Dielectrics and Polarisation

Conductor and dielectric in a field

Dielectrics are non-conducting substances, with no free charge carriers. In an external field they cannot move charge from place to place; instead the external field induces dipole moments in the molecules, by stretching them or by re-orienting them. The molecular dipoles together produce net charge on the surfaces of the dielectric, and the field of that surface charge opposes the external field.

The key sentence to write: the induced field only reduces the applied field, it does not cancel it.

Polar and non-polar molecules

• Non-polar: the centres of positive and negative charge coincide, so there is no permanent dipole moment. Examples $\text{O}_2$ and $\text{H}_2$, symmetric molecules. In a field the positive and negative charges are displaced in opposite directions until the electric force is balanced by the internal restoring force, giving an induced dipole moment along the field.
• Polar: the centres of positive and negative charge are separated even with no field, so the molecule has a permanent dipole moment. Examples $\text{HCl}$ and $\text{H}_2\text{O}$. Without a field, thermal agitation orients them randomly and the total moment is zero; a field tends to align them, and the degree of alignment is set by the competition between the dipole energy $pE$ and the thermal energy.
• Either way, a dielectric acquires a net dipole moment in an external field. That is what polarisation means.

Polarisation

Polarisation $\vec P$ is the dipole moment per unit volume, with SI unit $\text{C m}^{-2}$ and dimensions $[\text{L}^{-2}\text{A}\text{T}]$. For a linear isotropic dielectric it is proportional to the field:

$$\vec P=\varepsilon_0\chi_e\vec E$$

where $\chi_e$ is the electric susceptibility of the medium, a dimensionless constant characteristic of the material. (CBSE has also accepted the shorter statement $\vec P=\chi_e\vec E$, i.e. $\vec P\propto\vec E$; the NCERT form with $\varepsilon_0$ is the one to write when the marks are for the relation itself.)

Bound surface charge

Take a rectangular dielectric slab in a uniform external field $\vec E_0$ parallel to two of its faces. Each volume element $\Delta v$ acquires dipole moment $P\,\Delta v$ along the field.

• Deep inside, the positive end of one molecular dipole sits next to the negative end of the next, so every interior element is electrically neutral: there is no net volume charge.
• At the two faces normal to the field the cancellation fails. Positive ends are left unneutralised on one face and negative ends on the other.

So a uniformly polarised dielectric is equivalent to two charged sheets carrying induced surface densities $+\sigma_P$ and $-\sigma_P$. These are bound charges, not free charges — they cannot be drawn off by a wire. Their field opposes $\vec E_0$, so the net field inside is

$$E=\dfrac{\sigma-\sigma_P}{\varepsilon_0}=\dfrac{E_0}{K},\qquad K>1$$

Why the field inside a dielectric slab decreases

The one-line answer CBSE marks: the slab is polarised by the external field, and the field of the induced bound charges on its surfaces is directed opposite to the external field, so the net field inside is reduced — by the factor $K$, the dielectric constant.

Permittivity and dielectric constant

$$\varepsilon=\varepsilon_0K,\qquad K=\dfrac{\varepsilon}{\varepsilon_0}=\dfrac{E_0}{E}=\dfrac{C}{C_0}$$

• $\varepsilon$ is the permittivity of the medium, $\varepsilon_0$ the permittivity of vacuum. For vacuum $K=1$ and $\varepsilon=\varepsilon_0$.
• $K$ is dimensionless and always greater than one. It is also written $\varepsilon_r$, the relative permittivity.
• Related consequence used in a board question: introducing a medium of dielectric constant $5$ inside a Gaussian surface reduces the flux through it to one-fifth, $\phi'=\dfrac{Q}{5\varepsilon_0}=\dfrac{\phi}{5}$.

Diagram to describe in words

Two vertical plates with $+$ signs on the left and $-$ signs on the right, and horizontal field arrows pointing right between them. Fill the gap with a rectangle representing the slab, drawn full of small horizontal dipoles all pointing right. Show a column of $-$ signs along the slab's left face and $+$ signs along its right face — the bound charges — with short arrows inside the slab pointing left, opposing the applied field. Alongside, sketch the same plates with a metal block instead: the induced charge is enough to make the field inside the block exactly zero.

Standard results