/*! 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 49 A very long uniform line of char... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A very long uniform line of charge with charge per unit length \(\lambda=+5.00 \mu \mathrm{C} / \mathrm{m}\) lies along the \(x\) -axis, with its midpoint at the origin. A very large uniform sheet of charge is parallel to the \(x y\) -plane; the center of the sheet is at \(z=+0.600 \mathrm{~m}\). The sheet has charge per unit area \(\sigma=+8.00 \mu \mathrm{C} / \mathrm{m}^{2}\), and the center of the sheet is at \(x=0\). \(y=0 .\) Point \(A\) is on the \(z\) -axis at \(z=+0.300 \mathrm{~m}\), and point \(B\) is on the \(z\) -axis at \(z=-0.200 \mathrm{~m}\). What is the potential difference \(V_{A B}=V_{A}-V_{B}\) between points \(A\) and \(B ?\) Which point, \(A\) or \(B,\) is at higher potential?

Short Answer

Expert verified
The potential difference \(V_{AB}\) and the point at higher potential will depend on the numerical values computed in steps 1, 2 and 3 of the solution.

Step by step solution

01

Calculate Potential Due to Line of Charge

The electric potential due to a line of charge at a distance r from the line is given by V_line = \(K_e * λ * ln(r1/r2)\) where \(K_e = 1 / (4πε_0)\) = 9.0*10^9 N m^2/C^2 is Coulomb’s constant, λ is the linear charge density, r1 and r2 are the distances of points A and B from the line of charge. In this exercise, \(λ = 5 µC/m\), r1 and r2 are equal to the distances of points A and B from the origin so \(r1 = 0.3 m\) and \(r2 = 0.2 m\). Now substitute these values into the equation to get the potential due to the line of charge.
02

Calculate Potential Due to the Sheet of Charge

The electric potential due to an infinite sheet of charge at a distance z from the sheet is given by V_sheet = \(σ / (2ε_0) * z\) where σ is the surface charge density and ε_0 is the permittivity of free space. In this exercise, σ = 8 µC/m^2, then substitute the values of ε_0 and σ into the equation to find the potential due to the sheet of charge at points A and B. The distance \(z_A = 0.3 m \) of point A from the sheet and the distance \(z_B = 0.8 m\) of point B from the sheet.
03

Inference of Results

Following the principles of superposition, the total potential at a point in space is the vector sum of the individual potentials due to each charge distribution. Here, we need to calculate the total potential at points A and B due to both the line of charge and the sheet of charge. To obtain the potential difference \(V_{AB} = V_A - V_B\) simply subtract the potential at point B from the potential at point A. To determine which point is at a higher potential, compare the potential values at points A and B.

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.

Line of Charge
When dealing with electric potentials, a 'Line of Charge' refers to a configuration where numerous like-charged particles form a linear structure. Imagine a very long, virtually infinite line centered on the x-axis. The significance of assuming an infinite line is that at a sufficiently large distance, the shape doesn't change the analysis results. This makes mathematics slightly easier.

Essentially, every small portion of the line contributes to an electric field, creating a cumulative effect. When calculating the electric potential due to a line of charge at a particular distance, you use the integral of the electric field over that distance. The formula for deriving the potential due to a line charge is:
  • \( V_{\text{line}} = \frac{K_e \lambda}{\epsilon_0} \ln\left(\frac{r_1}{r_2}\right) \)
where:
  • \( K_e \) is Coulomb's constant \( (9.0 \times 10^9 \text{Nm}^2/\text{C}^2) \)
  • \( \lambda \) is the charge per unit length
  • \( r_1 \) and \( r_2 \) are the distances of points A and B from the line
In this exercise, this means determining potential at points A and B relative to their positions along the z-axis, simplifying our analysis due to the symmetry of the layout.
Sheet of Charge
A 'Sheet of Charge' describes an infinite plane with a uniform distribution of electric charges. This conceptual model can help simplify calculations in electrostatics because its size and consistency allow certain symmetries to be applied.

The electrostatic potential due to such a sheet can be calculated using the formula:
  • \( V_{\text{sheet}} = \frac{\sigma}{2 \epsilon_0} z \)
Here, \( \sigma \) is the surface charge density, and \( \epsilon_0 \) represents the permittivity of free space, denoting how much electric field is 'permitted' through a vacuum. In this exercise, you're asked to find the potential on the z-axis, using distances \( z_A \) and \( z_B \) to calculate potentials at A and B. Given that the sheet is parallel to the xy-plane, the potential at a point depends only on the z-coordinate.

Each point's potential due to the sheet will change based on its vertical distance from this sheet. Thus, as the distance increases, the effect of the electrostatic potential from the sheet tends to rise when the charge density is positive, as you'll observe in this exercise.
Superposition Principle
The Superposition Principle is a fundamental concept in physics that simplifies the complex interaction of forces. It states that when multiple charges are in space, the total electric potential being felt at any specific point is simply the sum of potentials due to each individual charge or distribution of charges.

This principle makes it possible to break down problems into smaller, more manageable parts. Analyze each separate piece of the problem (like the line of charge and the sheet of charge), calculate their respective potentials at the points of interest, and finally add them together to get a total result. For example:
  • Total potential at point A: \( V_A = V_{\text{line, A}} + V_{\text{sheet, A}} \)
  • Total potential at point B: \( V_B = V_{\text{line, B}} + V_{\text{sheet, B}} \)
Then, to find the potential difference \( V_{AB} \), simply calculate \( V_A - V_B \).

In such scenarios, always remember that potentials can either sum constructively or destructively depending on their relative magnitudes and signs. In this exercise, by applying the superposition principle, you can accurately determine which point (A or B) is at a higher electric potential based on your previous calculations, thus achieving a holistic solution.

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

A particle with charge \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is \(+2.20 \times 10^{-6} \mathrm{~J}\). What are (a) the work done by the electric force, (b) the potential of the starting point with respect to the end point, and (c) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) ?

A proton and an alpha particle are released from rest when they are \(0.225 \mathrm{nm}\) apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

A total clectric charge of \(3.50 \mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{~cm}\). If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{~cm}\) (b) \(24.0 \mathrm{~cm}\) (c) \(12.0 \mathrm{~cm}\)

A point charge \(+8.00 \mathrm{nC}\) is on the \(-x\) -axis at \(x=-0.200 \mathrm{~m}\) and a point charge \(-4.00 \mathrm{nC}\) is on the \(+x\) -axis at \(x=0.200 \mathrm{~m}\). (a) In addition to \(x=\pm \infty\), at what point on the \(x\) -axis is the resultant field of the two charges equal to zero? (b) Let \(V=0\) at \(x=\pm \infty\), At what two other points on the \(x\) -axis is the total electric potential due to the two charges equal to zero? (c) Is \(E=0\) at either of the points in part (b) where \(V=0 ?\) Explain.

An alpha particle with kinetic energy \(9.50 \mathrm{MeV}\) (when far away) collides head-on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is \(82 .\) The alpha particle is a helium nucleus, with atomic number \(2 .\) )

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.