The function of a dielectric in a capacitor is to

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

Two identical capacitors $A$ and $B$ are connected in series to a battery of $E$.$M$.$F$. $E$. Capacitor $B$ contains a slab of dielectric constant $K$. $Q_A$ and $Q_B$ are the charges stored in $A$ and $B$. When the dielectric slab is removed,the corresponding charges are $Q_A^{\prime}$ and $Q_B^{\prime}$. Then:

$A$ combination of parallel plate capacitors is maintained at a certain potential difference. When a $3 \, mm$ thick slab is introduced between all the plates,in order to maintain the same potential difference,the distance between the plates is increased by $2.4 \, mm$. Find the dielectric constant of the slab.

$A$ parallel plate air capacitor has a capacitance of $10 \ \mu F$. As shown in the figure,if this capacitor is divided into two equal parts and these parts are filled with dielectric materials of dielectric constants $K_1 = 2$ and $K_2 = 4$,then the capacitance of this arrangement is ............. $\mu F$.

$A$ 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?