AC dielectrics and ideal capacitor behavior: In a perfect capacitor, current density leads the electric field by 90°. Given J = ω * epsilon_0 * epsilon_r' * E0 * cos(ωt + 90°), and that dielectric losses are zero, evaluate the assertion–reason pair.

Introduction / Context:
In linear dielectrics driven by a sinusoidal field, displacement current density J_d is related to the time derivative of the electric displacement D. For an ideal (lossless) dielectric, the complex permittivity is purely real; hence current and field are in perfect quadrature (90° phase difference).

Given Data / Assumptions:

• Perfect capacitor: no conduction loss and no dielectric loss (loss tangent = 0).
• epsilon_r' denotes the real (in-phase permittivity) part used in the expression.
• E(t) = E0 cos(ωt) is the applied field.

Concept / Approach:

For a perfect capacitor, D = epsilon_0 * epsilon_r' * E, and J_d = dD/dt = omega * epsilon_0 * epsilon_r' * E0 * sin(ωt). Since sin(ωt) = cos(ωt + 90°), J leads E by 90°. Zero dielectric losses imply no in-phase component of current with voltage; all current is reactive, which explains the 90° lead and the stated form of J.

Step-by-Step Solution:

Verification / Alternative check:

Introducing dielectric loss adds an imaginary part to permittivity; the current then has a component in phase with E, reducing the phase lead below 90°. In the perfect case (loss = 0), the 90° lead is exact.

Why Other Options Are Wrong:

• If losses were not zero, the exact 90° relation would not hold.

Common Pitfalls:

• Confusing conduction current (σE) with displacement current (dD/dt).

Final Answer:

Both A and R are true and R is correct explanation of A