If a capacitor with air,a capacitor with a dielectric,and a capacitor with a conducting slab have capacitances $C_1$,$C_2$,and $C_3$ respectively,then:

$A$ parallel plate capacitor has a capacitance of $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $K$ of the oil is:

The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).

Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant $8$,and a copper plate of thickness $d/2$ is introduced between the plates of $N$ ($d$ is the distance between the plates). Then the potential differences across $M$ and $N$ are,respectively,in the ratio:

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

$A$ parallel plate capacitor with air between the plates has a capacitance of $1.0 \text{ pF}$. If the distance between the plates is doubled and the space between them is filled with a dielectric substance, the capacitance becomes $2.0 \text{ pF}$. Then the value of the dielectric constant of the dielectric substance is . . . . . . .