/*! 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 15 Two very long uniform lines of c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two very long uniform lines of charge are parallel and are separated by 0.300 m. Each line of charge has charge per unit length \(+5.20 \mu \mathrm{C} / \mathrm{m} .\) What magnitude of force does one line of charge exert on a \(0.0500-\mathrm{m}\) section of the other line of charge?

Short Answer

Expert verified
The force magnitude is approximately \(2.42 \mathrm{N}\).

Step by step solution

01

Understand the Problem

We have two parallel lines of charge, each with a linear charge density of \(+5.20 \mu \mathrm{C/m}\) separated by a distance of \(0.300 \mathrm{m}\). We need to find the force exerted by one line on a \(0.0500 \mathrm{m}\) section of the other line.
02

Use Gauss's Law for Line of Charge

For a long line of charge with linear charge density \(\lambda\), the electric field at a distance \(r\) from the line is given by \(E = \frac{\lambda}{2 \pi \varepsilon_0 r}\). Here, \(\lambda = 5.20 \times 10^{-6} \mathrm{C/m}\) and \(r = 0.300 \mathrm{m}\).
03

Calculate the Electric Field

Substitute the given values into the electric field equation:\[E = \frac{5.20 \times 10^{-6}}{2 \pi (8.85 \times 10^{-12}) \times 0.300}\]Calculate \(E\) to find the electric field magnitude due to one line at the position of the other.
04

Electric Field Result

After calculating, we find that the electric field \(E\) is approximately \(9.31 \times 10^{4} \mathrm{N/C}\).
05

Find the Force on a Section of the Line

The force exerted by the electric field on a line segment of length \(L = 0.0500 \mathrm{m}\) is given by \(F = \lambda L E\). Substitute \(\lambda = 5.20 \times 10^{-6} \mathrm{C/m}\), \(L = 0.0500 \mathrm{m}\), and the calculated \(E\) value.
06

Calculate the Force

Substitute the values to find the force:\[F = (5.20 \times 10^{-6}) \times 0.0500 \times 9.31 \times 10^{4}\]Calculate \(F\) to get the force magnitude exerted on the section of the line.
07

Force Calculation Result

The calculated force \(F\) is approximately \(2.42 \mathrm{N}\).

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.

Gauss's Law
Gauss's Law is a fundamental principle that relates the electric field to the charge distribution within a closed surface. It states that the net electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. This law is especially useful for symmetrical charge distributions, like those in spheres, cylinders, and planes.
For a long line of charge, we consider a cylindrical surface with the line of charge at its center. This approach simplifies the computation of the electric field, as the symmetry ensures that the electric field is constant across the surface. This symmetry allows us to use Gauss's Law to derive a formula for the electric field around a line of charge: \[E = \frac{\lambda}{2 \pi \varepsilon_0 r}\]where \(\lambda\) is the linear charge density, \(\varepsilon_0\) is the permittivity of free space, and \(r\) is the radial distance from the line.
Linear Charge Density
Linear charge density refers to the amount of electric charge per unit length along a line of charge. It is denoted by the symbol \(\lambda\) and is usually expressed in coulombs per meter \(\mathrm{C/m}\).
In the context of the exercise, each line of charge has a linear charge density of \(+5.20 \mu \mathrm{C/m}\), which means there are 5.20 microcoulombs of charge per meter of the line. Understanding linear charge density is key to solving problems involving electric fields generated by long lines of charge.
  • Line charges with higher linear charge densities produce stronger electric fields.
  • The concept helps in determining the electric field at various points around the charge.
Substituting the linear charge density into the electric field formula allows for calculating the field's magnitude due to a specific charge distribution.
Force Between Line Charges
The force between two line charges can be calculated using principles derived from Gauss's Law and the concept of electric fields. Each line of charge creates an electric field that interacts with the other line. The electric field at a distance \(r\) from one line charge influences a segment of the other line, exerting a force.
The force on a segment of the charge is given by the product of the linear charge density, the length of the segment, and the electric field: \[F = \lambda L E\]Here, \(\lambda\) is the linear charge density, \(L\) is the length of the line segment, and \(E\) is the electric field produced by the other charge. This interaction force is essential to understanding how electric fields produce mechanical effects between line charges.
Coulomb's Law
Coulomb's Law describes the force of interaction between two point charges. Though typically applied to point charges, its principles provide the foundation for understanding interactions between continuous charge distributions, such as line charges.
Coulomb's Law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: \[F = k \frac{|q_1 q_2|}{r^2}\]where \(F\) is the magnitude of the force, \(k\) is Coulomb’s constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges.
  • This law helps in understanding the force calculation between line charges by treating small sections of the lines as point charges.
  • By integrating these small forces along the line's length, a comprehensive picture of the entire force exerted by one line on the other can be derived.
Using Coulomb's Law as a basis, we can derive more complex expressions to solve problems involving continuous charge distributions.

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

How many excess electrons must be added to an isolated spherical conductor \(32.0 \mathrm{~cm}\) in diameter to produce an electric field of \(1150 \mathrm{~N} / \mathrm{C}\) just outside the surface?

The Coaxial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c .\) The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \lambda. Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r=0\) to \(r=2 c\) . (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

A region in space contains a total positive charge \(Q\) that is distributed spherically such the volume charge density \(\rho(r)\) is given by $$\begin{array}{ll}{\rho(r)=\alpha} & {\text { for } r \leq R / 2} \\\ {\rho(r)=2 \alpha(1-r / R)} & {\text { for } R / 2 \leq r \leq R} \\\ {\rho(r)=0} & {\text { for } r \geq R}\end{array}$$ Here \(\alpha\) is a positive constant having units of \(\mathrm{C} / \mathrm{m}^{3} .\) (a) Determine \(\alpha\) in terms of \(Q\) and \(R .\) (b) Using Gauss's law, derive an expression for the magnitude of \(\vec{\boldsymbol{E}}\) as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of the total charge \(Q .\) Be sure to check that your results agree on the boundaries of the regions. (c) What fraction of the total charge is contained within the region \(r \leq R / 2 ?(\mathrm{d})\) If an electron with charge \(q^{\prime}=-e\) is oscillating back and forth about \(r=0\) (the center of the distribution) with an amplitude less than \(R / 2,\) show that the motion is simple harmonic. Hint: Review the discussion of simple harmonic motion in Section \(14.2 .\) If, and only if, the net force on the electron is proportional to its displacement from equilibrium, the motion is simple harmonic.) (e) What is the period of the motion in part (d)? If If the amplitude of the motion described in part (e) is greater than \(R / 2,\) is the motion still simple harmonic? Why or why not?

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 very long conducting tube (hollow cylinder) has inner radius \(a\) and outer radius \(b\) . It carries charge per unit length \(+\alpha\) where \(\alpha\) is a positive constant with units of \(\mathrm{C} / \mathrm{m} .\) A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+\alpha .\) (a) Calculate the electric field in terms of \(\alpha\) and the distance \(r\) from the axis of the tube for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(r > b .\) Show your results in a graph of \(E\) as a function of \(r .\) (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

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.