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

A parallel plate capacitor with air between the plates has a capacitance of $8 \;pF \left(1 \;pF =10^{-12} \;F \right) .$ What will be the capacitance (in $pF$) if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6 ?$

Two identical parallel plate capacitors are connected in series to a battery of $100\,V$. A dielectric slab of dielectric constant $4.0$ is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

The radii of the inner and outer spheres of a condenser are $9\,cm$ and $10\,cm$ respectively. If the dielectric constant of the medium between the two spheres is $6$ and charge on the inner sphere is $18 \times {10^{ - 9}}\;coulomb$, then the potential of inner sphere will be, if the outer sphere is earthed........$volts$

A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of slab is given by