Energy of an induced dipole in a uniform electric field An atom with electric polarizability a is placed in a homogeneous electric field E. What is the energy stored in (or magnitude of the lowering of) the polarized atom due to the induced dipole?

Introduction / Context:In dielectrics, an applied electric field induces a dipole moment p proportional to the field (for linear response). The potential energy associated with this induced dipole is a core concept in polarization and dielectric heating.

Given Data / Assumptions:

• Linear polarizability: p = a * E.
• Uniform static field E; small-signal regime.
• a is a scalar (isotropic case).

Concept / Approach:The work done to polarize the atom from zero field to E equals the integral of p · dE′, leading to the factor 1/2. Specifically, U = − ∫_0^E p(E′) · dE′ = − ∫_0^E a E′ · dE′ = − (1/2) a E^2. Often problems ask for the magnitude of the stored energy change, which is |U| = (1/2) a E^2. Hence the correct choice is 1/2 * a * E^2.

Step-by-Step Solution:Use p = a E (linear induced dipole).Work to build field: dU = − p · dE′ → integrate from 0 to E.Compute integral: U = − (1/2) a E^2 → stored-energy magnitude = (1/2) a E^2.

Verification / Alternative check:The factor 1/2 mirrors the capacitor energy U = (1/2) C V^2, since C ∝ a for a single polarizable entity in a field; both arise from integrating a linearly increasing response versus the driving quantity.

Why Other Options Are Wrong:“a * E^2” misses the 1/2 integration factor. “a^2 * E” and “1/2 * a * E” have incorrect dimensions. “a * E” is linear in E, also dimensionally inconsistent.

Common Pitfalls:

• Forgetting the 1/2 factor that arises from integrating a linear response from zero to the final field.
• Confusing sign (negative potential energy) with the requested magnitude (stored energy).

Final Answer:1/2 * a * E^2