A charge Q `muc` is placed at the centre of cube, the flux coming out from any surfaces will be : -

Options

A. `Q/(6epsilon_0)xx10^(-6)`
B. `Q/(6epsilon_0)xx10^(-3)`
C. `Q/(2epsilon_0)`
D. `Q/(8epsilon_0)`

Answer

To solve the problem of finding the electric flux coming out from any surface of a cube with a charge \( Q \) (in microcoulombs) placed at its center, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Charge and the Cube**:
Given that a charge \( Q = Q_{\mu c} = Q \times 10^{-6} \) Coulombs is placed at the center of the cube, we can denote this charge as \( Q \).

3. **Calculate the Total Flux through the Cube**:
Since the charge is at the center of the cube, the total electric flux through the entire surface of the cube can be calculated as:
\[ \Phi_{\text{total}} = \frac{Q}{\epsilon_0} \]

4. **Determine the Number of Surfaces**:
A cube has 6 surfaces. Since the charge is symmetrically placed at the center, the flux will be uniformly distributed across all surfaces.

5. **Calculate the Flux through Each Surface**:
To find the flux through one surface of the cube, we divide the total flux by the number of surfaces:
\[ \Phi_{\text{each surface}} = \frac{\Phi_{\text{total}}}{6} = \frac{Q}{6 \epsilon_0} \]

6. **Final Expression**:
Substituting \( Q = Q \times 10^{-6} \) into the equation gives:
\[ \Phi_{\text{each surface}} = \frac{Q \times 10^{-6}}{6 \epsilon_0} \]

### Conclusion:
The flux coming out from any surface of the cube is given by:
\[ \Phi_{\text{each surface}} = \frac{Q \times 10^{-6}}{6 \epsilon_0} \]