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 charge \( q \) is placed at one of its vertices, we can use Gauss's law. Here’s a step-by-step solution:

### Step 1: Understanding Gauss's Law
Gauss's law states that the electric flux \( \Phi \) through a closed surface is proportional to the charge \( Q_{\text{enc}} \) enclosed by that surface. Mathematically, it is expressed as:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Identifying the Charge Distribution
In this problem, we have a charge \( q \) uniformly distributed throughout a very small sphere placed at one vertex of a cube. Since the charge is at the vertex, it is important to determine how much of this charge is effectively enclosed by the cube.

### Step 3: Visualizing the Cube and Charge
When the charge is at the vertex of the cube, it can be visualized that the cube occupies one of the eight equal parts (or octants) of the space around the charge.

### Step 4: Calculating the Enclosed Charge
Since the charge is uniformly distributed and the cube occupies only one octant, the charge enclosed by the cube can be calculated as:
\[ Q_{\text{enc}} = \frac{q}{8} \]
This is because the total charge \( q \) is divided equally among the eight octants.

### Step 5: Applying Gauss's Law
Now, we can apply Gauss's law to find the electric flux through the cube:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{q/8}{\epsilon_0} = \frac{q}{8\epsilon_0} \]

### Conclusion
Thus, the electric flux through the cube is:
\[ \Phi = \frac{q}{8\epsilon_0} \]

### Final Answer
The electric flux through the cube is \( \frac{q}{8\epsilon_0} \).

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