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

A parallel plate capacitor has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :

A parallel - plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :

When a dielectric material is introduced between the plates of a charged condenser then electric field between the plates

A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V\ volt$. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

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