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A material has conductivity sigma = 5.0 S/m and relative permittivity epsilon_r = 1. An electric field E = 250*sin(10¹⁰ * t) V/m exists in it. Identify the correct expressions for conduction current density Jc, displacement current density JD, and the frequency at which they are equal.

1. Jc = 1250 sin(10¹⁰ t) A/m²; JD = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz
2. Jc = 22.1 cos(10¹⁰ t) A/m²; JD = 1250 sin(10¹⁰ t) A/m²; f = 90 GHz
3. Jc = 1250 sin(10¹⁰ t) A/m²; JD = 22.1 sin(10¹⁰ t) A/m²; f = 50 GHz
4. Jc = 1250 cos(10¹⁰ t) A/m²; JD = 22.1 cos(10¹⁰ t) A/m²; f = 50 GHz

Correct answer: Jc = 1250 sin(10¹⁰ t) A/m²; JD = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz

Solution

Jc = sigma*E = 5*250*sin(10¹⁰ t) = 1250*sin(10¹⁰ t). JD = epsilon₀*(dE/dt) = 8.85e-12 * 250*10¹⁰ * cos(10¹⁰ t) = 22.1*cos(10¹⁰ t). Amplitudes are equal when sigma = omega*epsilon, giving omega = sigma/epsilon₀ = 5/(8.85e-12) ≈ 5.65e11 rad/s => f ≈ 90 GHz.

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