Between the plates of a parallel plate capacitor a dielectric plate is introduced just to fill the space between the plates. The capacitor is charged and later disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference across the plates and the length of the dielectric plate drawn out is

A capacitor with air as the dielectric is charged to a potential of $100\;volts$. If the space between the plates is now filled with a dielectric of dielectric constant $10$, the potential difference between the plates will be……$volts$

A capacitor of capacity $'C'$ is connected to a cell of $'V'\, volt$. Now a dielectric slab of dielectric constant ${ \in _r}$ is inserted in it keeping cell connected then

A parallel plate capacitor with air between the plates has a capacitance of $8 \;pF \left(1 \;pF =10^{-12} \;F \right) .$ What will be the capacitance (in $pF$) if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6 ?$

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