The length of each side of a cubical closed surface is L metre. If charge 48C is situated at one of the corners of the cube, Find the flux passing through the cube. (In Volt-metre)

Options

A. `(6)/(in_(0))`
B. `(3)/(in_(0))`
C. `(48)/(in_(0))`
D. `(8)/(in_(0))`

Answer

To find the electric flux passing through a cubical closed surface with a charge located at one of its corners, we can use Gauss's law. Gauss's law states that the electric flux (Φ) through a closed surface is equal to the charge (Q) enclosed by that surface divided by the permittivity of free space (ε₀).

### Step-by-Step Solution:

1. **Identify the Charge and the Surface**:
- We have a charge \( Q = 48 \, C \) located at one corner of a cube with side length \( L \).

2. **Determine the Enclosed Charge**:
- Since the charge is located at one corner of the cube, we need to determine how much of this charge is enclosed by the cube. A cube has 8 corners, and the charge at one corner is shared among 8 adjacent cubes.
- Therefore, the charge enclosed by the cube is:
\[ Q_{\text{enclosed}} = \frac{Q}{8} = \frac{48 \, C}{8} = 6 \, C \]

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

4. **Calculate the Electric Flux**:
- Substituting the values into the equation:
\[ \Phi = \frac{6 \, C}{8.85 \times 10^{-12} \, C^2/(N \cdot m^2)} \]
- Performing the calculation:
\[ \Phi \approx \frac{6}{8.85 \times 10^{-12}} \approx 6.78 \times 10^{11} \, N \cdot m^2/C \]

5. **Final Result**:
- The electric flux passing through the cube is approximately \( 6.78 \times 10^{11} \, N \cdot m^2/C \) or \( 6.78 \times 10^{11} \, V \cdot m \) (since \( 1 \, N \cdot m^2/C = 1 \, V \cdot m \)).

### Summary:
The electric flux through the cube is approximately \( 6.78 \times 10^{11} \, V \cdot m \).