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Question

A parallel plate capacitor with plate area \( A = 0.01 \, \text{m}^2 \) and separation \( d = 2 \,
\text{mm} \) is being charged at a rate of \( \frac{dQ}{dt} = 0.5 \, \text{A} \). What is the
displacement current between the plates? (Given \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}
\))

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Explanation

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Displacement current i_d = ε₀ (d Φ_E/dt) . Since Φ_E = (Q/ε₀) , we have (d Φ_E/dt) = (1/ε₀) (dQ/dt) . But in a capacitor, i_d = (dQ/dt) . Thus, i_d = 0.5 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 0.5 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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