A cube of side a is placed in a uniform electric field `E = E_(0) hat(i) + E_(0) hat(j) + E_(0) k`. Total electric flux passing through the cube would be

Options

A. `E_(0)a^(2)`
B. `2E_(0)a^(2)`
C. `6 E_(0)a^(2)`
D. None of these

Answer

To calculate the total electric flux passing through a cube placed in a uniform electric field, we can follow these steps:

### Step 1: Understand the Electric Field
The electric field is given as:
\[ \mathbf{E} = E_0 \hat{i} + E_0 \hat{j} + E_0 \hat{k} \]
This means the electric field has equal components in the x, y, and z directions.

### Step 2: Identify the Cube's Faces
A cube has six faces. We will denote the area of each face as \( A = a^2 \), where \( a \) is the side length of the cube.

### Step 3: Calculate the Electric Flux Through Each Face
The electric flux \( \Phi \) through a surface is given by:
\[ \Phi = \mathbf{E} \cdot \mathbf{A} \]
where \( \mathbf{A} \) is the area vector of the surface.

1. **For the face perpendicular to the x-axis (front and back faces)**:
- Area vector for the front face: \( \mathbf{A_1} = a^2 \hat{i} \)
- Area vector for the back face: \( \mathbf{A_2} = -a^2 \hat{i} \)

- Flux through front face:
\[ \Phi_1 = \mathbf{E} \cdot \mathbf{A_1} = (E_0 \hat{i} + E_0 \hat{j} + E_0 \hat{k}) \cdot (a^2 \hat{i}) = E_0 a^2 \]

- Flux through back face:
\[ \Phi_2 = \mathbf{E} \cdot \mathbf{A_2} = (E_0 \hat{i} + E_0 \hat{j} + E_0 \hat{k}) \cdot (-a^2 \hat{i}) = -E_0 a^2 \]

- Total flux through these two faces:
\[ \Phi_{1+2} = E_0 a^2 - E_0 a^2 = 0 \]

2. **For the face perpendicular to the y-axis (left and right faces)**:
- Area vector for the left face: \( \mathbf{A_3} = -a^2 \hat{j} \)
- Area vector for the right face: \( \mathbf{A_4} = a^2 \hat{j} \)

- Flux through left face:
\[ \Phi_3 = \mathbf{E} \cdot \mathbf{A_3} = (E_0 \hat{i} + E_0 \hat{j} + E_0 \hat{k}) \cdot (-a^2 \hat{j}) = -E_0 a^2 \]

- Flux through right face:
\[ \Phi_4 = \mathbf{E} \cdot \mathbf{A_4} = (E_0 \hat{i} + E_0 \hat{j} + E_0 \hat{k}) \cdot (a^2 \hat{j}) = E_0 a^2 \]

- Total flux through these two faces:
\[ \Phi_{3+4} = -E_0 a^2 + E_0 a^2 = 0 \]

3. **For the face perpendicular to the z-axis (top and bottom faces)**:
- Area vector for the bottom face: \( \mathbf{A_5} = -a^2 \hat{k} \)
- Area vector for the top face: \( \mathbf{A_6} = a^2 \hat{k} \)

- Flux through bottom face:
\[ \Phi_5 = \mathbf{E} \cdot \mathbf{A_5} = (E_0 \hat{i} + E_0 \hat{j} + E_0 \hat{k}) \cdot (-a^2 \hat{k}) = -E_0 a^2 \]

- Flux through top face:
\[ \Phi_6 = \mathbf{E} \cdot \mathbf{A_6} = (E_0 \hat{i} + E_0 \hat{j} + E_0 \hat{k}) \cdot (a^2 \hat{k}) = E_0 a^2 \]

- Total flux through these two faces:
\[ \Phi_{5+6} = -E_0 a^2 + E_0 a^2 = 0 \]

### Step 4: Calculate Total Electric Flux
Now, adding the flux through all six faces:
\[ \Phi_{\text{total}} = \Phi_{1+2} + \Phi_{3+4} + \Phi_{5+6} = 0 + 0 + 0 = 0 \]

### Final Answer
The total electric flux passing through the cube is:
\[ \Phi_{\text{total}} = 0 \]