Three identical capacitors $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ have a capacitance of $1.0 \mu \mathrm{F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $\mathrm{C}_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{\mathrm{r}}$. The cell electromotive force (emf) $V_0=8 \mathrm{~V}$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $\mathrm{C}_3$ is found to be $5 \mu \mathrm{C}$. The value of $\varepsilon_{\mathrm{r}}=$. . . . 

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Write the relation between $\vec P$ and $\vec E$ for a linear isotropic dielectric.

The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is

A combination of parallel plate capacitors is maintained at a certain potential difference When a $3\, mm$ thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by $2.4\, mm$. Find the dielectric constant of the slab. 

Between the plates of a parallel plate condenser there is $1\,mm$ thick paper of dielectric constant $4$. It is charged at $100\;volt$. The electric field in $volt/metre$ between the plates of the capacitor is

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