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

2. **Identify the Charge**:
In this case, the charge \( Q \) is given as \( Q = Q_{\mu c} = Q \times 10^{-6} \, \text{C} \).

3. **Calculate 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 is:
\[ \Phi_{\text{total}} = \frac{Q \times 10^{-6}}{\epsilon_0} \]

4. **Determine Flux through One Surface**:
A cube has 6 faces. Since the charge is symmetrically placed at the center, the flux will be evenly distributed through all six faces. Therefore, the flux through one face of the cube can be calculated by dividing the total flux by 6:
\[ \Phi_{\text{one face}} = \frac{\Phi_{\text{total}}}{6} = \frac{Q \times 10^{-6}}{6 \epsilon_0} \]

5. **Final Expression**:
Thus, the flux coming out from any surface of the cube is:
\[ \Phi_{\text{one face}} = \frac{Q \times 10^{-6}}{6 \epsilon_0} \]