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

If half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $4$,the capacitance is $C_1$. If one third of the space between the plates of the capacitor is filled with the medium of dielectric constant $4$,the capacitance is $C_2$. If in both cases,the dielectric is placed parallel to the plates of the capacitor,then $C_1: C_2=$

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)

When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

The outer sphere of a spherical air capacitor is earthed. For increasing its capacitance: