A charge Q is situated at the corner of a cube the electric flux passed through all the six faces of the cube is :

Options

A. `(Q)/(6 in_(0))`
B. `(Q)/(8 in_(0))`
C. `(Q)/(in_(0))`
D. `(Q)/(2 in_(0))`

Answer

To solve the problem of finding the electric flux through the six faces of a cube with a charge \( Q \) located at one of its corners, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a cube and a charge \( Q \) located at one of its corners.
- The cube has 8 corners in total.

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

3. **Determining the Enclosed Charge**:
- Since the charge \( Q \) is located at a corner of the cube, it is shared by 8 adjacent cubes (octants).
- Therefore, the charge enclosed by our cube is:
\[ Q_{\text{enclosed}} = \frac{Q}{8} \]

4. **Calculating the Electric Flux**:
- Substituting \( Q_{\text{enclosed}} \) into Gauss's Law:
\[ \Phi = \frac{Q/8}{\epsilon_0} = \frac{Q}{8\epsilon_0} \]

5. **Final Result**:
- Thus, the total electric flux passing through all six faces of the cube is:
\[ \Phi = \frac{Q}{8\epsilon_0} \]

### Conclusion:
The electric flux through all six faces of the cube is \( \frac{Q}{8\epsilon_0} \).