A charge +q is placed at the mid point of a cube of side L. The electric flux emerging from cube is

Options

A. `(6ql^(2))/(epsilon_(0))`
B. `(q)/(6l^(2)epsilon_(0))`
C. Zero
D. `(q)/(epsilon_(0))`

Answer

To solve the problem of finding the electric flux emerging from a cube with a charge +q placed at its midpoint, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a cube with side length \( L \).
- A charge \( +q \) is located exactly at the center of the cube.

2. **Recall Gauss's Law**:
- Gauss's Law states that the 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 within the surface and \( \epsilon_0 \) is the permittivity of free space.

3. **Identify the Enclosed Charge**:
- In this case, since the charge \( +q \) is located at the center of the cube, the entire charge is enclosed by the cube.
- Therefore, \( Q_{\text{enc}} = +q \).

4. **Calculate the Electric Flux**:
- Using Gauss's Law, we can substitute the enclosed charge into the formula:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{q}{\epsilon_0} \]

5. **Conclusion**:
- The electric flux emerging from the cube is:
\[ \Phi = \frac{q}{\epsilon_0} \]

### Final Answer:
The electric flux emerging from the cube is \( \frac{q}{\epsilon_0} \).