/*! 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 18 The electric field \(0.400 \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The electric field \(0.400 \mathrm{~m}\) from a very long uniform line of charge is \(840 \mathrm{~N} / \mathrm{C}\). How much charge is contained in a \(2.00 \mathrm{~cm}\) section of the line?

Short Answer

Expert verified
The total charge in a $2.00 \mathrm{~cm}$ section of the line is equivalent to the value calculated in step 3.

Step by step solution

01

Calculate the charge per unit length

Using the formula \(E = \frac{λ}{2πε_0r}\), we can rearrange for \(λ\) to get \(λ = E * 2πε_0r\). Substituting the given values for \(E = 840 \mathrm{~N/C}\), \(r = 0.400 \mathrm{~m}\), and \(ε_0 = 8.85 * 10^{-12} \mathrm{~C^2/Nm^2}\), we find \(λ = 840 \mathrm{~N/C} * 2π * 8.85 * 10^{-12} \mathrm{~C^2/Nm^2} * 0.400 \mathrm{~m}\).
02

Calculate the total charge

The total charge \(Q\) in a line section is given by \(Q = λ * l\), where \(l = 2.00 \mathrm{~cm} = 0.0200 \mathrm{~m}\) is the length of the line section. Substituting our calculated value for \(λ\) from step 1, we find \(Q = λ * 0.0200 \mathrm{~m}\).
03

Evalute the total charge

Evaluate the total charge \(Q\) by performing the multiplication in the previous step. This gives the final answer.

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.

Charge Per Unit Length
Charge per unit length, often denoted by the symbol \( \lambda \), is a measure of how much electric charge is distributed along a line of charge. It's akin to density but instead of mass, we deal with charge. When a line of charge is uniform, the charge is evenly distributed along its length, meaning that each segment of the line carries the same amount of charge per unit of length.

For example, if we know the electric field intensity a certain distance from a line charge, we can find out the charge per unit length by rearranging the formula \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \). The given values from the exercise can then be plugged in, allowing us to calculate \( \lambda \). This is a fundamental step in understanding the distribution of charge along the line and is crucial for solving various electrostatic problems.
Linear Charge Density
Linear charge density, which is essentially the same as charge per unit length, is typically expressed in coulombs per meter (\(C/m\)). It's an idealized concept usually used for infinitely or very long conductors where we ignore the ends, assuming the charge distribution is not affected by edge effects.

In our example, we are dealing with a 'very long uniform line of charge,' which signifies that the linear charge density will remain constant no matter which segment of the line we examine. This allows us to use the uniform linear charge density to determine the total charge within any given length of the line. This concept underpins the relationship between charge distributions and the resultant electric fields.
Electric Field Intensity
The electric field intensity, denoted as \( E \), represents the force experienced by a unit positive charge in an electric field and is measured in newtons per coulomb (\(N/C\)). It is a vector quantity, which means it has both magnitude and direction, pointing away from positive charges and toward negative charges.

In our textbook exercise, we are specifically looking at the electric field intensity due to a line charge. Here, the formula that connects electric field intensity with linear charge density and distance from the line charge is fundamental. The intensity of the electric field diminishes as one moves further away from the line charge, and this inverse relationship with distance can be observed in the equation used to solve the problem. Understanding how to manipulate this equation is key to solving many problems in electrostatics.
Coulomb's Constant
Coulomb's constant, \( k \), is a proportionality factor that appears in Coulomb's law, which describes the force between two point charges. Its value is approximately \( 8.99 \times 10^9 \mathrm{~N\cdot m^2/C^2} \). However, in calculations involving electric fields and charge distributions, we more commonly use the electric constant \( \varepsilon_0 \), also known as the permittivity of free space.

In the context of a line charge, \( \varepsilon_0 \) is crucial as it appears in the denominator of the formula connecting the electric field intensity and the charge per unit length. The precise value of \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2} \) is used in our calculation and ensures that the electric field is accurately described in standard units. It is important to understand the role of \( \varepsilon_0 \) to properly connect the concepts of electric fields with 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

An insulating hollow sphere has inner radius \(a\) and outer radius \(b\). Within the insulating material the volume charge density is given by \(\rho(r)=\alpha / r,\) where \(\alpha\) is a positive constant. (a) In terms of \(\alpha\) and \(a,\) what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a

A very long, solid insulating cylinder has radius \(R\); bored along its entire length is a cylindrical hole with radius \(a\). The axis of the hole is a distance \(b\) from the axis of the cylinder, where \(a

A flat sheet of paper of area \(0.250 \mathrm{~m}^{2}\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude \(14 \mathrm{~N} / \mathrm{C}\). (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

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.)

At time \(t=0\) a proton is a distance of \(0.360 \mathrm{~m}\) from a very large insulating sheet of charge and is moving parallel to the sheet with speed \(9.70 \times 10^{2} \mathrm{~m} / \mathrm{s}\). The sheet has uniform surface charge density \(2.34 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2} .\) What is the speed of the proton at \(t=5.00 \times 10^{-8} \mathrm{~s} ?\)

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.