When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

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

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$, then the potential difference across the other capacitor will become:

$A$ capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be: