An air capacitor has a capacitance of $1 \mu F$. Now,the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is ($d=$ distance between two plates,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

$A$ parallel plate capacitor has a plate separation of $d$. $A$ metal plate of thickness $d/2$ and the same area as the capacitor plates is inserted between them without touching the plates. The capacitance of the resulting capacitor becomes:

In a parallel plate capacitor,the separation between the plates is $3\,mm$ with air between them. Now,a $1\,mm$ thick layer of a material of dielectric constant $K = 2$ is introduced between the plates,due to which the capacity increases. In order to bring its capacity back to the original value,the separation between the plates must be increased to......$mm$. (in $.5$)

In the figure,a capacitor is filled with dielectrics of constants $K_1$,$K_2$,and $K_3$. The resultant capacitance is

$A$ 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 ......................

In the figure,a capacitor is filled with dielectrics. The resultant capacitance is