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

A parallel plate capacitor with air between the plates has a capacitance $C$. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant $6$ , then the capacitance will become

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$, the dielectric constant of the medium is

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 the plate has a capacitance of $15 pF$. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant $3.5.$ Then the capacitance becomes $\frac{ x }{4}\,pF$.The value of $x$ is $............$