$A$ parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor,which of the following does not change?

The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

In the given figure,there are two capacitors in series. $A$ rigid conducting slab of thickness $b$ can slide vertically. Find the equivalent capacitance of the system.

Two parallel plate capacitors of capacity $C$ and $3C$ are connected in parallel and charged to a potential difference of $18V$. The battery is then disconnected,and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $K=9$. The final potential difference across the combination of capacitors will be $V^{\prime}$. Find $V^{\prime}$. (in $V$)

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now,a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates,with the voltage source remaining connected to the capacitor.

$A$ 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 $V$)