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

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants ${k_1},{k_2}$ and ${k_3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by

A parallel plate capacitor Air filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to voltage source $V_0 = 30\,V$ and then connected to the first capacitor after disconnecting the voltage source. The charge and voltage on capacitor.

Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is

Write the capacitance of parallel plate capacitor with medium of dielectric of dielectric constant $\mathrm{K}$.

Putting a dielectric substance between two plates of condenser, capacity, potential and potential energy respectively