A parallel plate capacitor has a dielectric slab of a dielectric constant $K$ between its plates that covers $1/3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_1$. When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. Choose the correct option/options, ignoring edge effects.

A parallel plate capacitor having plates of area $S$ and plate separation $d$, has the capacitance $C_1$ in air. When two dielectrics of different relative permittivity $\left({\epsilon }_1=2 \text{and} {\epsilon }_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is, 

Three identical capacitors $C_1$, $C_2$ and $C_3$ have a capacitance of $1.0 \mathrm{\mu F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity ${\epsilon }_r$. The cell electromotive force $\left(\mathrm{emf}\right) V_0=8V$. 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 $C_3$ is found to be $5 \mathrm{\mu C}$. The value of ${\epsilon }_r$

In the given circuit, the charge $Q_2$ on the $2 \mathrm{\mu F}$ capacitor changes as $C$ is varied from $1 \mathrm{\mu F}$ to $3 \mathrm{\mu F}$. $Q_2$ as a function of $C$ is given properly by (figures are drawn schematically and are not to scale)