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

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

$A$ parallel plate capacitor having capacity $C_0$ is charged to $V_0$. With the battery disconnected,if the separation between the plates is doubled,then the energy stored in it is $E_1$. Instead,if the separation between the plates is doubled with the battery in connection,the energy stored in it is $E_2$. Then the value of $\frac{E_2}{E_1}$ is

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?