$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential difference $V$ and then the battery is disconnected. $A$ slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then:

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

$A$ parallel plate capacitor,partially filled with a dielectric slab of dielectric constant $K$,is connected to a cell of emf $V \text{ volt}$,as shown in the figure. The separation between the plates is $D$. Then:

$A$ 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]

$A$ parallel plate capacitor of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

The potential energy of a charged parallel plate capacitor is $U_0$. If a slab of dielectric constant $K$ is inserted between the plates,then the new potential energy will be