Chapter 32: Problem 14
A parallel-plate capacitor with circular plates of radius \(R\) is being charged. Show that the magnitude of the current density of the displacement current is \(J_{d}=\varepsilon_{0}(d E / d t)\) for \(r \leq R\)
Short Answer
Expert verified
The displacement current density is \( J_d = \varepsilon_0 \frac{dE}{dt} \) for \( r \leq R \).
Step by step solution
01
Understanding the Problem
We are given a parallel-plate capacitor with circular plates of radius \( R \), and we need to find the displacement current density, \( J_d \), during a charging process for a point inside the plate, i.e., where \( r \leq R \). This involves understanding how the electric field inside the capacitor changes with time.
02
Applying Gauss's Law
Consider the capacitor with a surface S through which the electric field \( E \) is uniform and perpendicular to the plates. Using Gauss's law for electricity, we find that the electric field between the plates is given by \( E = \frac{Q}{\varepsilon_0 A} \), where \( Q \) is the charge on the plates and \( A = \pi R^2 \) is the area of a plate.
03
Expressing Charge in Terms of the Electric Field
The charge \( Q \) is related to the electric field by the formula \( Q = \varepsilon_0 A E \). Thus, \( E \) can be expressed in terms of \( Q \) and vice-versa.
04
Finding the Displacement Current
The displacement current \( I_d \) is defined as \( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \), where \( \Phi_E \) is the electric flux. We know \( \Phi_E = E \times A \), so \( \frac{d\Phi_E}{dt} = A \frac{dE}{dt} \). Hence, \( I_d = \varepsilon_0 A \frac{dE}{dt} \).
05
Relating Displacement Current to Displacement Current Density
The displacement current density is the displacement current per unit area, \( J_d = \frac{I_d}{A} \). Substituting for \( I_d \) from the previous step gives us \( J_d = \varepsilon_0 \frac{dE}{dt} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parallel-Plate Capacitor
A parallel-plate capacitor is a basic component in electronic circuits that can store energy in the form of an electric field. It consists of two conducting plates separated by a small distance filled with a non-conducting medium or dielectric. When a voltage is applied across the plates, positive charge accumulates on one plate, while an equal amount of negative charge gathers on the opposite one.
This configuration creates a uniform electric field between the plates. The parallel-plate capacitor's ability to store charge makes it an essential element in filters, oscillators, and many other electronic circuits.
This configuration creates a uniform electric field between the plates. The parallel-plate capacitor's ability to store charge makes it an essential element in filters, oscillators, and many other electronic circuits.
- **Structure:** Typically comprises two metal plates parallel to each other.
- **Function:** Stores energy by maintaining a potential difference between the plates.
- **Uses:** Includes energy storage, signal coupling, and filtering noise in circuits.
Gauss's Law
Gauss's law is a fundamental principle in electromagnetism, establishing the relationship between electric charge and electric field. It asserts that the electric flux through a closed surface is proportionate to the charge enclosed by that surface. In a mathematical form, Gauss's law is expressed as \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \), where \( \vec{E} \) is the electric field, \( d\vec{A} \) is a differential area on the closed surface, and \( Q_{enc} \) is the enclosed charge.
In the context of a parallel-plate capacitor, Gauss's law helps us understand how the electric field is related to the charges on the plates and is crucial for deriving the expression for electric field \( E \) as \( E = \frac{Q}{\varepsilon_0 A} \).
In the context of a parallel-plate capacitor, Gauss's law helps us understand how the electric field is related to the charges on the plates and is crucial for deriving the expression for electric field \( E \) as \( E = \frac{Q}{\varepsilon_0 A} \).
- **Application:** Helps calculate electric fields in symmetric setups like planar, spherical, and cylindrical symmetry.
- **Importance:** Simplifies electric field calculations where symmetry plays a role.
Electric Field
An electric field is a region around a charged particle where an electric force is exerted on other charges. In a parallel-plate capacitor, this field is created between the plates when a voltage is applied. It's uniform in nature, and the direction of the electric field lines goes from the positive to the negative plate.
The magnitude of the electric field in a parallel-plate capacitor is given by \( E = \frac{Q}{\varepsilon_0 A} \), where \( Q \) is the charge on one plate, \( A \) is the area of the plate, and \( \varepsilon_0 \) is the permittivity of free space. This simple relationship shows how a change in charge or plate size affects the strength of the electric field.
The magnitude of the electric field in a parallel-plate capacitor is given by \( E = \frac{Q}{\varepsilon_0 A} \), where \( Q \) is the charge on one plate, \( A \) is the area of the plate, and \( \varepsilon_0 \) is the permittivity of free space. This simple relationship shows how a change in charge or plate size affects the strength of the electric field.
- **Uniformity:** In a parallel-plate capacitor, the electric field remains constant across the distance between the plates.
- **Direction:** Directed perpendicularly from one plate to the other.
- **Impact:** Influences how capacitors are used in applications like memory storage and power supplies.
Displacement Current Density
Displacement current density is a concept introduced by Maxwell to account for the changing electric field in situations like charging a capacitor. Although it's not an actual current of moving charges, it acts similarly by contributing to the total current in a loop.
In a charging parallel-plate capacitor, the displacement current density \( J_d \) is associated with the time variation of the electric field, expressed as \( J_d = \varepsilon_0 \frac{dE}{dt} \). This means that even in the absence of physical moving charges between the plates, a changing electric field results in a "current" that must be considered in the context of continuity across the entire circuit.
In a charging parallel-plate capacitor, the displacement current density \( J_d \) is associated with the time variation of the electric field, expressed as \( J_d = \varepsilon_0 \frac{dE}{dt} \). This means that even in the absence of physical moving charges between the plates, a changing electric field results in a "current" that must be considered in the context of continuity across the entire circuit.
- **Purpose:** Ensures consistency in Ampère's law when dealing with time-varying fields.
- **Current Analogy:** Acts like a real current even though actual charge movement isn't happening.
- **Expression:** \( J_d = \varepsilon_0 \frac{dE}{dt} \) highlights its dependence on the rate of change of the electric field.