A parallel plate capacitor has potential $20\,kV$ and capacitance $2\times10^{-4}\,\mu F$. If area of plate is $0.01\,m^2$ and distance between the plates is $2\,mm$ then find dielectric constant of medium

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 :-

In a capacitor of capacitance $20\,\mu \,F$, the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates, then the new capacitance is......$\mu \,F$

A parallel plate capacitor having a separation between the plates $d$ , plate area $A$ and material with dielectric constant $K$ has capacitance $C_0$. Now one-third of the material is replaced by another material with dielectric constant $2K$, so that effectively there are two capacitors one with area $\frac{1}{3}\,A$ , dielectric constant $2K$ and another with area $\frac{2}{3}\,A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$ then $\frac{C}{{{C_0}}}$ is

What is dielectric ?

If a dielectric substance is introduced between the plates of a charged air-gap capacitor. The energy of the capacitor will