$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. Which graph shows the variation of electric field $(E)$ with distance $(x)$ from the left plate?

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is:

The electrostatic force between two charges kept in air is $F$. If $30 \%$ of the space between the charges is filled with a medium,then the electrostatic force between the charges becomes $\frac{F}{2.56}$. The dielectric constant of the medium is

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 dielectric slabs of dielectric constants $K_1$ and $K_2$ are filled between the plates of a parallel plate capacitor as shown in the figure. If the area of each plate is $A$ and the separation between the plates is $d$,what is the equivalent capacitance of the capacitor?

$A$ parallel plate capacitor has a capacitance of $2\ \mu F$ with a separation of $0.4\ cm$ between the plates. If the distance between the plates is halved and the space is filled with a dielectric material of constant $K = 2.8$,the final capacitance of the capacitor will be .....$\mu F$.