Polar and Non-Polar molecules; Dielectric constant and Displacement Vector; Gauss law in Dielectric; Electric Polarization, Displacement Vector and Their Relation; Force and Energy in Dielectric syste

Polar and Non-Polar molecules:

Dielectrics:

The object through electric can’t pass is called an insulator or dielectric. There are two types of dielectrics, viz. non-polar and polar.

Non-polar molecules:

The dielectric of which molecules do not possess permanent electric dipole moment is called non-polar dielectric.

Nonpolar Covalent Bond:
A bond between 2 nonmetal atoms that have the same electronegativity and therefore have equal sharing of the bonding electron pair
The net dipole moment of non-polar molecule is zero.

Example: In H-H, each H atom has an electronegativity value of 2.1, therefore the covalent bond between them is considered non-polar. In H-Cl, the electronegativity of the Cl atom is 3.0, while that of the H atom is 2.1. The result is a bond where the electron pair is displaced toward the more electronegative atom. This atom then obtains a partial-negative charge while the less electronegative atom has a partial-positive charge. This separation of charge or bond dipole can be illustrated using an arrow with the arrowhead directed toward the more electronegative atom.
Depending on the relative electronegativities of the two atoms sharing electrons, there may be partial transfer of electron density from one atom to the other. When the electronegativities are not equal, electrons are not shared equally and partial ionic charges develop. A molecule is non-polar if it is diatomic and has only one kind of atom.

Polar molecules:

Polar means having electrical poles (i.e. electrical polarity). The molecules in which the arrangement or geometry of the atoms is such that one end of the molecule has a positive electrical charge and the other side has a negative charge are called as polar molecules. Examples of polar molecules are Water (H₂O) , Ammonia (NH₃), Hydrochloric acid (HCl), Sulfur Dioxide (SO₂), Hydrogen Sulfide (H₂S), Carbon Monoxide (CO) etc.

Behaviour of molecules in external electric field:

In nonpolar dielectric molecules, permanent electric dipoles are not present. However, after applying an external electric field to an insulator or dielectric, positive and negative electric charges within it undergo slight relative shift on opposite directions. Electric field distorts the negative cloud of electrons around positive atomic nuclei in a direction opposite to the field. This slight separation (stretching or shift of a fraction of an atomic diameter) of charge makes one side of the atom somewhat positive and the opposite side somewhat negative creating dipoles. These induced dipoles give rise to Polarization.

In polar dielectric molecules, permanent dipoles are present however, they are generally in random orientations due to thermal agitation, creating unpolarization condition. When an electric field is applied, the dipoles are oriented by rotation and aligned in the direction of the electric field so that one type of bound charges (+ve or -ve) appear on one surface and the opposite type on opposite surface. Thus electric field polarizes the material of polar molecules .

• Thus application of an electric field creates polarization either by inducing dipoles (in nonpolar) or reorienting and aligning them in external electric field (in polar) in both types of molecules.

• Polarization reduces the electric field intensity within the dielectric material by a factor known as the dielectric constant (or relative permittivity) of the material.

Dielectric constant and Displacement Vector

Dielectric constant (or relative permittivity) may be defined as the ratio of the electric field in free space to that in dielectric material (insulator). It is denoted by the symbol K or \(\epsilon_{r} \).

\begin{align*}K(\epsilon_{r} )=\frac{field\space in\space free\space space}{field\space in\space dielectric} \end{align*}\begin{align*} =\frac{\vec{E_0}}{\vec{E}}\rightarrow 1\end{align*}

If \(\epsilon_0 \) and\(\epsilon \) are the absolute permitivities of free space and dielectric respectively,
then electric fields are given by, \begin{align*}\vec{E_0}=\frac{1}{4\pi \epsilon_0}\frac{q\vec{r}}{r^3}\space and \space\vec{E}=\frac{1}{4\pi \epsilon}\frac{q\vec{r}}{r^3} \rightarrow 2 \end{align*}

Substituting these values in 1, we get \begin{align*}K=\frac{\epsilon }{\epsilon_0} \end{align*}

So, dielectric constant or relative permittivity may be defined as the ratio of the
absolute permittivity of the dielectric to that of free space.

