A cube of side 1 is placed in a uniform field E, where `E= E hat(i)`. The net electric flux through the cube is

Options

A. zero
B. `1^(2)E`
C. `41^(2)E`
D. `61^(2)E`

Answer

To find the net electric flux through a cube placed in a uniform electric field \( \mathbf{E} = E \hat{i} \), we can use Gauss's law, which states that the electric flux \( \Phi_E \) through a closed surface is given by:

\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]

where \( Q_{\text{enc}} \) is the charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the Electric Field**: The electric field is given as \( \mathbf{E} = E \hat{i} \). This means the electric field is uniform and directed along the x-axis.

2. **Consider the Cube**: The cube has a side length of 1 unit. The cube is placed in the uniform electric field, but we need to check if there is any charge enclosed within the cube.

3. **Check for Enclosed Charge**: According to the problem, there is no mention of any charge being present inside the cube. Therefore, we can conclude that:
\[ Q_{\text{enc}} = 0 \]

4. **Apply Gauss's Law**: Since there is no charge enclosed within the cube, we can substitute \( Q_{\text{enc}} \) into Gauss's law:
\[ \Phi_E = \frac{0}{\epsilon_0} = 0 \]

5. **Conclusion**: The net electric flux through the cube is:
\[ \Phi_E = 0 \]

### Final Answer:
The net electric flux through the cube is \( 0 \).

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