Physics, asked by rahulkumar8213128, 1 month ago

Deduce Claudius mosotti equation for dielectric constant in an electric field

Answers

Answered by kwanpooja
2

Answer:

P= N.Pav where N is the number of atoms or molecule per unit volume [20]. α is the polarizability of the molecule. By assuming the near field E3 is zero, Clausius and Mossotti derived a relation for the dielectric constant of a material under electronic and ionic polarization [21].

Answered by mukeshsharma4365
0

Answer:

The Clausius–Mossotti relation expresses the dielectric constant (relative permittivity, εr) of a material in terms of the atomic polarizability, α, of the material's constituent atoms and/or molecules, or a homogeneous mixture thereof. It is named after Ottaviano-Fabrizio Mossotti and Rudolf Clausius. It is equivalent to the Lorentz–Lorenz equation. It may be expressed as:[1][2]

{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}={\frac {N\alpha }{3\varepsilon _{0}}}}{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}={\frac {N\alpha }{3\varepsilon _{0}}}}

where

{\displaystyle \varepsilon _{r}=\epsilon /\epsilon _{0}}{\displaystyle \varepsilon _{r}=\epsilon /\epsilon _{0}} is the dielectric constant of the material, which for non-magnetic materials is equal to {\displaystyle n^{2}}n^{2} where {\displaystyle n}n is the refractive index

{\displaystyle \varepsilon _{0}}\varepsilon _{0} is the permittivity of free space

{\displaystyle N}N is the number density of the molecules (number per cubic meter), and

{\displaystyle \alpha }\alpha is the molecular polarizability in SI-units (C·m2/V).

In the case that the material consists of a mixture of two or more species, the right hand side of the above equation would consist of the sum of the molecular polarizability contribution from each species, indexed by i in the following form:[3]

{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}=\sum _{i}{\frac {N_{i}\alpha _{i}}{3\varepsilon _{0}}}}{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}=\sum _{i}{\frac {N_{i}\alpha _{i}}{3\varepsilon _{0}}}}

In the CGS system of units the Clausius–Mossotti relation is typically rewritten to show the molecular polarizability volume {\displaystyle \alpha '=\alpha /(4\pi \varepsilon _{0})}\alpha '=\alpha /(4\pi \varepsilon _{0}) which has units of volume (m3).[2] Confusion may arise from the practice of using the shorter name "molecular polarizability" for both {\displaystyle \alpha }\alpha and {\displaystyle \alpha '}\alpha ' within literature intended for the respective unit system.

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