Displacement vector :

As shown in Eq. 2, electric field depends upon the medium in which charge has been placed. So, let us assume new electric field vector (\(\vec{D} \)) known as the electric displacement vector or electric induction. This electric displacement vector (\(\vec{D} \)) is defined in such a way that it depends only on the distribution and magnitude of charges which produce electric field but is
independent of the medium.
The electric displacement vector (\(\vec{D} \)) due to a point charge q at a point r is given by, \begin{align*}\vec{D}=\frac{1}{4\pi }\frac{q\vec{r}}{r^3}=\epsilon (\frac{1}{4\pi \epsilon}\frac{q\vec{r}}{r^3} ) \end{align*}\begin{align*}\implies\vec{D}=\epsilon\vec{E} \end{align*}

For free space,\begin{align*} \vec{D}=\epsilon_0 \vec{E} \end{align*}

Gauss Law in Dielectric

The gauss law states that the net electric flux across an arbitary closed surface drawn in a electric field is equal to \(\frac{1}{epsilon_0} \) times the magnitude of charge enclosed by the surface.
Hence \(\oint_{s}\vec{E_0}.\hat{n}ds=\frac{q}{\epsilon_0}\rightarrow 1 \)
Where \(\hat{n}\) is a unit normal vector drawn outward normal to the surface.
Here \(\vec{E_0} \) and \(\hat{n}\) are always parallel to each other.
Then \(\oint_{s}\vec{E_0}ds=\frac{q}{\epsilon_0} \)
\begin{align*}or,oint_{s_1}\vec{E_0}ds+oint_{s_2}\vec{E_0}ds+oint_{s_3}\vec{E_0}ds+oint_{s_4}\vec{E_0}ds=\frac{q}{\epsilon_0} \end{align*}
Three out of L.H. side of vanishes then \begin{align*}E_0 A= \frac{q}{\epsilon_0}\implies E_0=\frac{q}{\epsilon_0 A}\rightarrow 2\end{align*}
where A is the area of each plates which is the gauss law in space. This is the magnitude of electric field in absemce of dielectric between the plates of conductor.

When we consider a case at which the dielectric is placed between the plates of the capacitor, then the charges on the dielectric are induced ude to charge or field given to the plates of the capacitor. Let q' be the induced charge on the dielectric due to the charge present on the upper plate of he capacitor and \(\vec{E} \) be the reduced value of electric field due to the presence of dielectric then from Gauss law,
\begin{align*}\oint_{s}\vec{E_0}.\hat{n}ds=\frac{q-q'}{\epsilon_0} \end{align*}
Where (q-q') is the net charge causing resulting field \(\vec{E}\)
\begin{align*}or, E=\frac{q-q'}{\epsilon_0 A}\rightarrow 3 \end{align*}

we know the dielectric constant
\begin{align*}K=\frac{E_0}{E}\implies E_0=KE \end{align*}\begin{align*}or,E_0=K\frac{q-q'}{\epsilon_0 A}\rightarrow 4 \end{align*}
From 2 and 4 we get
\begin{align*}q=K(q-q') \end{align*}\begin{align*}or,q'=q(\frac{K-1}{K})\rightarrow 5 \end{align*}
This is the magnitude of surface charge in terms of free charge. In the absence of dielectric K=1, then q'=0. Then the gauss law is similar to the previous case when the dielectric is not placed.
If several point charges \(q_1,q_2,...,q_n\) are enclosed by the surface 'S' then \(\rho\) be the volume charge density. V be the volume enclosed by surface 'S'.
\begin{align*}i.e. \rho=\frac{dq}{dV} \end{align*}\begin{align*}or,q=\oint_{V} \rho dV \rightarrow 6\end{align*}
Also we have Gauss law,
\begin{align*}\oint_{s}\vec{E_0}.d\vec{s}=\frac{q}{\epsilon_0} \end{align*}\begin{align*}or,\oint_{s}\vec{E_0}.d\vec{s}=\frac{1}{\epsilon_0}\oint_{V} \rho dV \end{align*}
But by applying Gauss divergence theorem,
\begin{align*}\oint_{s}\vec{E_0}.d\vec{s}=\oint_{V}(\bigtriangledown .\vec{E})dV=\frac{1}{\epsilon_0}\oint_{V} \rho dV \end{align*}\begin{align*} \end{align*}
\begin{align*}or,\oint_{V}(\bigtriangledown .\vec{E})dV- \frac{1}{\epsilon_0}\oint_{V} \rho dV=0\end{align*}\begin{align*}or,\oint_{V}[(\bigtriangledown .\vec{E})- \frac{1}{\epsilon_0}\rho]dV\end{align*}
Since Volume is arbitary \begin{align*}\bigtriangledown .\vec{E}=\frac{\rho}{\epsilon_0}\end{align*}\begin{align*}or,\bigtriangledown .\epsilon_0 \vec{E}=\rho \end{align*}
\begin{align*}\bigtriangledown .\vec{D}=\rho \end{align*}
This is the Gauss Law in Differential Form. Now, Gauss law in free space \(\bigtriangledown .\vec{D}=0 \).

