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

A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?.....$\mu F$

The capacity of an air condenser is $2.0\, \,\mu F$. If a medium is placed between its plates. The capacity becomes $ 12\, \,\mu F$. The dielectric constant of the medium will be

Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is

If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become

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