4. Use Gauss' Law to calculate the electric field a) inside and b) outside of a sphere of radius R with charge Q distributed uniformly throughout its volume.
4. Use Gauss' Law to calculate the electric field a) inside and b) outside of a sphere of radius R with charge Q distributed uniformly throughout its volume.
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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![### Application of Gauss' Law to a Uniformly Charged Sphere
**Problem Statement:**
4. Use Gauss' Law to calculate the electric field a) inside and b) outside of a sphere of radius \( R \) with charge \( Q \) distributed uniformly throughout its volume.
**Explanation:**
To tackle this problem using Gauss' Law, we need to consider two regions:
a) **Inside the Sphere**:
We consider a Gaussian surface within the sphere. For a radius \( r \) (where \( r < R \)), the charge enclosed by the Gaussian surface is proportional to the volume of the sphere inside radius \( r \).
b) **Outside the Sphere**:
Here, we consider a Gaussian surface outside the sphere. For a radius \( r \) (where \( r > R \)), the entire charge \( Q \) is enclosed within this Gaussian surface.
**Detailed Steps and Concepts:**
1. **Gauss' Law**: It states that the electric flux through a closed surface is proportional to the charge enclosed by the surface. Mathematically,
\[
\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
where \( \vec{E} \) is the electric field, \( d\vec{A} \) is the differential area on the closed surface, and \( Q_{\text{enc}} \) is the charge enclosed by the surface.
2. **Inside the Sphere**: For a Gaussian surface of radius \( r \), the enclosed charge \( Q_{\text{enc}} \) is:
\[
Q_{\text{enc}} = Q \left( \frac{r^3}{R^3} \right)
\]
Using Gauss' Law:
\[
E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{Q \left( \frac{r^3}{R^3} \right)}{\epsilon_0}
\]
Solving for \( E \):
\[
E = \frac{Q r}{4 \pi \epsilon_0 R^3}
\]
3. **Outside the Sphere**: For a Gaussian surface of radius \( r \), the enclosed](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F940fea84-314a-4cf2-b65e-4e6cbd95cd4c%2F9176982e-3209-409b-8237-b0a168bcaa8f%2F9d3dgk_processed.png&w=3840&q=75)
Transcribed Image Text:### Application of Gauss' Law to a Uniformly Charged Sphere
**Problem Statement:**
4. Use Gauss' Law to calculate the electric field a) inside and b) outside of a sphere of radius \( R \) with charge \( Q \) distributed uniformly throughout its volume.
**Explanation:**
To tackle this problem using Gauss' Law, we need to consider two regions:
a) **Inside the Sphere**:
We consider a Gaussian surface within the sphere. For a radius \( r \) (where \( r < R \)), the charge enclosed by the Gaussian surface is proportional to the volume of the sphere inside radius \( r \).
b) **Outside the Sphere**:
Here, we consider a Gaussian surface outside the sphere. For a radius \( r \) (where \( r > R \)), the entire charge \( Q \) is enclosed within this Gaussian surface.
**Detailed Steps and Concepts:**
1. **Gauss' Law**: It states that the electric flux through a closed surface is proportional to the charge enclosed by the surface. Mathematically,
\[
\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
where \( \vec{E} \) is the electric field, \( d\vec{A} \) is the differential area on the closed surface, and \( Q_{\text{enc}} \) is the charge enclosed by the surface.
2. **Inside the Sphere**: For a Gaussian surface of radius \( r \), the enclosed charge \( Q_{\text{enc}} \) is:
\[
Q_{\text{enc}} = Q \left( \frac{r^3}{R^3} \right)
\]
Using Gauss' Law:
\[
E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{Q \left( \frac{r^3}{R^3} \right)}{\epsilon_0}
\]
Solving for \( E \):
\[
E = \frac{Q r}{4 \pi \epsilon_0 R^3}
\]
3. **Outside the Sphere**: For a Gaussian surface of radius \( r \), the enclosed
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