Question:easy

\(\dfrac{n\alpha}{3\epsilon_0} = \dfrac{(\epsilon_r - 1)}{(\epsilon_r + 2)}\) is known as ____________ relation.

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The formula \(\tfrac{\epsilon_r-1}{\epsilon_r+2} = \tfrac{n\alpha}{3\epsilon_0}\) links dielectric constant to polarizability via the local field.
Updated On: Jul 2, 2026
  • Debye
  • Clausius-Mossotti
  • Einstein-Debye
  • Bose-Einstein
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: The equation connects a bulk, measurable property (the relative permittivity $\epsilon_r$) with an atomic-scale property (the polarizability $\alpha$) and the number density $n$ of dipoles.

Step 2: Such a micro-to-macro dielectric bridge is derived by inserting the Lorentz local field $E_{\text{loc}} = E + P/3\epsilon_0$ into the definition of polarization, which produces the factor $(\epsilon_r+2)$ in the denominator.

Step 3: The resulting expression $\dfrac{\epsilon_r-1}{\epsilon_r+2} = \dfrac{n\alpha}{3\epsilon_0}$ is the well-known Clausius-Mossotti relation for non-polar dielectrics.

Step 4: Debye's equation extends this to polar molecules by adding an orientational term, but the bare polarizability form written here is specifically Clausius-Mossotti. Hence option (B) is correct.\[\boxed{\text{Clausius-Mossotti}}\]
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