BOUNDARY CONDITIONS OF \(\vec{D}\) and \(\vec{E} \) INTERFACE, CLAUSIUS MOSOTTI FIELD EQUATION, LANGEVIN DEBYE EQUATION

BOUNDARY CONDITIONS OF \(\vec{D}\) and \(\vec{E} \) INTERFACE

There are many boundary lines between two medium. Two medium may be two dielectric or a dielectric and free surface belong to boundary conditions. In this case, we examine how the electric field changes at boundary between two different medium.
The electric field is changed in going from one medium to another is determined by the two basic ideas of electrostatics, they are
i) By Gauss law i.e. \(\oint \vec{E}.\hat{n}ds=\frac{q}{\epsilon_0}\space or\space \oint \vec{D}.d\vec{S}=q \)
ii) The electrostatics field is a electrostatic field is a conservative field i.e. \(\oint \vec{E}.d\vec{l}=0 \)

Let us consider teo dielectric media with permitivities \(\epsilon_1 \) and \(\epsilon_2 \) in contact as shown in figure. Let us take a cylindrical surface ABCD of height h, which intersects the interface and encloses an area \(\Delta S \) of the interface.
Let \(D_{1n}\) be the average normal component of displacement vector to the top of the box in medium 1. \(D_{2n}\) be the average normal component of displacement vector D on 2nd medium.\(D_{1n}\) is inwards normally to the surface \(\Delta S\) and \(D_{2n}\) is outwards normally to the surface \(\Delta S\).
In order to find the boundary conditions for \(\vec{D}\), we have to apply Gauss law to this surface of cylinder ABCD. If the height AD, BC of the cylinder is negligibly small in comparision with the diamete of the cylinder.
\begin{align*}\oint \vec{D}.d\vec{S}=q \end{align*} Since there are no free charges at the interface, q=0.
Let \(\hat{n_1}\) and \(\hat{n_2}\) be the unit vectors on the surface AB and CD.
\begin{align*}i.e.\oint \vec{D}.d\vec{S}=0 \end{align*}\begin{align*}D_{1n}\Delta S=D_{2n}\Delta S=0 \end{align*}
\begin{align*}Here,\space D_{1n}=\vec{D_1}.\hat{n_1} \end{align*}\begin{align*}and\space -D_{2n}=\vec{D_2}.\hat{n_2} \end{align*}
We know,
\begin{align*}\vec{D}=\epsilon \vec{E} \end{align*}\begin{align*}So\space \vec{D_{1n}}= \epsilon_1 \vec{E_{1n}},\space \vec{D_{2n}}= \epsilon_2 \vec{E_{2n}}\rightarrow 1 \end{align*}
\begin{align*}i.e.\epsilon_1 \vec{E_{1n}}=\epsilon_2 \vec{E_{2n}}\rightarrow 2 \end{align*}
The equation 1 shows that the normal component of displacement vector is continuous across the charge free boundary between the two dielectrics.
For the second boundary condition, i,e, for \(\vec{E}\), is found from the fact that no work is done in taking a unit test charge around a closed path. Let us consider a rectangular path EFGH of length \(\Delta l\) as in figure, with the EF side lying in the one medium of permittivity \(\epsilon_1\) and the side GH lying in the 2nd medium of permittivity \(\epsilon_2 \).
If te sides FG and EH are negligibly small compared with EF and HG, then the work done in taking a unit harge around this rectangle is \begin{align*}\oint \vec{E}.d\vec{l}=0 \end{align*}
\begin{align*}i.e.(E_{1t}-E_{2t})\Delta l=0 \end{align*}\begin{align*}\implies E_{1t}=E_{2t} \rightarrow 3 \end{align*}
where \(E_{1t} \) represents the tangential component of E in medium 1 and \(E_{2t} \) represents the tangential component of E in medium 2.

Hence the equation 3 shows that, the tangential component of electric intensity has the same value on both sides of the surface of separation.
Hence equation 1, 2 and 3 are two fundamental boundary conditions.

It follows from them that lines \(\vec{D}\space or\space \vec{E} \) will be refracted at the boundary between two dielectrics as shown in figure. If \(\theta_1 \space and \space \theta_2\) be the angle of refraction in medium 1 and medium 2 respectively i.e. NON' is the normal interface and \(\epsilon_1\space and \space \epsilon_2\) make the angle \(\theta_1 \space and \space \theta_2\) respectively with the normal.
Normal component of \(E_1=E_{1n}=E_1 cos\theta_1\)
Normal component of \(E_2=E_{2n}=E_2 cos\theta_2\)
So from equation 2, \begin{align*}\epsilon_1 E_1 cos\theta_1=\epsilon_2 E_2 cos\theta_2 \rightarrow 4\end{align*}
Tangential component of \(E_1=E{1t}=E_1 sin\theta_1\)
Tangential component of \(E_2=E{2t}=E_2 sin\theta_2\)
From 3,\begin{align*}E_1 sin\theta_1=E_2 sin\theta_2 \rightarrow 5 \end{align*}
From 4 and 5,
\begin{align*}\epsilon_1 cot\theta_1 = \epsilon_2 cot\theta_2 \rightarrow 6\end{align*}
Case-I:
If \(\epsilon_1>\epsilon_2\). Then the normal of electric field in medium 1 should be less than that of medium 2.
Case-II:
If one of the substances i.e. medium or dielectric is a conductor, suppose medium 1 is conductor, then the lectric field inside the conductor is zero. Therefore there are no permanent dipoles in the conductors. Hence, the polarization vector \(\vec{P} \) must be zero. According to the relation, \(\vec{D}=\epsilon_0 \vec{E}+\vec{P}\) if \(\vec{P} \) and \(\vec{E}\) both are zero, then the displacement vector \(\vec{D}\) is also zero.
\begin{align*}\therefore D_{1n}=0 \end{align*}
Hence from the boundary condition.
\begin{align*}\oint \vec{D}.d\vec{S}=q \end{align*}\begin{align*}or,-D_{1n}\Delta S+D_{2n}\Delta S=q \end{align*}
\begin{align*}or,0+D_{2n}\Delta S=q \end{align*}\begin{align*}or,D_{2n}=\frac{q}{\Delta S}=\sigma\rightarrow 7 \end{align*}
where \(\sigma \) is the surface charge density.
Hence the normal component of the displacement vector just outside the conductor is equal to the surface charge density on the conductor.
The second condition is obtained from the fact that no work is done in taking a unit test charge around a closed path a portion of which lies in the medium. Thus the line integral of \(\vec{E}\) around the rectangular path, we must get it equal to zero,i.e.\( E_{2t}\Delta l=0\)\begin{align*}\implies E_{2t}=0\rightarrow 8 \end{align*}
Thus the equation 7 and 8 are the boundary conditions for the displacement and electric field in dielectric just outside a conductor.

CLAUSIUS MOSOTTI FIELD EQUATION

The electric field which is responsible for polarizing a molecule of the dielectrics is called the molecular field. This is the electric field at a molecular position in the dielectrics. The molecular field is produced by all external sources and by all polarized molecules in the dielectric except one molecule under consideration. Molecular field may be calculated in following ways.

Let us cut a spherical cavity of radius ‘r’ (such that its dimensions are very great as compared to the molecular dimensions and very small as compared to the volume of the dielectrics) surrounding the point at which the molecular field is to be computed (shown in Fig.a ). The dielectric which is now left, will be treated as acontinuum. The cavity is put in its original position in the dielectric (molecule by molecule) except the molecule where the molecular field is to be computed. The molecules which are just replaced in the cavity are treated as individual dipoles and not as a continuum.

Let the dielectric be kept in placed in the uniform electric field between two parallel plates (of a capacitor) as shown in Figure (b). The dotted line shows the boundary of the dielectric. Let the surface charge density of real charges on the plates be \(\sigma_{x} \). Again let the surface of cavity has polarized charges of surface density \(\sigma_{p} \).

The electric field experienced by the molecule of the dielectric at the center of the cavity C,\( E_{in}\), given by,

\(E_{in} = E_1 + E_2+ E_3 + E_4\)

Where, (i) E1 is the field between two plates with no dielectric, so that E1=\(\sigma/\epsilon_0\),

(ii) E2 is the field at C due to polarized charges on the plane surfaces of the dielectric facing the capacitor plates and is given by \(E2=\sigma_{P}/\epsilon_0\),

(iii) E3 is the field at C due to polarized charges on the surface of cavity, to be calculated,

(iv) E4 is the field at C due to permanent dipoles. But in present case for nonpolar isotropic dielectrics E4 =0.

Thus, electric field experienced by the molecules of the dielectric at the center of the cavity C, Ein is given by,

\begin{align*}E_{in}=\frac{\sigma}{\epsilon_0}-\frac{\sigma_{p}}{\epsilon_0}+E_3\rightarrow 1 \end{align*}

Now evaluation E3:

Consider a small elemental area ds on the surface of cavity of an angular width \(d\theta \) and at an angle of \(\theta \) with the direction of field \(\vec{E} \). The vector \(\vec{P} \) shows the direction of displacement at the center of \(d\vec{S} \), Figure (a). The normal component of displacement is \begin{align*}\vec{P_{N}}=\vec{P}cos\theta \end{align*}

By definition of polarization, it is the surface charge per unit area. This charge on d𝑠to provide flux normal to d𝑠is\begin{align*} \vec{P}cos\theta ds\end{align*}

And the electric field intensity at C due to this charge is\begin{align*}=\frac{\vec{P}cos\theta ds }{4\pi \epsilon_0 r^2} \end{align*}

Where ‘r’ is the radius of cavity. The field is directed along the radius CA. Resolving the field along the applied field and perpendicular to it, we have the components:

Along the field, \(\frac{\vec{P}cos\theta ds }{4\pi \epsilon_0 r^2}cos\theta \)

Perpendicular to the field, \(\frac{\vec{P}cos\theta ds }{4\pi \epsilon_0 r^2}sin\theta \)

If the area ds be taken around 2\(\pi \) radians about LM, it will describe a ring, the surface area of which is given by, \begin{align*} =2\pi r^2 sin\theta d\theta\end{align*}

Therefore, electric field intensity at C due to ring in the field direction is,\begin{align*}=\frac{\vec{P}cos^2\theta }{4\pi \epsilon_0 r^2} 2\pi r^2 sin\theta d\theta\end{align*}\begin{align*}=\frac{\vec{P}cos^2\theta }{2 \epsilon_0 }sin\theta d\theta \end{align*}

Here, normal components for ring reduces to zero due to cancellation.

Now integrating the electric field intensity at C due to charges on the surface of cavity, we have \begin{align*}E_3=\frac{\vec{P} }{2 \epsilon_0 }\int_0^{\pi}cos^2\theta sin\theta d\theta\end{align*}\begin{align*}\implies E_3=\frac{\vec{P} }{2 \epsilon_0 }\times\frac{2}{3} \end{align*}\begin{align*}\implies E_3=\frac{\vec{P} }{3 \epsilon_0 }\rightarrow 2 \end{align*}

Now, substituting the value of E3 in equation (1) , it becomes,\begin{align*}E_{in}=\frac{\sigma}{\epsilon_0}-\frac{\sigma_{p}}{\epsilon_0}+ \frac{\vec{P} }{3 \epsilon_0 }\rightarrow 3\end{align*}

The resultant electric field, \(\vec{E}\) between plates can be given by \begin{align*}\vec{E}=\frac{\sigma}{\epsilon_0}-\frac{\sigma_{p}}{\epsilon_0} \end{align*}\begin{align*}So,\space E_{in}=\vec{E}+ \frac{\vec{P} }{3 \epsilon_0 }\rightarrow 4 \end{align*}

Also, we know that,\begin{align*}\vec{D}=\epsilon_0 \vec{E}+\vec{P} \end{align*}

Hence, \(\vec{E}=\frac{\vec{P}}{\epsilon -\epsilon_0}\)

Put this value in equation (4), \begin{align*}E_{in}=\frac{\vec{P}}{\epsilon -\epsilon_0}+ \frac{\vec{P} }{3 \epsilon_0 } \end{align*}\begin{align*}\implies E_{in}=\frac{\vec{P} }{3 \epsilon_0 } (\frac{{\epsilon +2\epsilon_0}}{{\epsilon -\epsilon_0}})\rightarrow 5\end{align*}

If n be the no. of molecules per unit volume, \(\alpha \) be the molecular polarizability, then polarization \(\vec{P}\), is defined as electric moment per unit volume. i.e.\begin{align*}\vec{P}=n\alpha \vec{E_{in}} \end{align*}\begin{align*}\vec{E_{in}}=\frac{\vec{P}}{n\alpha } \end{align*}

By putting the value in equation (5), it becomes,\begin{align*}\frac{\vec{P}}{n\alpha } =\frac{\vec{P} }{3 \epsilon_0 } (\frac{{\epsilon +2\epsilon_0}}{{\epsilon -\epsilon_0}})\end{align*}\begin{align*}\implies \frac{n\alpha}{\epsilon_0}=\frac{\frac{\epsilon}{\epsilon_0}-1}{\frac{\epsilon}{\epsilon_0}+2} \end{align*}\begin{align*}\implies \frac{n\alpha}{\epsilon_0}=\frac{\epsilon_{r}-1}{\epsilon_{r}+2} \rightarrow 6\end{align*}

where, \(\epsilon_{r} \) is the relative permittivity of dielectric.

Equation (6) represents well-known Clausius- Mossotti relation for molecular field in dielectric. Furthermore, Relative dielectric constant \(\epsilon_{r} \) is related with the refractive index of medium as,\begin{align*}\epsilon_{r}=\eta ^2 \end{align*}

Hence, equation (6) can be expressed in term of refractive index as,\begin{align*} \frac{n\alpha}{\epsilon_0}=\frac{\eta ^2-1}{\eta ^2+2} \rightarrow 7\end{align*}

Equation (7) is called Lorentz formula which is valid only when \(\epsilon_{r} \) is frequency independent.

LANGEVIN DEBYE EQUATION

While calculating the polarization of non-polar molecules, it has been observed that molecular polarizibility \(\alpha \) does not depend on the temperature. But this is not exactly in case of polar molecules. If polar molecules are present in an electric field then in addition to induced polarization, permanent polarization of molecules are also present. In absence of electric field, permanent dipole are randomly oriented and net polarization is zero. On application of electric field, the permanent dipole are forced to orient along the direction of electric field but orientation is continuously distributed due to thermal agitation. When an equilibrium is attained, the dipoles may make zero to \(\pi \) angles with the direction of field.

The polarization of such molecules with permanent dipole may therefore be resolved in the direction of electric field and average value of \(\vec{P_{p}}(=\vec{P_{p}}cos\theta)\) may be found using Fig.-1. If \(\vec{P_{p}}\) and \(\vec{P_{i}}\) are permanent and induced polarization then total polarization can be represented as,\begin{align*} \vec{P}=\vec{P_{p}}+\vec{P_{i}}\end{align*}

If \(E_{in}\) is the inner field acting on dielectric molecules then the potential energy of the permanent dipoles per unit volume of moment \(\vec{P_{p}} \) is expressed as,\begin{align*}U=-\vec{E_{in}}.\vec{P_{p}}=-E_{in}P_{p}cos\theta \rightarrow 1 \end{align*}

Acc. to Maxwell-Boltzmann law of statistical mechanics, the no. of molecules distributed with the axisof their dipoles oriented in all directions within a solid angle \(d\Omega \) at angle \(\theta \) is\begin{align*}dN=Ae^{-(\frac{U}{K_{B}T})}d\Omega \end{align*}

where A is constant, \(K_{B}\) be the Boltzmann constant and T be absolute temperature.

The dipole moment per unit volume in the direction of electric field is \(=\vec{P_{p}}cos\theta \). Now the dipole moment per unit volume of dN molecules in the direction of field is,\begin{align*}=dN \vec{P_{p}}cos\theta = Ae^{-(\frac{U}{K_{B}T})} \vec{P_{p}}cos\theta d\Omega =Ae^{(\frac{E_{in}P_{p}cos\theta }{K_{B}T})} \vec{P_{p}}cos\theta d\Omega \end{align*}

So, total permanent dipole moment per unit volume \(P_{p} \) is,\begin{align*}\sum P_{p}=\int_{0}^{\pi }Ae^{x cos\theta}P_{p}cos\theta d\Omega \rightarrow 3 \end{align*}\begin{align*}where\space x=\frac{P_{p}E}{K_{B}T} \end{align*}

Now thye average dipole moment\begin{align*}<P_{p}>=\frac{\int_{0}^{\pi }Ae^{x cos\theta}P_{p}cos\theta d\Omega }{\int_{0}^{\pi }Ae^{x cos\theta} d\Omega }\rightarrow 4 \end{align*}

But solid angle \(d\Omega =\frac{area\space of\space ring\space between\space \theta \space and\space\theta+d\theta}{r^2} \)\begin{align*}=\frac{2\pi r^2 sin\theta d\theta}{r^2} \end{align*}\begin{align*}=2\pi sin\theta d\theta \end{align*}

Let \(cos\theta =a \) in eqn 4, then \(sin\theta d\theta =-da \) and \begin{align*} <P_{p}>=\frac{\int_{-1}^{+1}e^{xa}P_{p}ada}{\int_{-1}^{+1}e^{xa}da}\end{align*}\begin{align*}\frac{<P_{p}>}{P_{p}}=\frac{[\frac{a}{x}e^{xa}-\frac{1}{x^2}e^{xa}]_{-1}^{+1}}{[\frac{1}{x}e^{xa}]_{-1}^{+1}} \end{align*}\begin{align*}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}-\frac{1}{x} \end{align*}\begin{align*}=cothx-\frac{1}{x}\rightarrow 5 \end{align*}\begin{align*} =L(x)\end{align*}

where L(x) is called Langevin function.

In general L(x) is very small hence\begin{align*}L(x)=\frac{1}{x}+\frac{x}{3}+...-\frac{1}{x}= \frac{x}{3}\end{align*}\begin{align*}or,<P_{p}>=P_{p}\frac{x}{3}=\frac{P_{p}^{2}E}{3K_{B}T}\rightarrow 6 \end{align*}

Hence the orientational polarizability is given by equation 6,\begin{align*} \therefore \alpha_{eff.}=\alpha +\alpha_{d}=\alpha + \frac{P_{p}^{2}E}{3K_{B}T} \rightarrow 7\end{align*}

This equation 7 is called Langevin Debye formula where \(\alpha_{eff.} \) depends upon the temperature T. So for polar molecules, the polarizability is inversely proportional to temperature. So, we can also say that K(permitivity) depends upon the temperature.

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