/*! 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 64 Field between two wires * Cons... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Field between two wires * Consider a high-voltage direct current power line that consists of two parallel conductors suspended 3 meters apart. The lines are oppositely charged. If the electric field strength halfway between them is \(15,000 \mathrm{~N} / \mathrm{C}\), how much excess positive charge resides on a \(1 \mathrm{~km}\) length of the positive conductor?

Short Answer

Expert verified
The amount of excess positive charge on a 1km length of the positive conductor is \(1.996*10^{-4} C\).

Step by step solution

01

Understand and Write down Given Information

Write down the given information, the distance between the two conductors is \(3m\), the electric field strength is \(15000 N/C\). and the length along which we need to find the charge is \(1km = 1000m\). The target of the problem is to find the excess positive charge on the \(1km\) length of positive conductor.
02

Use the Formula for Electric Field

The electric field \(E\) between two infinitely long parallel conductors with opposite charge per unit length \(\lambda\) at a distance \(r\) from one line to another is given by \(E = \frac{2 \lambda}{4 \pi \epsilon_0 r}\), where \(\epsilon_0\) is the electric constant \(\approx 8.85 × 10^{-12} C^2/N.m^2\).
03

Rearrange the formula

We rearrange the formula to solve for the charge per unit length \(\lambda\). \[\lambda = \frac{E 4 \pi \epsilon_0 r} {2}\].
04

Calculate the Charge per unit length

Substitute all values into the formula \(\lambda = \frac{15000 N/C * 4 \pi * 8.85 × 10^{-12} C^2/N.m^2 * 1.5m}{2}\) to find \(\lambda = 1.996*10^{-7} C/m\).
05

Calculate the Charge over 1 km

Multiply \(\lambda\) by the length of the wire to find the total charge on the positive conductor. \[\Lambda = 1.996*10^{-7} C/m * 1000m = 1.996*10^{-4} C\].
06

Write the conclusion

So, the amount of excess positive charge on a \(1km\) length of the positive conductor is about \(1.996*10^{-4} C\).

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 91Ó°ÊÓ!

Key Concepts

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

Electric Field Strength
Understanding electric field strength is fundamental when studying the forces that charged objects exert on each other. Electric field strength, often symbolized as \(E\), can be thought of as the force per unit charge experienced by a positive test charge positioned within the field. In the given exercise, the concept of electric field strength helps us determine how much excess positive charge resides on a conductor.

Specifically, when dealing with parallel conductors, the electric field strength halfway between them can be interpreted as the sum of the field contributions from both wires, considering their charges are opposite. In our example, the electric field strength is given as \(15,000 N/C\), which indicates a high force experienced per coulomb of charge. This high value suggests a significant amount of charge on the conductors, directly influencing the electrical and magnetic fields surrounding them.
Charge per Unit Length
The charge per unit length, denoted as \(\lambda\), indicates how much electrical charge is distributed along a conductor and is critical in computing the electric field generated by the conductor. In our power line scenario, finding the charge per unit length facilitates the assessment of subsequent phenomena, such as the electric field and the potential energy stored between the conductors.

To obtain the charge per unit length, we manipulate the electric field formula specifically tailored for two parallel conductors. Once we have \(\lambda\), we can easily calculate the total charge on a section of the conductor by multiplying the charge density by the length of the conductor section considered. For the 1 km length of the power line, the step-by-step solution detailed how to apply the formula and derive the charge per unit length, which ultimately led to finding the amount of charge distributed along the positive conductor.
Electric Constant
The electric constant, represented as \(\epsilon_0\) and also known as the permittivity of free space, is a fundamental quantity that appears in the equations of electromagnetism. Its value, approximately \(8.85 \times 10^{-12} C^2/N.m^2\), characterizes the amount of resistance encountered when forming an electric field in a vacuum. The electric constant also relates the units of electric charge to mechanical quantities such as force and distance.

When calculating the electric field strength and charge per unit length as done in this exercise, the electric constant plays a pivotal role as it modifies how electric charges interact across a vacuum or air. By incorporating \(\epsilon_0\) in our calculations, we ensure that the derived values reflect the real-world scenario where the conductors operate in the presence of air or another medium.
Direct Current Power Line
Direct current (DC) power lines are a critical component of our electrical infrastructure, utilized to transmit electricity over long distances with minimal loss of power. Unlike alternating current (AC), direct current maintains a constant flow of electricity, making it especially suitable for high-voltage power transmission. In the example problem, the DC power line consists of two parallel conductors that carry opposite charges, creating a strong electric field between them.

This setup is not only relevant in terms of understanding electrical transmission but also serves as an excellent practical illustration of the principles of electric fields. As discussed in this exercise, the properties of the electric field such as the field strength, charge per unit length, and the role of the electric constant all contribute to how effectively a DC power line can operate and are important considerations in the design and analysis of such systems.

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

Fields at the surfaces Consider the electric field at a point on the surface of (a) a sphere with radius \(R\), (b) a cylinder with radius \(R\) whose length is infinite, and (c) a slab with thickness \(2 R\) whose other two dimensions are infinite. All of the objects have the same volume charge density \(\rho\). Compare the fields in the three cases, and explain physically why the sizes take the order they do.

Field at the end of a cylinder (a) Consider a half-infinite hollow cylindrical shell (that is, one that extends to infinity in one direction) with radius \(R\) and uniform surface charge density \(\sigma .\) What is the electric field at the midpoint of the end face? (b) Use your result to determine the field at the midpoint of a half-infinite solid cylinder with radius \(R\) and uniform volume charge density \(\rho\), which can be considered to be built up from many cylindrical shells.

Building a sheet from rods ** An infinite uniform sheet of charge can be thought of as consisting of an infinite number of adjacent uniformly charged rods. Using the fact that the electric field from an infinite rod is \(\lambda / 2 \pi \epsilon_{0} r\), integrate over these rods to show that the field from an infinite sheet with charge density \(\sigma\) is \(\sigma / 2 \epsilon_{0}\)

Field from a spherical shell, right and wrong ** The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius \(R\) and surface charge density \(\sigma\), is given by Eq. (1.42) as \(\sigma / \epsilon_{0}\). Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of \(\sigma / 2 \epsilon_{0}\) (b) Why isn't the result correct? Explain how to modify it to obtain the correct result of \(\sigma / \epsilon_{0} .\) Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what's going on for points on the shell that are very close to the point in question?

Zero field \(?\) Four charges, \(q,-q, q\), and \(-q\), are located at equally spaced intervals on the \(x\) axis. Their \(x\) values are \(-3 a,-a, a\), and \(3 a\), respectively. Does there exist a point on the \(y\) axis for which the electric field is zero? If so, find the \(y\) value.

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.