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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### 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
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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