallel plate capacitor of plate area $A$ and plate seperation $d$ is charged to potential difference $V$ and then the battery is disconnected. Aslab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q, E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then 

A parallel plate capacitor has potential $20\,kV$ and capacitance $2\times10^{-4}\,\mu F$. If area of plate is $0.01\,m^2$ and distance between the plates is $2\,mm$ then find dielectric constant of medium