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A parallel-plate capacitor with circular plates of radius \(0.10 \mathrm{~m}\) is being discharged. A circular loop of radius \(0.20 \mathrm{~m}\) is concentric with the capacitor and halfway between the plates. The displacement current through the loop is \(3.0 \mathrm{~A}\). At what rate is the electric field between the plates changing?

Short Answer

Expert verified
The electric field changes at a rate of approximately \(3.42 \times 10^{11}\) V/m/s.

Step by step solution

01

Understand the Concept

In a charging or discharging capacitor, a changing electric field produces a displacement current. The relevant equation for the displacement current is based on Ampère-Maxwell law, which relates the displacement current to the change of the electric field over time.
02

Establish the Relevant Equation

The displacement current density \( J_d \) is given by the equation \( J_d = \varepsilon_0 \cdot \frac{dE}{dt} \), where \( \varepsilon_0 \) is the permittivity of free space and \( \frac{dE}{dt} \) is the rate of change of the electric field. The total displacement current \( I_d \) is the product of the displacement current density and the area \( A \) of the capacitor plates, i.e., \( I_d = J_d \times A \).
03

Calculate the Area of the Capacitor Plate

The area \( A \) of circular plates with radius \( r = 0.10 \text{ m} \) is given by \( A = \pi r^2 = \pi (0.10)^2 = 0.0314 \text{ m}^2 \).
04

Relate Displacement Current to Electric Field Rate of Change

Since the displacement current \( I_d = \varepsilon_0 \cdot \frac{dE}{dt} \cdot A \), we can solve for \( \frac{dE}{dt} \) as follows: \( \frac{dE}{dt} = \frac{I_d}{\varepsilon_0 \cdot A} \).
05

Calculate the Rate of Change of the Electric Field

Substitute the known values: \( I_d = 3.0 \text{ A}, \varepsilon_0 = 8.854 \times 10^{-12} \text{ F/m}, \) and \( A = 0.0314 \text{ m}^2 \) into the expression: \[ \frac{dE}{dt} = \frac{3.0}{8.854 \times 10^{-12} \times 0.0314} \approx 3.42 \times 10^{11} \text{ V/m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field
The electric field is a crucial concept in physics, especially when dealing with capacitors. It is essentially a force field that surrounds electric charges. This field exerts a force on any other charges present in the field's vicinity. Understanding this helps in comprehending how capacitors work. In a capacitor, the electric field is generated between two plates when they are charged.
For a parallel-plate capacitor, the electric field is uniform between the plates and can be calculated using the relation:
  • The electric field strength, denoted as \( E \).
  • The relation \( E = \frac{V}{d} \), where \( V \) is the potential difference between the plates and \( d \) is the separation distance between the plates.
  • When the capacitor discharges, this electric field decreases at a rate which affects how the displacement current behaves.
If this field is changing, it can induce a displacement current even in the absence of a conduction current, which is a very intriguing aspect at the intersection of classical and modern electromagnetic theory.
Parallel-Plate Capacitor
A parallel-plate capacitor is a simple electrical device that's essential for storing energy in the field between two parallel plates. The plates themselves are conductive and have equal but opposite charges. Between the plates, an electric field forms and stores potential energy.
The configuration for this type of capacitor is straightforward:
  • The plates have a specific shape, usually circular or rectangular. The exercise above considers circular plates with radius \(0.10 ext{ m}\).
  • The space between them can include a vacuum or an insulating material often referred to as a dielectric.
  • The plate's area affects the capacitance, or the ability of the capacitor to store charge, which is given by \(C = \varepsilon_0 \frac{A}{d}\), where \(A\) is the area and \(d\) is the separation.
In practical applications, they are key components in many electronic circuits, affecting how these circuits operate by influencing capacitance and charge storage.
Ampère-Maxwell Law
The Ampère-Maxwell law is a modified version of Ampère's Law that includes the concept of displacement current. This rendition of the law is part of Maxwell’s equations, fundamental laws which describe electromagnetism.
In its essence, the Ampère-Maxwell law states:
  • The magnetic field, around a closed loop, is related to both the electric current through the loop and the change in electric flux through the loop.
  • The introduction of displacement current, \(I_d\), is crucial when analyzing currents in circuits with capacitors.
  • Mathematically, it can be represented as: \[oint \ **B **** \cdot \**dgsi**\ = \ \mu_0 \left( I + \varepsilon_0 \ rac{dee}{d t} \right)\] where \(I_d = \varepsilon_0 \frac{dE}{dt} \cdot A\).
This law underscores the concept that a changing electric field can create a magnetic field, even if no actual current is flowing through a wire. Thus, it deftly ties together electricity and magnetism into a single coherent framework.

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Most popular questions from this chapter

The induced magnetic field at radial distance \(6.0 \mathrm{~mm}\) from the central axis of a circular parallel-plate capacitor is \(1.2 \times 10^{-7} \mathrm{~T}\). The plates have radius \(4.0 \mathrm{~mm}\). At what rate \(d \vec{E} / d t\) is the electric field between the plates changing?

Measurements in mines and boreholes indicate that Earth's interior temperature increases with depth at the average rate of \(30 \mathrm{C}^{\circ} / \mathrm{km}\). Assuming a surface temperature of \(10^{\circ} \mathrm{C}\), at what depth does iron cease to be ferromagnetic? (The Curie temperature of iron varies very little with pressure.)

You place a magnetic compass on a horizontal surface, allow the needle to settle, and then give the compass a gentle wiggle to cause the needle to oscillate about its equilibrium position. The oscillation frequency is \(0.312 \mathrm{~Hz}\). Earth's magnetic field at the location of the compass has a horizontal component of \(18.0 \mu \mathrm{T}\). The needle has a magnetic moment of \(0.760 \mathrm{~mJ} / \mathrm{T}\). What is the needle's rotational inertia about its (vertical) axis of rotation?

A magnet in the form of a cylindrical rod has a length of 5.00 cm and a diameter of 6.00 mm. It has a uniform magnetization of 5.30 103 A/m.What is its magnetic dipole moment?

The exchange coupling mentioned in Module \(32-8\) as being responsible for ferro- magnetism is not the mutual magnetic intermagnetism is not the mutual magnetic interaction between two elementary magnetic dipoles. To show this, calculate (a) the magnitude of the magnetic field a distance of \(10 \mathrm{~nm}\) away, along the dipole axis, from an atom with magnetic dipole moment \(1.5 \times 10^{-25} \mathrm{~J} / \mathrm{T}\) (cobalt), and (b) the minimum energy required to turn a second identical dipole end for end in this field. (c) By comparing the latter with the mean translational kinetic energy of \(0.040 \mathrm{eV}\), what can you conclude?

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