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    The Clausius and Mossotti equation is given by:
    Question

    The Clausius and Mossotti equation is given by:

    A.

    ϵ+1ϵ1=(n0α/3ϵ0)\frac{\epsilon+1}{\epsilon-1}=\left(n_0 \alpha / 3 \epsilon_0\right)​​

    B.

    ϵ1ϵ+2=(noα/3ϵo)\frac{\epsilon-1}{\epsilon+2}=\left(n_{\mathrm{o}} \alpha / 3 \epsilon_{\mathrm{o}}\right)​​

    C.

    ϵ+2ϵ1=(n0α/3ϵ0)\frac{\epsilon+2}{\epsilon-1}=\left(n_0 \alpha / 3 \epsilon_0\right)​​

    D.

    ϵ+1ϵ2=(n0α/3ϵo)\frac{\epsilon+1}{\epsilon-2}=\left(n_0 \alpha / 3 \epsilon_{\mathrm{o}}\right)​​

    Correct option is B

    εr1εr+2=(nα3ε0)Where: εr=ε/ε0 is the relative permittivity or dielectric constant of the material. ε0 is the permittivity of free space. n is the number density of the molecules (i.e., number of molecules per cubic meter). α is the molecular polarizability (a measure of how easily the electron cloud of a molecule can be distorted by an electric field).\begin{aligned}&\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \left( \frac{n \alpha}{3 \varepsilon_0} \right) \\\\\text{Where:} \\&\bullet\ \varepsilon_r = \varepsilon / \varepsilon_0\ \text{is the relative permittivity or} \ \textbf{dielectric constant} \ \text{of the material.} \\&\bullet\ \varepsilon_0\ \text{is the} \ \textbf{permittivity of free space}\text{.} \\&\bullet\ n\ \text{is the} \ \textbf{number density} \ \text{of the molecules (i.e., number of molecules per cubic meter).} \\&\bullet\ \alpha\ \text{is the} \ \textbf{molecular polarizability} \ \text{(a measure of how easily the electron cloud of a molecule can be} \\&\quad\ \text{distorted by an electric field).}\end{aligned}​​

    ​​

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