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

In a medium of dielectric constant $K$, the electric field is $\vec E$ . If ${ \varepsilon _0}$ is permittivity of the free space, the electric displacement vector is

A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is

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

The area of the plates of a parallel plate condenser is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it, one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the condenser is