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Ch 31: Electromagnetic Fields and Waves
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 31: Electromagnetic Fields and WavesProblema 8a
Capitolo 31, Problema 8a

A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at the rate 1.0×106 V/m s. What is the magnetic field strength on the axis?

Guida verificata passo dopo passo
1
Step 1: Understand the problem. The problem involves a parallel-plate capacitor with a changing electric field. This changing electric field induces a magnetic field according to Maxwell's equations. Specifically, we will use the concept of displacement current and Ampere-Maxwell law to find the magnetic field strength on the axis.
Step 2: Write down the given values. The diameter of the capacitor plates is 10 cm (convert to meters: 0.1 m), the spacing between the plates is 1.0 mm (convert to meters: 0.001 m), and the rate of change of the electric field is 1.0×10^6 V/m/s.
Step 3: Calculate the displacement current density. The displacement current density \( J_d \) is given by \( J_d = \epsilon_0 \frac{dE}{dt} \), where \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)) and \( \frac{dE}{dt} \) is the rate of change of the electric field.
Step 4: Use Ampere-Maxwell law to find the magnetic field strength. The magnetic field strength \( B \) on the axis of the capacitor is given by \( B = \frac{\mu_0 J_d A}{2 \pi r} \), where \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \text{T·m/A} \)), \( A \) is the area of the capacitor plates (\( A = \pi r^2 \)), and \( r \) is the radial distance from the axis (here, \( r = 0 \) since we are calculating on the axis).
Step 5: Substitute the values into the formula. Calculate \( J_d \) using \( \epsilon_0 \frac{dE}{dt} \), then substitute \( J_d \), \( \mu_0 \), and \( A \) into the formula for \( B \). Simplify the expression to find the magnetic field strength on the axis.

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Capacitance

Capacitance is the ability of a system to store electric charge per unit voltage. For a parallel-plate capacitor, it is determined by the area of the plates and the distance between them, given by the formula C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the area of one plate, and d is the separation between the plates.
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Capacitors & Capacitance (Intro)

Electric Field

The electric field (E) is a vector field that represents the force per unit charge experienced by a positive test charge placed in the field. In a parallel-plate capacitor, the electric field is uniform and can be calculated using E = V/d, where V is the voltage across the plates and d is the distance between them. The problem states that this electric field is increasing over time.
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Intro to Electric Fields

Magnetic Field Induction

According to Maxwell's equations, a changing electric field induces a magnetic field. This phenomenon is described by Faraday's law of electromagnetic induction and the displacement current concept. In this scenario, the increasing electric field between the capacitor plates generates a magnetic field, which can be calculated using the relationship B = (μ₀/2)(dE/dt) for the magnetic field strength at the axis of the capacitor.
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Magnetic Fields and Magnetic Dipoles
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