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

Define dielectric constant.

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2\,\mu \,F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu \,F$

Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)

A parallel plate capacitor of plate area $A$ and plate seperation $d$ is charged to potential difference $V$ and then the battery is disconnected. Aslab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q, E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then 

Due to which the surface charge density arises on the surface of a dielectric slab, when it is placed in a uniform electric field ?