a) Find E for 84P20 Հ show all Gaussiaan surfaces 92² b) Plot E vs r c) Find charges on the surfaces of the conducting shell genc SELA d A= q t
a) Find E for 84P20 Հ show all Gaussiaan surfaces 92² b) Plot E vs r c) Find charges on the surfaces of the conducting shell genc SELA d A= q t
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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![**Exercise: Analyzing a Concentric Sphere and Conducting Shell System**
In this exercise, you are provided with a solid sphere of radius \( a = 2.00 \, \text{cm} \), which is concentric with a spherical conducting shell. The shell's inner radius is \( b = 2.00a \) and its outer radius is \( c = 2.40a \). The sphere holds a net uniform charge \( q_1 = +5.00 \, \text{fC} \), while the shell has a net charge \( q_2 = -q_1 \).
### Tasks:
1. **Find \(\mathbf{E}\) for \(0 < r < \infty\)**
- Begin by showing all Gaussian surfaces involved.
2. **Plot \(\mathbf{E}\) vs \(r\)**
- Illustrate how the electric field varies with radius.
3. **Find charges on the surfaces of the conducting shell**
### Mathematical Derivations and Expressions:
- **Gauss's Law Application:**
\[
\frac{q_{\text{enc}}}{\varepsilon_0} = \oint E \, dA
\]
- For a spherical symmetry:
\[
\oint E \, dA = E \cdot 4\pi r^2
\]
- Inside the conducting material: \( E = 0 \)
### Graphical Explanation:
The diagram presented illustrates concentric spherical regions. Charge distributions are depicted with \( q_1 = +5.00 \, \text{fC} \) at the center and opposite charge on the shell, maintaining charge neutrality overall. A plot of \( E \) vs \( r \) should visually represent where the electric field is non-zero outside the charged regions and zero within the conductor.
### Key Observations:
- As the distance \( r \) increases from 0 past \( c \), the electric field \( E \) is expected to change following the inverse-square law until the conducting shell begins, where \( E \) reaches zero within the shell.
- The process involves conceptual understanding of charge distributions and how they affect electric field behavior in and around conducting materials.
---
This exercise applies Gauss's Law within the context of electrostatics, focusing on practical use of symmetry and charge distribution concepts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf04c643-8994-40f5-bafb-98e6c5614a89%2F3b882c92-48bb-419d-8ced-cf3b5559e3b4%2Fzwa9sy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exercise: Analyzing a Concentric Sphere and Conducting Shell System**
In this exercise, you are provided with a solid sphere of radius \( a = 2.00 \, \text{cm} \), which is concentric with a spherical conducting shell. The shell's inner radius is \( b = 2.00a \) and its outer radius is \( c = 2.40a \). The sphere holds a net uniform charge \( q_1 = +5.00 \, \text{fC} \), while the shell has a net charge \( q_2 = -q_1 \).
### Tasks:
1. **Find \(\mathbf{E}\) for \(0 < r < \infty\)**
- Begin by showing all Gaussian surfaces involved.
2. **Plot \(\mathbf{E}\) vs \(r\)**
- Illustrate how the electric field varies with radius.
3. **Find charges on the surfaces of the conducting shell**
### Mathematical Derivations and Expressions:
- **Gauss's Law Application:**
\[
\frac{q_{\text{enc}}}{\varepsilon_0} = \oint E \, dA
\]
- For a spherical symmetry:
\[
\oint E \, dA = E \cdot 4\pi r^2
\]
- Inside the conducting material: \( E = 0 \)
### Graphical Explanation:
The diagram presented illustrates concentric spherical regions. Charge distributions are depicted with \( q_1 = +5.00 \, \text{fC} \) at the center and opposite charge on the shell, maintaining charge neutrality overall. A plot of \( E \) vs \( r \) should visually represent where the electric field is non-zero outside the charged regions and zero within the conductor.
### Key Observations:
- As the distance \( r \) increases from 0 past \( c \), the electric field \( E \) is expected to change following the inverse-square law until the conducting shell begins, where \( E \) reaches zero within the shell.
- The process involves conceptual understanding of charge distributions and how they affect electric field behavior in and around conducting materials.
---
This exercise applies Gauss's Law within the context of electrostatics, focusing on practical use of symmetry and charge distribution concepts.
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