A cube of side l is p laced in a uniform field E, where `E=E hati`. The net electric flux through the cube is

Options

A. zero
B. `l^(2)E`
C. `4l^(2)E`
D. `6l^(2)E`

Explanation

To find the net electric flux through a cube placed in a uniform electric field \( \mathbf{E} = E \hat{i} \), we can follow these steps:

### Step 1: Understand the Geometry of the Cube
Consider a cube of side length \( l \). The cube has six faces, and we need to analyze the orientation of these faces with respect to the electric field direction.

### Step 2: Identify the Direction of the Electric Field
The electric field is given as \( \mathbf{E} = E \hat{i} \), which means it is directed along the positive x-axis.

### Step 3: Determine the Area Vectors of the Cube Faces
Each face of the cube has an area \( A = l^2 \). The area vectors for the faces of the cube are as follows:
- Face 1 (front face): Area vector \( \mathbf{A}_1 = l^2 \hat{i} \)
- Face 2 (back face): Area vector \( \mathbf{A}_2 = -l^2 \hat{i} \)
- Face 3 (left face): Area vector \( \mathbf{A}_3 = -l^2 \hat{j} \)
- Face 4 (right face): Area vector \( \mathbf{A}_4 = l^2 \hat{j} \)
- Face 5 (top face): Area vector \( \mathbf{A}_5 = l^2 \hat{k} \)
- Face 6 (bottom face): Area vector \( \mathbf{A}_6 = -l^2 \hat{k} \)

### Step 4: Calculate the Electric Flux Through Each Face
The electric flux \( \Phi \) through a surface is given by the formula:
\[ \Phi = \mathbf{E} \cdot \mathbf{A} = EA \cos \theta \]
where \( \theta \) is the angle between the electric field vector and the area vector.

- For Face 1 (front face):
\[ \Phi_1 = E \cdot (l^2 \hat{i}) = E l^2 \cdot 1 = E l^2 \]

- For Face 2 (back face):
\[ \Phi_2 = E \cdot (-l^2 \hat{i}) = E l^2 \cdot (-1) = -E l^2 \]

- For Face 3 (left face):
\[ \Phi_3 = E \cdot (-l^2 \hat{j}) = E l^2 \cdot 0 = 0 \]

- For Face 4 (right face):
\[ \Phi_4 = E \cdot (l^2 \hat{j}) = E l^2 \cdot 0 = 0 \]

- For Face 5 (top face):
\[ \Phi_5 = E \cdot (l^2 \hat{k}) = E l^2 \cdot 0 = 0 \]

- For Face 6 (bottom face):
\[ \Phi_6 = E \cdot (-l^2 \hat{k}) = E l^2 \cdot 0 = 0 \]

### Step 5: Sum the Electric Fluxes
Now, we can find the total electric flux through the cube by summing the fluxes through all six faces:
\[ \Phi_{\text{net}} = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 \]
Substituting the values we calculated:
\[ \Phi_{\text{net}} = E l^2 - E l^2 + 0 + 0 + 0 + 0 = 0 \]

### Conclusion
The net electric flux through the cube is:
\[ \Phi_{\text{net}} = 0 \]