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Ch. 31 - Maxwell's Equations and Electromagnetic Waves
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 31 - Maxwell's Equations and Electromagnetic WavesProblema 7b
Capitolo 30, Problema 7b

Suppose that a circular parallel-plate capacitor has radius r₀ = 3.0 cm and plate separation d = 5.0 mm. A sinusoidal potential difference V = V₀ sin (2𝝅ft) is applied across the plates, where V₀ = 180 V and f = 60 Hz. Determine the expression for the amplitude B₀(r) of this time-dependent (sinusoidal) field when r ≤ r₀ and when r > r₀.

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1
Understand the problem: We are tasked with finding the amplitude of the magnetic field B₀(r) generated by a sinusoidal potential difference applied to a circular parallel-plate capacitor. The problem involves two regions: inside the capacitor (r ≤ r₀) and outside the capacitor (r > r₀). The magnetic field is related to the displacement current, which arises due to the time-varying electric field between the plates.
Step 1: Write the expression for the electric field between the plates. The electric field E(t) between the plates is given by the potential difference divided by the plate separation: E(t) = V(t) / d. Substituting V(t) = V₀ sin(2πft), we get E(t) = (V₀ / d) sin(2πft).
Step 2: Calculate the displacement current density. The displacement current density J_d is related to the time derivative of the electric field: J_d = ε₀ (∂E/∂t). Substituting E(t) = (V₀ / d) sin(2πft), we find J_d = ε₀ (V₀ / d) (2πf) cos(2πft).
Step 3: Use Ampère's law with Maxwell's correction to find the magnetic field. Inside the capacitor (r ≤ r₀), the magnetic field B(r) is related to the displacement current enclosed within a radius r. The enclosed displacement current I_d is given by I_d = J_d × πr². Using Ampère's law, B(r) = (μ₀ I_d) / (2πr). Substituting I_d, we get B(r) = (μ₀ ε₀ V₀ 2πf r) / (2πd) cos(2πft).
Step 4: Determine the magnetic field outside the capacitor (r > r₀). In this region, the enclosed displacement current is constant and equal to the total displacement current through the entire capacitor. The total displacement current I_d,total is given by I_d,total = J_d × πr₀². Using Ampère's law, B(r) = (μ₀ I_d,total) / (2πr). Substituting I_d,total, we get B(r) = (μ₀ ε₀ V₀ 2πf r₀²) / (2πd r) cos(2πft).

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Capacitance of a Parallel-Plate Capacitor

The capacitance of a parallel-plate capacitor is defined as the ability to store electric charge per unit voltage. It is given by the formula C = ε₀(A/d), where A is the area of the plates, d is the separation between them, and ε₀ is the permittivity of free space. This concept is crucial for understanding how the capacitor behaves under an applied voltage.
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Parallel Plate Capacitors

Electric Field in a Capacitor

The electric field (E) between the plates of a capacitor is uniform and can be expressed as E = V/d, where V is the potential difference across the plates and d is the separation. In the case of a sinusoidal voltage, the electric field will also vary sinusoidally with time, which is essential for determining the amplitude of the electric field at different distances from the center of the plates.
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Intro to Capacitors

Radial Dependence of Electric Field

In a circular parallel-plate capacitor, the electric field's amplitude can vary with the radial distance (r) from the center of the plates. For r ≤ r₀, the field is uniform, while for r > r₀, the field decreases with distance. Understanding this radial dependence is key to deriving the expression for the amplitude of the electric field in the given scenario.
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Intro to Electric Fields
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In an EM wave traveling west, the B field oscillates up and down vertically and has a frequency of 85.0 kHz and an rms strength of 7.75 x 10⁻⁹ T. Determine the frequency and rms strength of the electric field. What is the direction of the electric field oscillations?

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Suppose that a circular parallel-plate capacitor has radius r₀ = 3.0 cm and plate separation d = 5.0 mm. A sinusoidal potential difference V = V₀ sin (2𝝅ft) is applied across the plates, where V₀ = 180 V and f = 60 Hz. In the region between the plates, show that the magnitude of the induced magnetic field is given by B = B₀(r) cos (2𝝅ft), where B₀(r) is a function of the radial distance r from the capacitor’s central axis.

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Suppose an air-gap capacitor has circular plates of radius r = 2.5 cm and separation d = 1.6 mm. A 68.0-Hz emf, ε = ε₀ cos ωt, is applied to the capacitor. The maximum displacement current is 35 μA. Determine the maximum value of dΦE/dt between the plates. Neglect fringing.

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(III) (a) When a circular parallel-plate capacitor is being charged as in Example 31–1, show that the Poynting vector S→\(\overrightarrow{S}\) points radially inward toward the center of the capacitor, parallel to the plates.

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(a) When a circular parallel-plate capacitor is being charged as in Example 31–1, show that the Poynting vector S→\(\overrightarrow{S}\) points radially inward toward the center of the capacitor, parallel to the plates.

(b) Integrate S→\(\overrightarrow{S}\) over the cylindrical boundary of the capacitor gap to show that the rate at which energy enters the capacitor is equal to the rate at which electrostatic energy is being stored in the electric field of the capacitor (Section 24–4). Ignore fringing of E→\(\overrightarrow{E}\).

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Suppose an air-gap capacitor has circular plates of radius r = 2.5 cm and separation d = 1.6 mm. A 68.0-Hz emf, ε = ε₀ cos ωt, is applied to the capacitor. The maximum displacement current is 35 μA. Determine the value of ε₀. Neglect fringing.

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