Dielectrics under AC Field – Role of Complex Permittivity A material under alternating electric stress is characterized by a complex relative permittivity εr* = εr′ − j εr″. The dielectric loss under AC excitation is proportional to which quantity?

Introduction / Context:Real dielectrics exhibit energy dissipation when subjected to alternating electric fields. This loss manifests as heat and is captured by a complex permittivity, whose imaginary part encodes the out-of-phase component of polarization responsible for power dissipation. Understanding this link is central to high-frequency capacitor design, insulation grading, and RF heating analysis.

Given Data / Assumptions:

• Complex relative permittivity εr* = εr′ − j εr″.
• Angular frequency ω and RMS field E are finite and uniform in the dielectric.
• Linear, isotropic response assumed.

Concept / Approach:

The time-average volumetric power loss in a lossy dielectric can be written as P_loss = ω ε0 εr″ E^2 for the small-loss case, directly proportional to the imaginary part εr″. An equivalent description uses the loss tangent tan δ = εr″/εr′, where the reactive (stored) energy scales with εr′ and the dissipative part scales with εr″; the product leads to P_loss ∝ ω ε0 εr′ tan δ E^2 = ω ε0 εr″ E^2.

Step-by-Step Solution:

Verification / Alternative check:

Using phasor analysis of current density J = j ω ε0 εr* E shows a real component J_real = ω ε0 εr″ E, whose product with E gives average power density, again emphasizing εr″.

Why Other Options Are Wrong:

• Both εr′ and εr″: only εr″ controls dissipation for fixed ω and E (εr′ affects storage).
• εr′: relates to capacitance, not loss.
• (εr″)^2: loss is first order in εr″, not quadratic.

Common Pitfalls:

Confusing loss tangent tan δ with εr″ itself; thinking higher εr′ necessarily means higher losses, which is not true unless tan δ is fixed.

Final Answer:

εr″ (the imaginary part)