A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\left.\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$

A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$

The capacitance of an air capacitor is $15\,\mu F$ the separation between the parallel plates is $6\,mm$. A copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$

What is linear isotropic dielectric ?

If the dielectric constant and dielectric strength be denoted by $k$ and $x$ respectively, then a material suitable for use as a dielectric in a capacitor must have

Explain polarisation of polar molecule in uniform electric field.