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Question

A parallel plate capacitor with plate area \( A = 0.02 \, \text{m}^2 \) and separation \( d = 5 \,
\text{mm} \) is connected to a circuit with a current of \( 2 \, \text{A} \). What is the rate of change
of electric flux between the plates? (Given \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \))

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Explanation

**Production of EM waves** requires accelerated charge, oscillating LC circuit produces changing E and B, antenna radiates when charge accelerates, frequency determined by L and C, f=1/(2π√(LC)). Hertz used spark gap with inductor and capacitor, produced ~10⁸ Hz radio waves, detected with loop, confirmed transverse nature, reflection, refraction, polarization, speed c. Displacement current i_d = ε₀ (d Φ_E/dt) . For a capacitor, i_d = i . Given i = 2 A , we have (d Φ_E/dt) = (i/ε₀) = (2/8.85 × 10⁻¹²) ≈ 2.26 × 10¹¹ Vm/s . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f,

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