$A$ parallel plate capacitor having plates of area $S$ and plate separation $d$,has capacitance $C_1$ in air. When two dielectrics of different relative permittivities $(\varepsilon_1=2$ and $\varepsilon_2=4)$ are introduced between the two plates as shown in the figure,the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

The plates of a parallel plate capacitor are separated by a distance $d.$ Two slabs of different dielectric constants $K_1$ and $K_2$ with thicknesses $\frac{3}{8} d$ and $\frac{d}{2}$ respectively are inserted into the capacitor. Due to this,the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25 K_2,$ find the value of $K_1.$

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

Two identical condensers are joined as shown in the figure. When the switch $S$ is closed,the total energy of the system is $U_1$. If the switch is opened and both the condensers are filled with a dielectric of dielectric constant $K = 3$,then the energy of the system becomes $U_2$. The value of $\frac{U_1}{U_2}$ is