Charge of 2 C is placed at the centre of a cube. What is the electric flux passing through one face ?

Options

A. `(1)/((3 epsi_(0)))`
B. `((1)/(4)) epsi_(0)`
C. `(2)/(epsi_(0))`
D. `(3)/(epsi_(0))`

Answer

To solve the problem of finding the electric flux passing through one face of a cube with a charge of 2 C placed at its center, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Electric Flux**:
Electric flux (Φ) through a closed surface is given by Gauss's Law, which states that the total electric flux through a closed surface is equal to the charge enclosed (Q) divided by the permittivity of free space (ε₀):
\[ \Phi = \frac{Q}{\epsilon_0} \]

2. **Identify the Total Charge**:
In this case, the total charge (Q) placed at the center of the cube is 2 C.

3. **Calculate the Total Electric Flux through the Cube**:
Using Gauss's Law:
\[ \Phi_{\text{total}} = \frac{Q}{\epsilon_0} = \frac{2}{\epsilon_0} \]

4. **Determine the Number of Faces of the Cube**:
A cube has 6 faces.

5. **Calculate the Electric Flux through One Face**:
Since the charge is symmetrically placed at the center of the cube, the electric flux will be uniformly distributed across all 6 faces. Therefore, the electric flux through one face of the cube (Φ_face) can be calculated as:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{2}{6\epsilon_0} = \frac{1}{3\epsilon_0} \]

### Final Answer:
The electric flux passing through one face of the cube is:
\[ \Phi_{\text{face}} = \frac{1}{3\epsilon_0} \]