/*! 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 43 The electric field in a region i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 43

The electric field in a region is pointed away from the z-axis and the magnitude depends upon the distance \(s\) from the axis. The magnitude of the electric field is given as \(E=\frac{\alpha}{s}\) where \(\alpha\) is a constant. Find the potential difference between points \(P_{1}\) and \(P_{2},\) explicitly stating the path over which you conduct the integration for the line integral.

Short Answer

Expert verified
The potential difference between points \(P_1\) and \(P_2\), along the chosen path parallel to the x-axis, can be found using the line integral of the electric field and is given by the expression: \[ \Delta V = -\alpha \left(\ln(x_2) - \ln(x_1)\right) \] where \(x_1\) and \(x_2\) are the x-coordinates of \(P_1\) and \(P_2\), respectively, and \(\alpha\) is a constant.

Step by step solution

01

Choose a path

To simplify the calculations, we choose a path that is parallel to the x-axis, starting at point \(P_1\) and ending at point \(P_2\). This path makes it easier to perform the integration, as we will only have to integrate along one axis.
02

Express the electric field as a vector

Given the magnitude of the electric field, \(E = \frac{\alpha}{s}\), we can express the electric field vector as \[ \vec{E} = E \hat{r} \] where \(\hat{r}\) is the unit vector pointing radially away from the z-axis. In cylindrical coordinates, the electric field vector can be expressed as \[ \vec{E} = \frac{\alpha}{s} \hat{s} \]
03

Compute the line integral along the chosen path

Using the chosen path (parallel to the x-axis), we can calculate the line integral of the electric field: \[ \begin{aligned} \Delta V = -\int_{P_1}^{P_2} \vec{E} \cdot d \vec{l} &= -\int_{P_1}^{P_2} \frac{\alpha}{s} \hat{s} \cdot (dx \, \hat{x} + dy \, \hat{y} + dz \, \hat{z})\\ &= -\int_{P_1}^{P_2} \frac{\alpha}{s} \hat{s} \cdot (dx \, \hat{x}) \\ \end{aligned} \]
04

Integrate

Now integrate the expression above: \[ \Delta V = -\alpha \int_{P_1}^{P_2} \frac{1}{s} ds \] Since the integration is with respect to \(s\), and we are only moving along the x-axis, \(s = x\). Therefore, we can rewrite the integral as: \[ \Delta V = -\alpha \int_{x_1}^{x_2} \frac{1}{x} dx \] where \(x_1\) and \(x_2\) are the x-coordinates of \(P_1\) and \(P_2\), respectively. Integration of \(\frac{1}{x}\) with respect to \(x\) gives \(\ln(x)\). Hence, \[ \Delta V = -\alpha \left[\ln(x)\right]_{x_1}^{x_2} \]
05

Evaluate the integral

Finally, we evaluate the integral to find the potential difference: \[ \Delta V = -\alpha \left(\ln(x_2) - \ln(x_1)\right) \] The potential difference between points \(P_1\) and \(P_2\), along the chosen path parallel to the x-axis, is given by the expression above.

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 where an electric force is exerted on a charged particle. Imagine an invisible force field surrounding charged objects. The force that a particle experiences at any point in this field is called the electric field at that point.

Key characteristics include:
  • A vector quantity 鈥� it has both magnitude and direction.
  • Represented by the symbol \( \vec{E} \).
  • Defined mathematically as the force \( \vec{F} \) per unit charge \( q \): \( \vec{E} = \frac{\vec{F}}{q} \).
In this exercise, the electric field is directed away from the z-axis, and its magnitude is dependent on the distance from this axis. Specifically, it diminishes as \( \frac{\alpha}{s} \), implying the farther from the axis, the weaker the field.
Line Integral
A line integral helps us find how an electric field behaves along a certain path. Imagine walking along a path in the electric field, and collecting data on how the field behaves at each step. This accumulated data is the line integral.

Here is how we use it:
  • It adds up the effects of a field along a path.
  • Used to calculate work done by or against an electric field.
  • In this case, we calculate the integral from \( P_1 \) to \( P_2 \).
For our problem, the line integral is used to determine the potential difference between two points. By integrating the electric field along the x-axis, the exercise simplifies the problem, focusing only on how the electric field varies with distance \( s \), which translates to x-axis movement.
Cylindrical Coordinates
Cylindrical coordinates are a 3D coordinate system often used when a problem involves symmetry around an axis. Think of it as a way to describe a point in space using the wire-frame model of a cylinder.

Key pointers include:
  • Coordinates are \( (s, \theta, z) \).
  • \( s \) measures distance from the z-axis.
  • \( \theta \) measures angle around the z-axis, similar to polar coordinates.
  • \( z \) measures height along the z-axis.
In our exercise, the electric field is expressed in cylindrical coordinates as \( \vec{E} = \frac{\alpha}{s} \hat{s} \). This helps simplify the mathematics by focusing on radial dependence without needing to consider variations with \( \theta \) or \( z \).
Integration
Integration is like finding the entire landscape given the topography of a path. In simple terms, it helps us find quantities like area under a curve or, in our case, potential difference across points, based on a given rate of change.

Here's what to understand about our integration process:
  • We start with an integral along the x-axis 鈥� \( \Delta V = -\alpha \int_{x_1}^{x_2} \frac{1}{x} dx \).
  • In our scenario, since the path is parallel to the x-axis, the integration limits are \( x_1 \) to \( x_2 \).
  • The integral of \( \frac{1}{x} \) is \( \ln(x) \).
Thus, by calculating this integral, we find the potential difference between points \( P_1 \) and \( P_2 \), resulting in the expression \( \Delta V = -\alpha (\ln(x_2) - \ln(x_1)) \). This expression succinctly captures the change in potential energy due to the electric field over the chosen path.

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

Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non- relativistic final speeds. Take the mass of the hydrogen ion to be \(1.67 \times 10^{-27} \mathrm{kg}\)

(a) Find the voltage near a \(10.0 \mathrm{cm}\) diameter metal sphere that has \(8.00 \mathrm{C}\) of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?

If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force? Explain.

In a Geiger counter, a thin metallic wire at the center of a metallic tube is kept at a high voltage with respect to the metal tube. Ionizing radiation entering the tube knocks electrons off gas molecules or sides of the tube that then accelerate towards the center wire, knocking off even more electrons. This process eventually leads to an avalanche that is detectable as a current. A particular Geiger counter has a tube of radius \(R\) and the inner wire of radius \(a\) is at a potential of \(V_{0}\) volts with respect to the outer metal tube. Consider a point \(P\) at a distance \(s\) from the center wire and far away from the ends. (a) Find a formula for the electric field at a point \(P\) inside using the infinite wire approximation. (b) Find a formula for the electric potential at a point P inside. (c) Use \(V_{0}=900 \mathrm{V}, a=3.00 \mathrm{mm}, R=2.00 \mathrm{cm}, \quad\) and find the value of the electric field at a point \(1.00 \mathrm{cm}\) from the center.

Can a particle move in a direction of increasing electric potential, yet have its electric potential energy decrease? Explain
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Physics of Motion

Read Explanation

Force

Read Explanation

Quantum Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Measurements

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.