Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant of a material,which when fully filled between the plates produces the same capacitance?

In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

$A$ parallel plate capacitor with plate separation $5 \ mm$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \ mm$,while keeping the battery connections intact,the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is . . . . . . .

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

$A$ parallel plate capacitor of capacitance $200 \,\mu F$ is connected to a battery of $200 \,V$. $A$ dielectric slab of dielectric constant $2$ is now inserted into the space between the plates of the capacitor while the battery remains connected. The change in the electrostatic energy in the capacitor will be ...... $J$.

Two identical capacitors $1$ and $2$ are connected in series. Capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0 \text{ volts}$. The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charges stored in the capacitors before removing the slab,and $Q'_1$ and $Q'_2$ be the values after removing the slab. Then: