In one design of a capacitor,thin sheets of metal of area $80 \ mm \times 80 \ mm$ sandwich between them a piece of paper whose thickness is $40 \ \mu m$. The relative permittivity of the paper is $4.0$ and its dielectric strength is $20 \ MVm^{-1}$. Calculate the maximum charge that can be put on the capacitor. [Permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \ Fm^{-1}$]

Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant of a material,which when fully filled between the plates produces the same capacitance?

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 $(II)$):

The potential energy of a charged parallel plate capacitor is $U_0$. If a slab of dielectric constant $K$ is inserted between the plates,then the new potential energy will be

$A$ parallel plate capacitor has plates of area $0.4 \pi \,m^2$ and spacing of $0.5 \,mm$. If a slab of thickness $0.5 \,mm$ and dielectric constant $4.5$ is introduced in between the plates of the capacitor, then the capacitance of the capacitor is

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?