With the rise in temperature, the dielectric constant $K$ of a liquid

Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)

Two identical parallel plate capacitors, of capacitance $C$ each, have plates of area $A$, separated by a distance $d$. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants $K_1$ , $K_2$ and $K_3$ . The first capacitor is filled as shown in fig. $I$, and the second one is filled as shown in fig. $II$. If these two modified capacitors are charged by the same potential $V$, the ratio of the energy stored in the two, would be ( $E_1$ refers to capacitor $(I)$ and $E_2$ to capacitor $(II)$) 

In the adjoining figure, capacitor $(1)$ and $(2)$ have a capacitance $‘C’$ each. When the dielectric of dielectric consatnt $K$ is inserted between the plates of one of the capacitor, the total charge flowing through battery is

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

When a dielectric material is introduced between the plates of a charged condenser then electric field between the plates