There is uniform electric field of `8xx10^(3)hati` N/C. What is the net flux (in Sl units) of the uniform electric field through a cube of side 0.3 m oriented so that its faces are parallel to the coordinate plane ?

Options

A. `2xx8xx10^(3)`
B. `0.3xx8xx10^(3)`
C. Zero
D. `8xx10^(6)xx6`

Answer

To find the net electric flux through a cube placed in a uniform electric field, we can follow these steps:

### Step 1: Understand the Electric Flux Concept
Electric flux (Φ) through a surface is given by the formula:
\[ \Phi = \vec{E} \cdot \vec{A} = E \cdot A \cdot \cos(\theta) \]
where:
- \( \vec{E} \) is the electric field vector,
- \( \vec{A} \) is the area vector of the surface,
- \( \theta \) is the angle between the electric field and the area vector.

### Step 2: Identify the Given Values
From the problem, we have:
- The electric field \( \vec{E} = 8 \times 10^3 \hat{i} \) N/C.
- The side length of the cube \( a = 0.3 \) m.

### Step 3: Calculate the Area of Each Face of the Cube
The area \( A \) of one face of the cube is given by:
\[ A = a^2 = (0.3 \, \text{m})^2 = 0.09 \, \text{m}^2 \]

### Step 4: Determine the Orientation of the Cube
The cube is oriented such that its faces are parallel to the coordinate planes. Therefore, the area vectors for the faces will point in the directions of the coordinate axes.

### Step 5: Analyze Each Face of the Cube
1. **Top and Bottom Faces**:
- Area vectors point in the \( \hat{k} \) direction (upward and downward).
- The angle \( \theta = 90^\circ \) (electric field is along \( \hat{i} \)).
- Flux through these faces:
\[ \Phi_{\text{top}} = E \cdot A \cdot \cos(90^\circ) = 0 \]
\[ \Phi_{\text{bottom}} = E \cdot A \cdot \cos(90^\circ) = 0 \]

2. **Front and Back Faces**:
- Area vectors point in the \( \hat{j} \) direction.
- The angle \( \theta = 90^\circ \).
- Flux through these faces:
\[ \Phi_{\text{front}} = E \cdot A \cdot \cos(90^\circ) = 0 \]
\[ \Phi_{\text{back}} = E \cdot A \cdot \cos(90^\circ) = 0 \]

3. **Left and Right Faces**:
- Area vectors point in the \( \hat{i} \) direction (left and right).
- For the left face (area vector pointing in the negative \( \hat{i} \) direction):
- The angle \( \theta = 180^\circ \).
- Flux through this face:
\[ \Phi_{\text{left}} = E \cdot A \cdot \cos(180^\circ) = -E \cdot A = - (8 \times 10^3) \cdot (0.09) = -720 \, \text{Nm}^2/\text{C} \]
- For the right face (area vector pointing in the positive \( \hat{i} \) direction):
- The angle \( \theta = 0^\circ \).
- Flux through this face:
\[ \Phi_{\text{right}} = E \cdot A \cdot \cos(0^\circ) = E \cdot A = (8 \times 10^3) \cdot (0.09) = 720 \, \text{Nm}^2/\text{C} \]

### Step 6: Calculate the Net Flux
Now, we can sum the fluxes through all the faces:
\[ \Phi_{\text{net}} = \Phi_{\text{top}} + \Phi_{\text{bottom}} + \Phi_{\text{front}} + \Phi_{\text{back}} + \Phi_{\text{left}} + \Phi_{\text{right}} \]
\[ \Phi_{\text{net}} = 0 + 0 + 0 + 0 - 720 + 720 = 0 \]

### Final Answer
The net electric flux through the cube is:
\[ \Phi_{\text{net}} = 0 \, \text{Nm}^2/\text{C} \]

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