If the charge is placed at the center of one side, flux through the cube is `(q)/(n epsilon_(0))` where .n. is

Answer

To solve the problem, we will use Gauss's law, which states that the electric flux through a closed surface is proportional to the charge enclosed by that surface.

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a cube and a charge \( Q \) placed at the center of one of its faces. We need to find the flux through the entire cube.

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

3. **Determining the Charge Enclosed**:
Since the charge \( Q \) is located at the center of one face of the cube, we need to determine how much of this charge is effectively enclosed by the cube. The cube can be thought of as being divided into six equal parts, with only one part containing the charge.

4. **Calculating the Flux through the Cube**:
The flux through the entire cube can be calculated as:
\[ \Phi_{\text{cube}} = \frac{Q}{6\epsilon_0} \]
This is because only one-sixth of the charge \( Q \) is effectively enclosed by the cube.

5. **Relating to the Given Expression**:
According to the problem, the flux through the cube is also given by:
\[ \Phi_{\text{cube}} = \frac{Q}{n\epsilon_0} \]
where \( n \) is the value we need to find.

6. **Setting the Equations Equal**:
Now we can set the two expressions for the flux equal to each other:
\[ \frac{Q}{6\epsilon_0} = \frac{Q}{n\epsilon_0} \]

7. **Solving for \( n \)**:
By simplifying the equation, we can cancel \( Q \) and \( \epsilon_0 \) (assuming \( Q \neq 0 \) and \( \epsilon_0 \neq 0 \)):
\[ \frac{1}{6} = \frac{1}{n} \]
Therefore, we find:
\[ n = 6 \]

### Final Answer:
The value of \( n \) is \( 6 \).