How does the polarised dielectric modify the original external field inside it ?
When polar or non-polar molecules placed in an external electric field a dielectric develops a net dipole moment.
The dipole moment per unit volume is called polarisation and is denoted by $\overrightarrow{\mathrm{P}}$.
For linear isotropic dielectrics,
$\overrightarrow{\mathrm{P}} \propto \overrightarrow{\mathrm{E}}$
$\therefore\overrightarrow{\mathrm{P}}=\chi_{e} \overrightarrow{\mathrm{E}}$
where $\chi_{e}$ is a constant characteristic of the dielectric and is known as the electric susceptibility of the dielectric medium.
Consider a rectangular dielectric slab placed in a uniform external field $\overrightarrow{\mathrm{E}}_{0}$ parallel to two of its faces as shown in figure.
This field causes a uniform polarization $\overrightarrow{\mathrm{P}}$ of the dielectric.
Every volume element $\Delta \mathrm{V}$ of the slab has a dipole moment $\overrightarrow{\mathrm{P}} \Delta \mathrm{V}$ in the direction of the field. The volume element $\Delta \mathrm{V}$ is macroscopically small but contains a very large number of molecula dipoles. It has no net charge (Total charge on dipole is zero) but has net dipole moment. The positive charge of one dipole sits close to the negative charge of the adjacent dipole in rectangular slab.
In a parallel plate condenser, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. There is a glass slab of $3\,mm$ thick and of radius $12\,cm$ with dielectric constant $6$ between its plates. The capacity of the condenser will be
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 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.......$\mu \,F$
A force $F$ acts between sodium and chlorine ions of salt (sodium chloride) when put $1\,cm$ apart in air. The permittivity of air and dielectric constant of water are ${\varepsilon _0}$ and $K$ respectively. When a piece of salt is put in water electrical force acting between sodium and chlorine ions $1\,cm$ apart is
If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become