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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Question
![### Problem Statement
A particle of charge \( Q \) sits at the center of a cube. Find the electric flux through one side.
### Explanation
To solve this problem, we need to use Gauss's Law, which states:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where:
- \(\Phi_E\) is the electric flux through a closed surface,
- \(Q_{\text{enc}}\) is the charge enclosed by the surface, and
- \(\epsilon_0\) is the permittivity of free space.
Since the charge \( Q \) is at the center of the cube, the flux is uniformly distributed through each of the six faces of the cube.
#### Step-by-Step Calculation:
1. **Calculate the Total Electric Flux:**
According to Gauss's Law, the total electric flux through the entire surface of the cube is:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
2. **Determine the Flux Through One Face:**
Since the cube has six sides and the charge is symmetrically located, the electric flux is equally distributed among the six sides. Hence, the flux through one side of the cube is:
\[ \Phi_{\text{one side}} = \frac{\Phi_E}{6} = \frac{Q}{6\epsilon_0} \]
### Conclusion
The electric flux through one side of the cube is given by:
\[ \Phi_{\text{one side}} = \frac{Q}{6\epsilon_0} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a85083f-c711-4fe4-a0ea-d798364bd78d%2Fff4c5e3e-e6f1-4abc-9cb4-94820deda692%2Fk47pzvo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A particle of charge \( Q \) sits at the center of a cube. Find the electric flux through one side.
### Explanation
To solve this problem, we need to use Gauss's Law, which states:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where:
- \(\Phi_E\) is the electric flux through a closed surface,
- \(Q_{\text{enc}}\) is the charge enclosed by the surface, and
- \(\epsilon_0\) is the permittivity of free space.
Since the charge \( Q \) is at the center of the cube, the flux is uniformly distributed through each of the six faces of the cube.
#### Step-by-Step Calculation:
1. **Calculate the Total Electric Flux:**
According to Gauss's Law, the total electric flux through the entire surface of the cube is:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
2. **Determine the Flux Through One Face:**
Since the cube has six sides and the charge is symmetrically located, the electric flux is equally distributed among the six sides. Hence, the flux through one side of the cube is:
\[ \Phi_{\text{one side}} = \frac{\Phi_E}{6} = \frac{Q}{6\epsilon_0} \]
### Conclusion
The electric flux through one side of the cube is given by:
\[ \Phi_{\text{one side}} = \frac{Q}{6\epsilon_0} \]
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