Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distances as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k = 2$. The capacitance of the system between $1$ and $3$ and between $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{C_1}{C_2}$ is

$A$ parallel plate capacitor has a capacitance of $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $K$ of the oil is:

$A$ parallel plate capacitor has a dielectric slab of 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.
$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{K}{2}$ $(D)$ $\frac{C}{C_1}=\frac{2+K}{K}$

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is:

When a dielectric material is introduced between the plates of a charged capacitor,the electric field between the plates:

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