/*! 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 5 A uniform electric field makes a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 5

A uniform electric field makes an angle of \(60.0^{\circ}\) with a flat surface. The area of the surface is \(6.66 \times 10^{-4} \mathrm{~m}^{2}\). The resulting electric flux through the surface is \(4.44 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\). Calculate the magnitude of the electric field.

Short Answer

Expert verified
The magnitude of the electric field is \(13320 \, \mathrm{N/C}\).

Step by step solution

01

Identify Given Values

In the problem, it is given that: \n\n The resultant electric flux 桅 = 4.44 N.m^2/C \n\n The area of the surface A = 6.66 x 10^-4 m^2 \n\n The angle (胃) between the electric field direction and the area normal = 60.0 degrees
02

Expressing Electric Flux

The electric flux through a surface is given by 桅 = E * A * cos(胃), where E is the magnitude of the electric field. Here, 胃 is the angle between the electric field and the normal (perpendicular) to the surface. Re-arrange the equation to solve for E and substitute the given values.
03

Calculate the Magnitude of Electric Field

Substitute the values into the rearranged equation: E = 桅 / (A * cos胃) = 4.44 N.m^2/C / (6.66 x 10^-4 m^2 * cos(60)) = 4.44 N.m^2/C / (6.66 x 10^-4 m^2 * 0.5) = 13320 N/C

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 91影视!

Key Concepts

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

Electric Field
An electric field is a region around a charged object where other charged objects experience a force. It is a vector quantity, meaning it has both magnitude and direction. The magnitude of the electric field (\(E\)) refers to the strength of this influence and is measured in newtons per coulomb (\(N/C\)). Electric fields can be uniform or non-uniform depending on the charge distribution.
  • A uniform electric field has the same strength and direction at every point.
  • Charged particles within the field experience force proportional to the field's strength and their charge.
Understanding electric fields involves looking at forces acting on charges. These concepts help us calculate how much force a charge would experience in the vicinity of electric fields. This force calculation is crucial for many applications, from determining electric potential energy to designing electric circuits.
Surface Area
Surface area plays a critical role when discussing electric flux, which is the amount of electric field passing through a specific area. The area of a surface, denoted as (\(A\)), is essential in calculating electric flux. In physics and engineering, understanding surface area helps in analyzing how fields interact with objects.
  • The surface can be flat or curved, impacting how an electric field crosses it.
  • In this scenario, the area is given as flat, making computations straightforward with basic geometry.
  • For the problem scenario, the flat surface area is given as \(6.66 \times 10^{-4} \text{ m}^2\).
When you know the surface area and other variables, you can accurately compute electric flux and associated electric field magnitudes. Calculations like these are common, especially when dealing with capacitors and electric shielding.
Angle of Incidence
The angle of incidence is the angle between the direction of the electric field and the normal (perpendicular) to the surface it interacts with. In this context, the term often gets associated with the effectiveness of electric field penetration through the surface.
  • The angle is typically denoted by \(\theta\) and plays an integral role in determining the effective component of the electric field acting through a surface.
  • At \(0^{\circ}\), the field is perpendicular to the surface, maximizing the flux, while at \(90^{\circ}\), the field is parallel, and no flux passes through.
  • In our example problem, the angle is \(60^{\circ}\), affecting the field component along the normal to become \(E \cdot \cos(60)\).
Understanding how this angle influences computations is essential, as it directly impacts the resulting electric flux through a given area. Solving problems related to the electric field often involves adjusting calculations for this angle.

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

An electron is released from rest at a distance of \(0.300 \mathrm{~m}\) from a large insulating sheet of charge that has uniform surface charge density \(+2.90 \times 10^{-12} \mathrm{C} / \mathrm{m}^{2}\). (a) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point \(0.050 \mathrm{~m}\) from the sheet? (b) What is the speed of the electron when it is \(0.050 \mathrm{~m}\) from the sheet?

A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho(r)\) given as follows: $$ \begin{array}{ll} \rho(r)=\rho_{0}\left(1-\frac{r}{R}\right) & \text { for } r \leq R \\ \rho(r)=0 & \text { for } r \geq R \end{array} $$ where \(\rho_{0}=3 Q / \pi R^{3}\) is a positive constant. (a) Show that the total charge contained in the charge distribution is \(Q .\) (b) Show that the electric field in the region \(r \geq R\) is identical to that produced by a point charge \(Q\) at \(r=0 .\) (c) Obtain an expression for the electric field in the region \(r \leq R .\) (d) Graph the electric-field magnitude \(E\) as a function of \(r\) (e) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field.

You measure an electric field of \(1.25 \times 10^{6} \mathrm{~N} / \mathrm{C}\) at a distance of \(0.150 \mathrm{~m}\) from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius \(0.150 \mathrm{~m} ?\) (b) What is the magnitude of this charge?

A very long insulating cylinder with radius \(R_{\text {cylinder }}\) has nonuniform positive charge density \(\rho=\left(1-r / R_{\text {cylinder }}\right) \rho_{0}\) where \(\rho_{0}\) is constant and \(r\) is measured radially from the axis of the cylinder. A particle with charge \(-Q\) and mass \(M\) orbits the cylinder at a constant distance \(R_{\text {orbit }}>R_{\text {cylinder }}\) (a) What is the linear charge density \(\lambda\) of the tube, in terms of \(R_{\text {cylinder }}\) and \(\rho_{0} ?\) (b) Determine the period of the motion in terms of \(R_{\text {orbit }}\) (Hint: Use Gauss's law to determine the electric field, and therefore the electric force felt by the particle, that acts centripetally.)

Electric Fields in an Atom. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{~m} .\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{~m} ?\) (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Oscillations

Read Explanation

Torque and Rotational Motion

Read Explanation

Electromagnetism

Read Explanation

Particle Model of Matter

Read Explanation

Space Physics

Read Explanation
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.