$A$ parallel plate capacitor with square plates of side $L$ is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be

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 $(II)$):

$A$ parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu F$. The area of the capacitor is divided into two equal halves and filled with two media (as shown in the figure) having dielectric constants $K_1=2$ and $K_2=4$. The capacitance of the system will be (in $\mu F$)

$A$ parallel plate capacitor has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)

The force between two point charges kept with a separation of $9 \ cm$ in air is $98 \ N$. If a dielectric slab of constant $4$,thickness $6 \ cm$ and another dielectric slab of constant $9$,thickness $3 \ cm$ are introduced between the two charges,then the new force becomes (in $N$)

Two point charges are kept in air with a separation $r$ between them. The force between them is $F_1$. If half of the space between the charges is filled with a dielectric of dielectric constant $K=4$,the force between them becomes $F_2$. If $1/3$ rd of the space between the charges is filled with a dielectric of dielectric constant $K=9$,then the ratio $F_1/F_2$ is: