--- title: "Dipole moment and Clausius-Mossotti equation" subject: "Electrical Engineering Material" chapter: 3 key: "semester_3/electrical_engineering_material:dipole-moment-clausius-mossotti" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/dipole-moment-clausius-mossotti site: BCT Engineering --- # Dipole moment and Clausius-Mossotti equation Electrical Engineering Material · Chapter 3: Dielectric Materials - Asked in: 2080 Baisakh, 8 marks; 2068 Chaitra, 3+7 marks (asked 2 times) - Video: [Derive Clausius Mossotti relation](https://www.youtube.com/watch?v=OZchTd4fxo0) — Physics mee - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/dipole-moment-clausius-mossotti - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3 ## Question Define electric dipole moment and local electric field. Derive the Clausius-Mossotti equation for electronic polarization, relating polarizability with permittivity. ## Short answer Dipole moment p = Qa (C·m). Local field E_loc = E + P/(3ε0). With P = Nα_e E_loc and P = ε0(εr-1)E, the Clausius-Mossotti equation is (εr-1)/(εr+2) = Nα_e/(3ε0). ## Answer #### Electric dipole moment Two equal and opposite charges $+Q$ and $-Q$ separated by a small distance $a$ form an electric dipole. Its **dipole moment** is a vector from $-Q$ to $+Q$: $$ \mathbf{p} = Q\mathbf{a} \qquad (\text{unit: C m}) $$ When an atom is placed in a field, its electron cloud shifts relative to the nucleus, creating an **induced dipole moment** proportional to the field acting on it: $$ p = \alpha_e E_{loc} $$ where $\alpha_e$ is the **electronic polarizability**. The **polarization** $P$ is the dipole moment per unit volume, $P = Np$. ```text no field: ( +) cloud centred on nucleus, p = 0 field E --> (-- +) cloud shifted left <-a-> p = Q a (points along E) ``` #### Local electric field The **local field** $E_{loc}$ is the actual field experienced by one atom inside a dielectric. It is the applied field plus the fields of all the other induced dipoles around it. In a solid it is larger than the applied macroscopic field $E$. #### Lorentz expression for the local field Cut an imaginary sphere around the reference atom, large compared with atomic size. The field at its centre is $$ E_{loc} = E + E_{sph} + E_{in} $$ - $E$: macroscopic field (from plate charges and outer surface polarization charges). - $E_{sph}$: field from bound charges on the inner surface of the spherical cavity. - $E_{in}$: field from dipoles inside the sphere; zero for cubic symmetry or random arrangement. ```text E --> .------. +' '- cavity surface charge + atom - magnitude P cos(theta), + . -> - theta measured from E +. .- '-....-' ``` The bound charge on the cavity wall at angle $\theta$ (measured from $E$) has density of magnitude $P\cos\theta$. It is positive on the wall facing the incoming field and negative on the far wall, so its field at the centre points along $E$. A ring between $\theta$ and $\theta + d\theta$ has area $dA = 2\pi r^2\sin\theta\,d\theta$. The field component along $E$ at the centre is $$ dE_{sph} = \frac{P\cos\theta \cdot 2\pi r^2\sin\theta\, d\theta}{4\pi\varepsilon_0 r^2}\cos\theta = \frac{P}{2\varepsilon_0}\cos^2\theta\sin\theta\, d\theta $$ $$ E_{sph} = \frac{P}{2\varepsilon_0}\int_0^{\pi}\cos^2\theta\sin\theta\, d\theta = \frac{P}{2\varepsilon_0}\left[-\frac{\cos^3\theta}{3}\right]_0^{\pi} = \frac{P}{3\varepsilon_0} $$ Hence $$ E_{loc} = E + \frac{P}{3\varepsilon_0} $$ #### Clausius-Mossotti equation With $N$ atoms per unit volume: $$ P = N\alpha_e E_{loc} = N\alpha_e\left(E + \frac{P}{3\varepsilon_0}\right) \quad\cdots (1) $$ From electrostatics, $D = \varepsilon_0E + P = \varepsilon_0\varepsilon_rE$, so $$ P = \varepsilon_0(\varepsilon_r - 1)E \quad\Rightarrow\quad E = \frac{P}{\varepsilon_0(\varepsilon_r - 1)} \quad\cdots (2) $$ Put (2) in (1) and divide by $P$: $$ \begin{aligned} 1 &= N\alpha_e\left[\frac{1}{\varepsilon_0(\varepsilon_r - 1)} + \frac{1}{3\varepsilon_0}\right] = \frac{N\alpha_e}{3\varepsilon_0}\cdot\frac{3 + (\varepsilon_r - 1)}{\varepsilon_r - 1} \end{aligned} $$ $$ \boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0}} $$ #### Meaning and use - It relates the microscopic polarizability $\alpha_e$ to the macroscopic, measurable $\varepsilon_r$. - It gives $E_{loc} = \frac{\varepsilon_r + 2}{3}E$; for Si ($\varepsilon_r = 11.9$, $N = 5 \times 10^{28}$ m⁻³) it gives $\alpha_e = 4.17 \times 10^{-40}$ F m² and $E_{loc} = 4.63E$. - At optical frequency $\varepsilon_r = n^2$, giving the Lorentz-Lorenz equation for refractive index. - Valid for non-polar dielectrics with cubic symmetry, gases and non-polar liquids; not for polar materials where permanent dipoles interact strongly. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The dipole moment of charges +Q and -Q separated by a is: A. p = Q/a, pointing from + to - B. p = Qa, pointing from - to + C. p = Qa, pointing from + to - D. p = Qa^2, pointing from - to + Answer: B. p = Qa is a vector from the negative to the positive charge, in C m. (Key point: Dipole moment) 2. In the Lorentz sphere method, E_in (field from dipoles inside the sphere) is zero because: A. The sphere is empty vacuum B. The applied field cancels it C. P is zero inside the sphere D. Of cubic symmetry or random arrangement of the dipoles Answer: D. For cubic symmetry the fields of the dipoles inside the sphere cancel at the centre. (Key point: Cavity field contributions) 3. Integrating surface charge P cos(theta) over the cavity wall gives E_sph equal to: A. P/(3 eps0) B. P/(2 eps0) C. P/eps0 D. P/(6 eps0) Answer: A. The integral of cos^2(theta) sin(theta) from 0 to pi is 2/3, so E_sph = (P/2eps0)(2/3) = P/(3 eps0). (Key point: Sphere field integral) 4. Which two relations are equated to obtain the Clausius-Mossotti equation? A. P = eps0 E and P = N alpha E B. P = eps0(eps_r - 1)E and P = N alpha_e E_loc C. D = eps0 E and E_loc = E D. P = N alpha_e E and E_loc = P/eps0 Answer: B. Macroscopic P = eps0(eps_r - 1)E is equated with microscopic P = N alpha_e E_loc, with E_loc = E(eps_r + 2)/3. 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