Electric Polarization, Displacement Vector and Their Relation

Electric displacement: When a dielectric slab is placed between the plates of the capacitor, then the charges on dielectric are induced or polarized. The charge given to the capacitor is known as free charge and the induced of dielectric is known as bound charge. Let q be the free charge given to the plates of the capacitor and q' be the induced charge on the dielectric. Then the electric intensity of the field is given by
\begin{align*}E=\frac{q-q'}{\epsilon_0 A} \end{align*}\begin{align*}\epsilon_0 E=\frac{q}{ A}-\frac{q'}{ A}\rightarrow 1 \end{align*}
The last term of R.H.S. of 1 is known as induced surface charge per unit area and is known as polarization vector and it is denoted by \( \vec{P}\). Then from 1,
\begin{align*}\epsilon_0\vec{E} =\frac{q}{ A}-\vec{P} \end{align*}\begin{align*}or,\frac{q}{ A}=\epsilon_0\vec{E}+\vec{P}\rightarrow 2 \end{align*}
The term \(\epsilon_0\vec{E} =\frac{q}{ A}-\vec{P} \) is known as electric displacement and is denoted by \(\vec{D}\). hence \(\vec{D}=\frac{q}{ A}-\vec{P} \), is a vector quantity since electric field and polarization vectors.
The vector \(\vec{D}\) only depends on free charges , the quantity \(\vec{P}\) depends upon induced or bound charges of the dielectric and \(\vec{E}\) depends upon the both bound and free charges. Hence the above relation gives the relation between free and induced charges of dielectric.

Relations of vectors \(\vec{D}\) and \(\vec{P}\) with \(\vec{E}\):

The vectors \(\vec{D}\) and \(\vec{P}\) are separately connected with vector \(\vec{E}\). We know, the vector \(\vec{D}\) is defined as
\begin{align*}\vec{D}=\frac{q}{A}=\frac{q}{K\epsilon_0 A}.K\epsilon_0 \end{align*}
But from gauss law in dielectric \begin{align*}\frac{q}{K\epsilon_0 A}=E \end{align*}
\begin{align*}So\space \vec{D}=K\epsilon_0 \vec{E}\rightarrow 1 \end{align*}\begin{align*}\vec{D}=\epsilon \vec{E} \end{align*}
where \(K.\epsilon_0=\epsilon \), which is the permitivity of the medium.
Also,
\begin{align*}\vec{D}=\epsilon_0\vec{E}+\vec{P} \end{align*}\begin{align*}or,\epsilon \vec{E}=\epsilon_0\vec{E}+\vec{P} \end{align*}
\begin{align*}\epsilon = \epsilon_0+\frac{\vec{P}}{\vec{E}}\end{align*}is the permitivity of the medium.
Similarly, we can define electric polarization \(\vec{P}\) in terms of \(\vec{E}\) as
\begin{align*}\vec{P}=\frac{q'}{A} \end{align*}
\begin{align*}But\space q'=q(1-\frac{1}{K}) \end{align*}\begin{align*}\therefore \vec{P}=\frac{q}{A}(1-\frac{1}{K})=\vec{D}(1-\frac{1}{K}) \end{align*}
\begin{align*}or,\vec{P}=\epsion \vec{E}(1-\frac{1}{K}) \end{align*}\begin{align*}Again,\space \vec{P}=\frac{\epsion \vec{E}}{K}(K-1) \end{align*}
\begin{align*}\vec{P}=\epsilon_0 (K-1) \vec{E}\end{align*}
In case of free space, K=1, then \(\vec{P}=0\). i.e. no polarization occurs in absence of dielectric.

