Polarization of a rare gas in an electric field A rare (noble) gas with number density N (atoms per m^3) is subjected to a uniform electric field E. In the linear regime, the macroscopic electric dipole moment per unit volume (polarization P) is proportional to which quantities?

Difficulty: Easy

Correct Answer: Proportional to N and to E (P ∝ N E)

Explanation:


Introduction / Context:
In dielectric media, polarization P gives the dipole moment per unit volume. For dilute gases such as noble gases, induced dipoles arise via electronic polarizability, and the Clausius–Mossotti/Lorentz–Lorenz relations connect microscopic polarizability to macroscopic P.


Given Data / Assumptions:

  • Dilute gas, non-interacting atoms with electronic polarizability αe.
  • Linear response (small fields).
  • Number density N (atoms/m^3) and applied field E.


Concept / Approach:

Each atom acquires an induced dipole p_induced = αe E (in SI, αe may be written with ε0 depending on convention). The polarization is P = N * p_induced, so P ∝ N * E. Hence, doubling the density or the field doubles P in the linear regime.


Step-by-Step Solution:

Write single-atom response: p = αe E.Sum over N atoms in unit volume: P = N αe E.Conclude P ∝ N and P ∝ E.


Verification / Alternative check:

Macroscopic dielectric constant εr relates to microscopic αe via Lorentz–Lorenz relation; for small αe N, the linear dependence is clear.


Why Other Options Are Wrong:

Independence from N or E contradicts linear polarization theory; E^2 dependence appears in nonlinear optics, not in this small-signal regime.


Common Pitfalls:

Confusing microscopic polarizability with relative permittivity, or assuming strong-field nonlinear effects at ordinary field strengths.


Final Answer:

Proportional to N and to E (P ∝ N E)

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