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:

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 has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

Three parallel plate capacitors each with area $A$ and separation $d$ are filled with two dielectrics ($k_1$ and $k_2$) in the following fashion. Which of the following is true? $(k_1 > k_2)$