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

An air capacitor has a capacitance of $1 \mu F$. Now,the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is ($d=$ distance between two plates,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. 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 $6$?

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:

The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. To maintain the same potential difference,the distance between the capacitor plates is increased by $1.6\,mm$. The dielectric constant of the plate is:

$A$ container has a base of $50 \text{ cm} \times 5 \text{ cm}$ and height $50 \text{ cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \text{ cm} \times 50 \text{ cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250 \text{ cm}^3 \text{ s}^{-1}$. What is the value of the capacitance of the container after $10 \text{ s}$ (in $\text{ pF}$)? [Given: Permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]