The total electric flux through a cube when a charge 8q is placed at one corner of the cube is

Options

A. `epsilon_(0)q`
B. `(epsilon_(0))/q`
C. `q/(epsilon_(0))`
D. `q/(4piepsilon_(0))`

Answer

To solve the problem of finding the total electric flux through a cube when a charge of \(8Q\) is placed at one corner of the cube, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have a cube, and a charge of \(8Q\) is placed at one of its corners.

2. **Apply Gauss's Law**:
- According to Gauss's Law, the total electric flux (\(\Phi\)) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Here, \(Q_{\text{enc}}\) is the total charge enclosed by the surface, and \(\epsilon_0\) is the permittivity of free space.

3. **Consider the Symmetry**:
- The charge \(8Q\) at the corner of the cube is shared by eight identical cubes that can be imagined to surround the charge. Each of these cubes will have an equal share of the total flux.

4. **Calculate the Total Flux for All Cubes**:
- The total electric flux through all eight cubes can be calculated as:
\[ \Phi_{\text{total}} = \frac{8Q}{\epsilon_0} \]

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

6. **Final Result**:
- Therefore, the total electric flux through the cube when a charge \(8Q\) is placed at one corner is:
\[ \Phi = \frac{Q}{\epsilon_0} \]

### Conclusion:
The total electric flux through the cube is \(\frac{Q}{\epsilon_0}\).