The plates of a parallel plate capacitor are charged up to $100 \,V$. $A$ $2 \,mm$ thick insulator sheet is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by $1.6 \,mm$. The dielectric constant of the insulator is

An air-filled parallel plate capacitor has capacity $C$. If the distance between the plates is doubled and it is immersed in a liquid,the capacity becomes twice. The dielectric constant of the liquid is:

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

$A$ parallel plate capacitor is charged to a potential difference of $100\,V$ and disconnected from the source of $emf$. $A$ slab of dielectric is then inserted between the plates. Which of the following three quantities change?
$(i)$ The potential difference
$(ii)$ The capacitance
$(iii)$ The charge on the plates

The distance between two plates of a capacitor is $d$ and its capacitance is $C_1$,when air is the medium between the plates. If a metal sheet of thickness $\frac{2d}{3}$ and of same area as the plate is introduced between the plates,the capacitance of the capacitor becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is: (in $:1$)

$A$ parallel-plate capacitor is connected to a resistanceless circuit with a battery until the capacitor is fully charged. The battery is then disconnected from the circuit and the plates of the capacitor are moved to half of their original separation using insulated gloves. Let $V_{new}$ be the potential difference across the capacitor plates when the plates have moved. Let $V_{old}$ be the potential difference across the capacitor plates when they were connected to the battery. Find the ratio $\frac{V_{new}}{V_{old}}$.