Consider a thin, spherical shell of radius 14.5 cm with a total charge of 29.9 µC distributed uniformly on its surface. (a) Find the electric field 10.0 cm from the center of the charge distribution. (b) Find the electric field 24.5 cm from the center of the charge distribution.

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)...
icon
Related questions
Question
---

### Problem Statement: Electric Field Calculation

Consider a thin, spherical shell of radius \( 14.5 \) cm with a total charge of \( 29.9 \) μC distributed uniformly on its surface.

#### (a) Find the electric field 10.0 cm from the center of the charge distribution.

#### (b) Find the electric field 24.5 cm from the center of the charge distribution.

---

### Explanation

To solve these problems, we will apply Gauss's Law, which relates the electric field to the charge enclosed by a Gaussian surface. The electric field \( E \) due to a spherical charge distribution can be calculated differently depending on whether the point of interest is inside or outside the spherical shell:

1. **Inside the Spherical Shell ( \( r < R \) )**:
   - For any point inside a uniformly charged spherical shell, the electric field is zero because the charges on the shell produce a net zero electric field inside.

2. **Outside the Spherical Shell ( \( r \geq R \) )**:
   - The electric field due to a spherical shell of charge is as if all the charge were concentrated at the center of the shell. Thus, it can be calculated using Coulomb's Law:
     \[
     E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}
     \]
     where \( \epsilon_0 \) is the permittivity of free space, \( Q \) is the total charge, and \( r \) is the distance from the center of the sphere.

Let's solve the problems step-by-step.

### Solution

#### (a) Electric Field at 10.0 cm from the Center

Since \( 10.0 \) cm is less than the radius \( 14.5 \) cm (i.e., \( r < R \)), according to Gauss's Law, the electric field inside a spherical shell is zero.

\[
E_{\text{inside}} = 0
\]

Thus, the electric field 10.0 cm from the center of the charge distribution is:
\[
\boxed{0}
\]

#### (b) Electric Field at 24.5 cm from the Center

Since \( 24.5 \) cm is greater than the radius \( 14.5 \) cm (i.e., \( r \geq R \)), we
Transcribed Image Text:--- ### Problem Statement: Electric Field Calculation Consider a thin, spherical shell of radius \( 14.5 \) cm with a total charge of \( 29.9 \) μC distributed uniformly on its surface. #### (a) Find the electric field 10.0 cm from the center of the charge distribution. #### (b) Find the electric field 24.5 cm from the center of the charge distribution. --- ### Explanation To solve these problems, we will apply Gauss's Law, which relates the electric field to the charge enclosed by a Gaussian surface. The electric field \( E \) due to a spherical charge distribution can be calculated differently depending on whether the point of interest is inside or outside the spherical shell: 1. **Inside the Spherical Shell ( \( r < R \) )**: - For any point inside a uniformly charged spherical shell, the electric field is zero because the charges on the shell produce a net zero electric field inside. 2. **Outside the Spherical Shell ( \( r \geq R \) )**: - The electric field due to a spherical shell of charge is as if all the charge were concentrated at the center of the shell. Thus, it can be calculated using Coulomb's Law: \[ E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2} \] where \( \epsilon_0 \) is the permittivity of free space, \( Q \) is the total charge, and \( r \) is the distance from the center of the sphere. Let's solve the problems step-by-step. ### Solution #### (a) Electric Field at 10.0 cm from the Center Since \( 10.0 \) cm is less than the radius \( 14.5 \) cm (i.e., \( r < R \)), according to Gauss's Law, the electric field inside a spherical shell is zero. \[ E_{\text{inside}} = 0 \] Thus, the electric field 10.0 cm from the center of the charge distribution is: \[ \boxed{0} \] #### (b) Electric Field at 24.5 cm from the Center Since \( 24.5 \) cm is greater than the radius \( 14.5 \) cm (i.e., \( r \geq R \)), we
Expert Solution
steps

Step by step

Solved in 2 steps with 3 images

Blurred answer
Knowledge Booster
Electric field
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON