A point charge Q is placed at the corner of a cube. How much electric flux will be associated with the cube?

Options

A. `q//epsilon_0`
B. `q//2epsilon_0`
C. `q//8epsilon_0`
D. `q//4epsilon_0`

Explanation

To solve the problem of finding the electric flux associated with a cube that has a point charge \( Q \) placed at one of its corners, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- A point charge \( Q \) is placed at one corner of a cube.
- Since the charge is at the corner, it is shared by multiple cubes. Specifically, it is shared by 8 cubes that can be imagined to surround the corner where the charge is located.

2. **Applying Gauss's Law**:
- Gauss's Law states that the total electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q_{enc} \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q_{enc}}{\epsilon_0} \]
- In this case, since the charge \( Q \) is at the corner of the cube, it is not fully enclosed by just one cube. Instead, it is effectively enclosed by 8 cubes.

3. **Calculating the Total Flux for All Cubes**:
- The total electric flux through all 8 cubes that share the charge \( Q \) is:
\[ \Phi_{total} = \frac{Q}{\epsilon_0} \]

4. **Finding the Flux Through One Cube**:
- Since the total flux is distributed equally among the 8 cubes, the electric flux \( \Phi_{cube} \) through one cube is:
\[ \Phi_{cube} = \frac{\Phi_{total}}{8} = \frac{Q}{8 \epsilon_0} \]

5. **Final Answer**:
- Therefore, the electric flux associated with the cube is:
\[ \Phi = \frac{Q}{8 \epsilon_0} \]

### Conclusion:
The electric flux associated with the cube is \( \frac{Q}{8 \epsilon_0} \).

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