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

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The distance between the plates is $d$. Now,the space between the plates is filled with two dielectrics. One dielectric has a dielectric constant $K_1 = 3$ and thickness $d/3$,while the other has a dielectric constant $K_2 = 6$ and thickness $2d/3$. Find the new capacitance of the capacitor in $pF$.

Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.
Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.

Two identical parallel plate capacitors,of capacitance $C$ each,have plates of area $A$,separated by a distance $d$. The space between the plates of the two capacitors is filled with three dielectrics,of equal thickness and dielectric constants $K_1$,$K_2$,and $K_3$. The first capacitor is filled as shown in fig. $I$,and the second one is filled as shown in fig. $II$. If these two modified capacitors are charged by the same potential $V$,the ratio of the energy stored in the two would be ($E_1$ refers to capacitor $(I)$ and $E_2$ to capacitor $(II)$):