The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

The variation of the electric field between the plates of a capacitor is shown in the figure. Find the relation between $K_1$,$K_2$,and $K_3$,if $2x = y$.

$A$ parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $U$ as $\varepsilon = \alpha U$,where $\alpha = 2 \ V^{-1}$. $A$ similar capacitor with no dielectric is charged to $U_0 = 78 \ V$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors. (in $V$)

$A$ parallel-plate capacitor of plate area $10 \text{ cm}^2$ and plate separation $3 \text{ mm}$ is charged to a potential difference $12 \text{ V}$ and then the battery is disconnected. $A$ slab of dielectric constant $3$ is then inserted between the plates. The work done on the system in the process of inserting the slab is $\alpha \varepsilon_0$. The value of $\alpha$ is (Take $\varepsilon_0$ as the permittivity of free space).

$A$ parallel plate capacitor with air between the plates has a capacitance of $4 \ pF$. If the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant $K = 6$,then the value of capacitance will be . . . . . . . (in $pF$)