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.
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.
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4571419-cef3-467f-82d2-89ad06819ab1%2Fc6d6f843-6d34-4603-938a-afff07ca06d0%2Fq40gn2g.png&w=3840&q=75)
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
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