In some experiments, very tiny distances or spaces ( ≈ nm ) can be measured by using capacitance. Consider forming an LC circuit using a parallel-plate capacitor with plate area A, and a known inductance L. If f is on the order of 1 MHz and can be measured to a precision of ∆f = 1 Hz, with what percent accuracy can x be determined? Assume fringing effects at the capacitor’s edges can be neglected.
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Inductors in AC Circuits
Problem 7b
Textbook Question
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.

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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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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