**Problem Statement:**A charge of uniform density (29 nC/m) is distributed along the x-axis from the origin to the point \( x = 20.9 \text{ m} \). What is the electric potential (relative to zero at infinity) at a point \( x = 79 \text{ m} \) on the x-axis?**Explanation for the Graphs/ Diagrams:**There are no graphs or diagrams provided in this image. If a detailed explanation or solution involving graphical representation or a diagram is required, one could consider visualizing the problem as follows:1. **Coordinate System:** - Draw the x-axis, marking the origin at \( x = 0 \) and extending to \( x > 20.9 \text{ m} \). 2. **Charge Distribution:** - Illustrate the uniformly distributed charge segment from \( x = 0 \text{ m} \) to \( x = 20.9 \text{ m} \). 3. **Point of Interest:** - Indicate the point \( x = 79 \text{ m} \) where the electric potential is to be calculated.This setup would help visualize the scenario for deriving the electric potential due to a line charge. **Solution Outline:**To solve the given problem, understand the following steps:1. **Break the Uniform Line Charge:** - Consider a small element of charge \( dq \) at a distance \( x \) from the origin within the line segment \( 0 \leq x \leq 20.9 \text{ m} \).2. **Expression for \( dq \):** - Let \( λ \) be the charge density (λ = 29 nC/m), then \( dq = λ \, dx = 29 \times 10^{-9} \text{ C/m} \times dx \).3. **Electric Potential Contribution \( dV \):** - The potential due to a small charge element \( dq \) at \( x = 79 \text{ m} \) is given by \( dV = \frac{kdq}{r} \) where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ N·m}^2/\text{C}^2 \), and \( r \) is the distance between the

The change in electric potential can be given as

A charge of uniform density (29 nC/m) is distributed along the x axis from the origin to the point x = 20.9 m. What is the electric potential (relative to zero at infinity) at a point, x = 79 m, on the x axis?

A charge of uniform density (29 nC/m) is distributed along the x axis from the origin to the point x = 20.9 m. What is the electric potential (relative to zero at infinity) at a point, x = 79 m, on the x axis?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

**Problem Statement:**

A charge of uniform density (29 nC/m) is distributed along the x-axis from the origin to the point \( x = 20.9 \text{ m} \). What is the electric potential (relative to zero at infinity) at a point \( x = 79 \text{ m} \) on the x-axis?

**Explanation for the Graphs/ Diagrams:**

There are no graphs or diagrams provided in this image. If a detailed explanation or solution involving graphical representation or a diagram is required, one could consider visualizing the problem as follows:

1. **Coordinate System:**
- Draw the x-axis, marking the origin at \( x = 0 \) and extending to \( x > 20.9 \text{ m} \).

2. **Charge Distribution:**
- Illustrate the uniformly distributed charge segment from \( x = 0 \text{ m} \) to \( x = 20.9 \text{ m} \).

3. **Point of Interest:**
- Indicate the point \( x = 79 \text{ m} \) where the electric potential is to be calculated.

This setup would help visualize the scenario for deriving the electric potential due to a line charge.

**Solution Outline:**

To solve the given problem, understand the following steps:

1. **Break the Uniform Line Charge:**
- Consider a small element of charge \( dq \) at a distance \( x \) from the origin within the line segment \( 0 \leq x \leq 20.9 \text{ m} \).

2. **Expression for \( dq \):**
- Let \( λ \) be the charge density (λ = 29 nC/m), then \( dq = λ \, dx = 29 \times 10^{-9} \text{ C/m} \times dx \).

3. **Electric Potential Contribution \( dV \):**
- The potential due to a small charge element \( dq \) at \( x = 79 \text{ m} \) is given by \( dV = \frac{kdq}{r} \) where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ N·m}^2/\text{C}^2 \), and \( r \) is the distance between the

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