Two parallel plates of area $A$ are separated by two different dielectrics as shown in the figure. Find the resultant capacitance.

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:

$A$ parallel plate capacitor filled with a medium of dielectric constant $10$ is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of the capacitor will ......................

$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

The distance between the plates of a parallel plate capacitor is $d$. $A$ metal plate of thickness $d/2$ is introduced between the plates. The capacitance will then be

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