When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery,then the charge on the plates relative to the earlier charge:

Two identical capacitors $A$ and $B$ are connected as shown in the circuit. Initially,the switch $S$ is closed. Now,the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $K = 3$. The ratio of total electrostatic energy stored in the capacitors before and after the introduction of the dielectric is

When a dielectric slab is inserted between the plates of a capacitor while it remains connected to a battery,which of the following occurs during this process?

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

$A$ 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

In a parallel plate capacitor,the separation between the plates is $3\,mm$ with air between them. Now,a $1\,mm$ thick layer of a material of dielectric constant $K = 2$ is introduced between the plates,due to which the capacity increases. In order to bring its capacity back to the original value,the separation between the plates must be increased to......$mm$. (in $.5$)