A charge of `2xx10^(-9) C` is placed on a corner of a cube of side 1m. Find the electric flux passing throguh a face of the given cube ?

Explanation

To find the electric flux passing through a face of a cube when a charge is placed at one of its corners, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charge and Cube Dimensions**:
- The charge \( Q = 2 \times 10^{-9} \, \text{C} \).
- The side length of the cube \( a = 1 \, \text{m} \).

2. **Use Gauss's Law**:
- According to Gauss's Law, the total electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\varepsilon_0} \]
- Here, \( Q_{\text{enc}} \) is the charge enclosed by the surface, and \( \varepsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

3. **Determine the Effective Charge Enclosed**:
- Since the charge is located at a corner of the cube, it is shared by 8 adjacent cubes. Therefore, the effective charge enclosed by our cube is:
\[ Q_{\text{enc}} = \frac{Q}{8} = \frac{2 \times 10^{-9}}{8} = 2.5 \times 10^{-10} \, \text{C} \]

4. **Calculate Total Electric Flux through the Cube**:
- Using Gauss's Law:
\[ \Phi_{\text{total}} = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{2.5 \times 10^{-10}}{8.85 \times 10^{-12}} \approx 28.24 \, \text{N m}^2/\text{C} \]

5. **Determine the Flux through One Face of the Cube**:
- Since the total flux is distributed equally through the three faces that are adjacent to the corner charge, the flux through one face is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{3} = \frac{28.24}{3} \approx 9.41 \, \text{N m}^2/\text{C} \]

### Final Answer:
The electric flux passing through one face of the cube is approximately \( 9.41 \, \text{N m}^2/\text{C} \).

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