Between circular capacitor plates of radius R = 0.10 m, the electric field varies as E(r,t) = αt[1 − r²/R²], where α = 2×10⁸ V·m⁻¹·s⁻¹. Take ε₀ = 9×10⁻¹² F/m. If the total displacement current is Nπ μA, find N.
Physics · Electromagnetic wave · displacement current and properties of electromagnetic wave
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Question

Between circular capacitor plates of radius R = 0.10 m, the electric field varies as E(r,t) = αt[1 − r²/R²], where α = 2×10⁸ V·m⁻¹·s⁻¹. Take ε₀ = 9×10⁻¹² F/m. If the total displacement current is Nπ μA, find N.

Physics · Electromagnetic wave · displacement current and properties of electromagnetic wave

G1 Physics · Accepted Answer

Because E varies with radius, the electric flux must be integrated over the plate area. Displacement current: I = ε₀ dΦₑ/dt = ε₀ ∫(∂E/∂t)dA. Here, ∂E/∂t = α(1 − r²/R²) and for a thin circular ring, dA = 2πr dr. Therefore, I = 2πε₀α ∫₀ᴿ [r − r³/R²]dr. Now, ∫₀ᴿ r dr = R²/2 and ∫₀ᴿ (r³/R²)dr = R²/4. Hence, I = 2πε₀α(R²/2 − R²/4) = πε₀αR²/2. Substitute: I = π(9×10⁻¹²)(2×10⁸)(0.10²)/2 = 9π×10⁻⁶ A = 9π μA. Therefore N = 9.