/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A parallel-plate capacitor with ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 2.0 A. At what rate is the electric field between the plates changing?

Short Answer

Expert verified
The rate of change of the electric field is approximately \( 7.18 \times 10^{11} \mathrm{~V/m/s} \).

Step by step solution

01

Understand the Displacement Current

The displacement current, which we will denote as \( I_d \), is given in the problem as 2.0 A. This current is associated with a changing electric field in the capacitor.
02

Use the Displacement Current Formula

We know that the displacement current is given by the equation \( I_d = \varepsilon_0 \cdot A \cdot \frac{dE}{dt} \), where \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \mathrm{~F/m}) \), \( A \) is the area of the capacitor plates, and \( \frac{dE}{dt} \) is the rate of change of the electric field.
03

Calculate the Area of the Capacitor Plates

The area \( A \) of the plates can be calculated using the formula for the area of a circle: \( A = \pi r^2 \). Here, \( r = 0.10 \mathrm{~m} \), so \( A = \pi (0.10)^2 \).
04

Solve for Rate of Change of Electric Field

Rearrange the displacement current formula to solve for \( \frac{dE}{dt} \):\[ \frac{dE}{dt} = \frac{I_d}{\varepsilon_0 A} \]Plug in \( I_d = 2.0 \mathrm{~A} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~F/m} \), and \( A = \pi (0.10)^2 \). Calculate \( \frac{dE}{dt} \).
05

Calculate Numerical Value

Perform the numerical computation to find:\[ \frac{dE}{dt} = \frac{2.0}{8.85 \times 10^{-12} \cdot \pi \times (0.10)^2} \]Compute the value to find \( \frac{dE}{dt} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 simple electrical component that consists of two conducting plates separated by a dielectric or insulating material. The plates are typically parallel to each other and have a uniform area, which can be circular, like in the given exercise. Such capacitors are used in various electronic devices to store energy in an electric field, providing a steady electric supply even if current flow is interrupted for short periods.
The key to a parallel-plate capacitor's functionality lies in its ability to store and release electric energy efficiently. This transfer of energy occurs as electric charge accumulates on one plate and moves through the dielectric to the other plate when a potential difference is applied.
  • The area of the plates (denoted as A) directly affects the capacitance, which is the ability of the capacitor to hold an electric charge.
  • The separation between the plates also plays a critical role. A smaller distance increases the capacitance.
By understanding these foundational ideas, the calculation of the electric field within a capacitor becomes more intuitive, as seen in the context of the given problem.
Electric Field
The concept of an electric field is essential in understanding how capacitors operate. An electric field is a region around charged particles or objects where an electric force is exerted on other charged objects. In a parallel-plate capacitor, this field is established between the plates when they are connected to a potential difference.
The electric field (E) within a parallel-plate capacitor is uniform and directed from the positive to the negative plate, leading to a consistent potential across the plates. This characteristic is because each plate produces an electric field that adds up in the space between them.
  • The strength or magnitude of the electric field is determined by the charge density on the plates.
  • The formula for the electric field within a capacitor is given by E = \( \frac{V}{d} \), where V is the potential difference, and d is the distance between the plates.
In cases where the electric field is changing, such as when a capacitor is being charged or discharged, the concept of displacement current becomes significant. This is especially relevant in our exercise where the rate of change of the electric field (\( \frac{dE}{dt} \)) is a crucial value to solve.
Permittivity of Free Space
The permittivity of free space, also known as the vacuum permittivity, is a fundamental physical constant denoted by \( \varepsilon_0 \). It quantifies the ability of a vacuum to permit electric field lines and is crucial in the calculation of many electromagnetic formulas, including the capacitance of capacitors.
In the formula for displacement current (\( I_d = \varepsilon_0 \cdot A \cdot \frac{dE}{dt} \)), \( \varepsilon_0 \) represents the ease with which an electric field can exist in vacuum conditions between the capacitor plates.
  • The standard value of \( \varepsilon_0 \) is approximately 8.85 \( \times 10^{-12} \text{ F/m} \) (farads per meter).
  • This constant is essential for determining how strongly the electric field interacts with the material between the plates.
Understanding \( \varepsilon_0 \) is important for accurately computing relationships between electric fields, displacement currents, and the resulting changes in electric field strength as seen in capacitor applications similar to our exercise context.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Abar magnet lies near a paper cylinder. (a) Sketch the magnetic field lines that pass through the surface of the cylinder. (b) What is the sign of \(\vec{B} \cdot d \vec{A}\) for every area \(d \vec{A}\) on the surface? (c) Does this contradict Gauss' law for magnetism? Explain.

The magnitude of the magnetic dipole moment of Earth is \(8.0 \times 10^{22} \mathrm{~J} / \mathrm{T}\). (a) If the origin of this magnetism were a magnetized iron sphere at the center of Earth, what would be its radius? (b) What fraction of the volume of Earth would such a sphere occupy? Assume complete alignment of the dipoles. The density of Earth's inner core is \(14 \mathrm{~g} / \mathrm{cm}^{3} .\) The magnetic dipole moment of an iron atom is \(2.1 \times 10^{-23} \mathrm{~J} / \mathrm{T}\). (Note: Earth's inner core is in fact thought to be in both liquid and solid forms and partly iron, but a permanent magnet as the source of Earth's magnetism has been ruled out by several considerations. For one, the temperature is certainly above the Curie point.)

A 0.50 T magnetic field is applied to a paramagnetic gas whose atoms have an intrinsic magnetic dipole moment of \(1.0 \times\) \(10^{-23} \mathrm{~J} / \mathrm{T}\). At what temperature will the mean kinetic energy of translation of the atoms equal the energy required to reverse such a dipole end for end in this magnetic field?

An electron is placed in a magnetic field \(\vec{B}\) that is directed along a \(z\) axis. The energy difference between parallel and antiparallel alignments of the \(z\) component of the electron's spin magnetic moment with \(\vec{B}\) is \(6.00 \times 10^{-25} \mathrm{~J}\). What is the magnitude of \(\vec{B} ?\)

Assume the average value of the vertical component of Earth's magnetic field is \(43 \mu \mathrm{T}\) (downward) for all of Arizona, which has an area of \(2.95 \times 10^{5} \mathrm{~km}^{2}\). What then are the (a) magnitude and (b) direction (inward or outward) of the net magnetic flux through the rest of Earth's surface (the entire surface excluding Arizona)?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.