FORCE AND ENERGY IN DIELECTRIC SYSTEM

Energy:

The basic energy and pressure laws are modified with dielectrics. When a dielectric is placed in an external field, it is polarized. Let the induced charge +q appears in one surface and -q appears on the opposite surface. +q and -q are separated by distance d is the case of parallel plate capacitor with dielectric i.e. the thickness of the dielectric slab, Dipole moment \(\vec{P}=qd\). SUppose that while the field \(\vec{E}\) is present the charges are further displaced by distance \(\Delta d \). The change in dipole moment is given by,
\begin{align*}dp=q(d+\Delta d)-qd \end{align*}
\begin{align*}or,\space dp=q\Delta d\rightarrow 1 \end{align*}
The force on the polarization charge is qE. So the work done for the displacement \(\Delta d\) is given by\begin{align*}dW=qE\Delta d\rightarrow 2 \end{align*}
\(\Delta d\) is the displacement in the direction of \(\vec{E}\) inserting the value dP given above, we get
\(dW=\vec{E}.d\vec{P}\space as\space \vec{E}\space and\space \vec{P} \) point in the same direction. This work done is stored in the form of energy in the dielectric. The supplier of the work is the source of the field, the battery that maintained the constant potential difference between the plates. This stored energy increases the internal energy of the molecules.
But \( \vec{P}=\epsilon_0 (K-1) \vec{E}\), for linear dielectric, where \(\vec{P}\) is the dipole moment per unit volume.
Therefore,
\begin{align*}d\vec{P}=\epsilon_0 (K-1) d\vec{E} \end{align*}
where \(d\vec{P}\) is the change in dipole moment per unit volume. So, work done is given by\begin{align*}\vec{E}.d\vec{P}= \vec{E}.\epsilon_0 (K-1) d\vec{E}\end{align*}
Since \(\vec{E}.d\vec{E}=EdE=\frac{1}{2}d(E^2)\), therefore,\begin{align*}\vec{E}.d\vec{P}=\frac{\epsilon_0}{2}(K-1)d(E^2) \end{align*}
\begin{align*}or,\int_{o}^{P}\vec{E}.d\vec{P}=\frac{\epsilon_0}{2}(K-1)E^2 \rightarrow 3 \end{align*}
This equation gives the energy change due to polarization P. Therefore the total energy density is equal to the sum of energy density in free space and the energy density in dielectric. Therefore total energy density \(=\frac12 \epsilon_0 E^2 +\frac{\epsilon_0}{2}(K-1)E^2 \)
\begin{align*}=\frac12 \epsilon_0 KE^2 \rightarrow 4 \end{align*}
That is energy increases by K times than that of without dielectric. Energy per unit volume represented by U is given by \begin{align*}U=\frac12 \epsilon_0 KE^2 \end{align*}
Since \(\vec{D}=K\epsilon_0 \vec{E}\) and \(\vec{D}\) is in the direction of \(\vec{E}\), therefore we can write,\begin{align*}U=\frac12\vec{E}.\vec{D}\rightarrow 5 \end{align*}
For any volume,\begin{align*}U=\frac12 \epsilon_0\int_{V} KE^2 dV \rightarrow 6 \end{align*}
\begin{align*}or,U=\frac12\int_{V}\vec{E}.\vec{D}dV\rightarrow 7 \end{align*}
If dielectric is isotropic then,\begin{align*}U=\frac12 \epsilon_0 K\int_{V} E^2 dV \rightarrow 8 \end{align*}
Although we have derived equation 8 for parallel plate capacitor but it is applicable in all cases.

Force:

An expression for force can be obtained by principle of virtual work. If the work is done by an external agent against the force of attraction between the plates in moving them further apart through a distance \(\Delta d \),
\begin{align*}W=\vec{F}.\Delta \vec{d} \end{align*}\begin{align*}or,W=-F\Delta d \end{align*}
The change in volume due to this additional separation (\(\Delta d\))is A\(\Delta d\), where A is the area of the plate. Change in energy due to this change in volume equals \(\frac12 \epsilon_0 K E^2 A\Delta d \)
work done = change in enrgy\begin{align*} \end{align*}\begin{align*} \end{align*}
\begin{align*}\frac12 \epsilon_0 K E^2 A\Delta d = \vec{F}.\Delta \vec{d}\end{align*}
\begin{align*}\frac12 (\vec{E}.\vec{D}\) A\Delta d = \vec{F}.\Delta \vec{d}[\because \vec{D}=\epsilon_0 K\vec{E}] \end{align*}
\begin{align*}or,\space F=\frac12 (\vec{E}.\vec{D}) A \end{align*}
\begin{align*}Pressure=\frac{\vec{F}}{A}= \frac12 (\vec{E}.\vec{D})\end{align*}
This is force of attraction between the plates tending to pull them together though the pressure exists at any point in an electric field.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Griffiths, D.J., Introduction to Electrodynamics, PHI Learning Private Limited, 2013