Dielectrics under AC field – dependence of dielectric loss on complex permittivity A dielectric material is subjected to an alternating electric field. The complex relative permittivity is written as ε_r = ε' – j ε'. The dielectric power loss in the medium is primarily proportional to which component(s) of the dielectric constant?

Introduction:
When a dielectric is placed in an alternating electric field, it not only stores electric energy but may also dissipate energy as heat. This dissipation is called dielectric loss and is crucial in capacitors, cables, microwave substrates, and high-frequency insulators where heating and efficiency matter.

Given Data / Assumptions:

• Time-harmonic electric field at angular frequency ω.
• Complex permittivity representation ε = ε′ − j ε′′ (real storage part ε′, loss part ε′′).
• Linear, isotropic dielectric behavior in small-signal regime.

Concept / Approach:

The average volumetric power dissipation in a dielectric is proportional to ω * ε0 * ε′′ * |E|^2. Here ε′′ is the imaginary component of permittivity and directly measures lossy polarization mechanisms (orientation lag, ionic and electronic relaxation, and conductivity-equivalent loss). The loss tangent is tan δ = ε′′ / ε′, but the loss itself scales with ε′′, not ε′ alone.

Step-by-Step Solution:

Verification / Alternative check:

Using tan δ, the reactive stored energy is tied to ε′ while the resistive component causing heat is tied to ε′′. Measured dielectric heating in RF ovens and capacitors follows ε′′ trends across frequency and temperature.

Why Other Options Are Wrong:

Real part ε′ governs energy storage and capacitance; saying “both” or “either” overstates the role of ε′; independence from ε_r is incorrect because loss is explicitly linked to ε′′.

Common Pitfalls:

Confusing tan δ (ratio) with the absolute loss term; neglecting conduction loss which is often included effectively within ε′′ at AC.

Final Answer:

imaginary part of dielectric constant