How does the polarised dielectric modify the original external field inside it ?

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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.

898-s115

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