QB_PH_SEC7_013: Dielectric, Magnetic Properties & Superconductivity

The Clausius-Mossotti relation connects the macroscopic dielectric constant $\varepsilon_r$ of a non-polar dielectric solid to its microscopic electronic polarizability $\alpha$ through

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Lorentz Local Electric Field**: In a cubic or isotropic dielectric: $E_{\text{loc}} = E + \frac{P}{3\varepsilon_0}$ 2. **Microscopic Polarization Definition**: $P = N p_{\text{induced}} = N \alpha E_{\text{loc}} = N \alpha \left(E + \frac{P}{3\varepsilon_0}\right)$ 3. **Macroscopic Susceptibility Relation**: $P = \varepsilon_0 (\varepsilon_r - 1) E \implies E = \frac{P}{\varepsilon_0(\varepsilon_r - 1)}$ 4. **Substitution & Rearrangement**: $P = N \alpha \left(\frac{P}{\varepsilon_0(\varepsilon_r - 1)} + \frac{P}{3\varepsilon_0}\right)$ $1 = \frac{N\alpha}{\varepsilon_0} \left(\frac{1}{\varepsilon_r - 1} + \frac{1}{3}\right) = \frac{N\alpha}{\varepsilon_0} \left(\frac{3 + \varepsilon_r - 1}{3(\varepsilon_r - 1)}\right) = \frac{N\alpha}{3\varepsilon_0} \left(\frac{\varepsilon_r + 2}{\varepsilon_r - 1}\right)$ $\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha}{3\varepsilon_0}$ 5. **Conclusion**: Option (A) is correct.