A very small sphere having a charge q, uniformly distributed throughout its volume, is placed at the vertex of a cube of side a. The electric flux through the cube is

Options

A. `(q)/(epsilon_(0))`
B. `(q)/(3epsilon_(0))`
C. `(q)/(6epsilon_(0))`
D. `(q)/(8epsilon_(0))`

Answer

To solve the problem of finding the electric flux through a cube when a small sphere with charge \( q \) is placed at one of its vertices, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
The electric flux \( \Phi \) through a closed surface is given by Gauss's law:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

2. **Charge Position**:
In this case, the charge \( q \) is located at one vertex of the cube. Since the charge is at the vertex, it is not fully enclosed by the cube.

3. **Visualizing the Cube**:
Imagine the cube and the charge at one of its vertices. If we consider the symmetry and the distribution of the charge, we can think about how many identical cubes can be formed around the charge.

4. **Dividing the Space**:
The charge \( q \) at the vertex of the cube is shared by 8 identical cubes (since the cube can be divided into 8 smaller cubes around the vertex). Each of these smaller cubes will contain an equal portion of the total charge.

5. **Charge Enclosed by One Cube**:
Since the charge is shared equally among the 8 cubes, the charge enclosed by one cube is:
\[ Q_{\text{enc}} = \frac{q}{8} \]

6. **Calculating Electric Flux**:
Now, applying Gauss's law to find the electric flux through the cube:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{q/8}{\epsilon_0} = \frac{q}{8\epsilon_0} \]

7. **Final Result**:
Therefore, the electric flux through the cube is:
\[ \Phi = \frac{q}{8\epsilon_0} \]