A parallel plate capacitor is connected to a battery and a dielectric slab is inserted between the plates, then which quantity increase

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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A parallel - plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery. The graph which shows the variation of electric field $(E)$ and distance $(x)$ from left plate.

A parallel plate capacitor $\mathrm{C}$ with plates of unit area and separation $\mathrm{d}$ is filled with a liquid of dielectric constant $\mathrm{K}=2$. The level of liquid is $\frac{\mathrm{d}}{3}$ initially. Suppose the liquid level decreases at a constant speed $V,$ the time constant as a function of time $t$ is Figure: $Image$