Lossy (leaky) dielectrics and current components A leaky dielectric has complex relative permittivity εr* = εr′ − j εr″. When an electric field E is applied, is the conduction-like current component that is in phase with E proportional to εr″?

Introduction / Context:
In AC fields, dielectrics exhibit both displacement (reactive) and loss (resistive) currents. The complex permittivity εr* = εr′ − j εr″ compactly represents this behavior. Understanding which part controls the in-phase (dissipative) current is vital for estimating dielectric heating and specifying materials for RF and power applications.

Given Data / Assumptions:

• Time-harmonic electric field with angular frequency ω.
• Linear, isotropic material with small to moderate losses.
• Engineering sign convention: negative imaginary part indicates loss.

Concept / Approach:

The current density in a linear dielectric can be written as J = ∂D/∂t = j ω ε0 εr* E = j ω ε0 (εr′ − j εr″) E. Separating components: J = j ω ε0 εr′ E + ω ε0 εr″ E. The first term (with j) is 90° out of phase with E (purely reactive, displacement current), while the second term is in phase with E and represents power dissipation. Hence, the in-phase current component magnitude is proportional to εr″ (and to ω and ε0). Power loss density is p_loss = Re{E · J*} = ω ε0 εr″ |E|^2, also showing linear proportionality to εr″.

Step-by-Step Solution:

Verification / Alternative check:

Equivalent-circuit modeling (capacitor with a parallel resistance) yields identical results: the resistive branch current is in phase with voltage and corresponds to εr″ in the permittivity representation.

Why Other Options Are Wrong:

(b) contradicts the algebra; (c) proportionality holds at all frequencies where linear material behavior is valid; (d) “perfect insulators” have εr″ → 0, thus negligible in-phase current; (e) εr′ governs reactive displacement, not dissipative in-phase current.

Common Pitfalls:

Mixing sign conventions or confusing εr″ with the loss tangent tan δ = εr″/εr′; both relate to loss but play different roles in formulas.

Final Answer:

True