A point charge Q is placed at one of the vertices of a cubical block. The electric flux flowing through this cube is

Options

A. `Q/(6 epsilon_(0))`
B. `Q/(4 epsilon_(0))`
C. `Q/(8epsilon_(0))`
D. `Q/(epsilon_(0))`

Explanation

To find the electric flux flowing through a cube when a point 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 equal to the charge \( Q_{\text{enc}} \) enclosed by that surface divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]

### Step 2: Visualizing the Charge and the Cube
In this problem, the charge \( Q \) is placed at one of the vertices of a cube. Since the charge is at the vertex, it is not entirely enclosed by the cube. Instead, it is shared among several cubes.

### Step 3: Considering the Symmetry
When the charge \( Q \) is at the vertex of the cube, it can be thought of as being shared by 8 identical cubes that meet at that vertex. Therefore, the charge \( Q \) contributes to the flux in each of these 8 cubes.

### Step 4: Calculating the Flux for One Cube
Since the total charge \( Q \) is shared equally among the 8 cubes, the effective charge enclosed by one cube is:
\[ Q_{\text{enc}} = \frac{Q}{8} \]
Now, applying Gauss's Law for one cube:
\[ \Phi_{\text{cube}} = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{Q/8}{\epsilon_0} = \frac{Q}{8\epsilon_0} \]

### Step 5: Conclusion
Thus, the electric flux flowing through the cube is:
\[ \Phi = \frac{Q}{8\epsilon_0} \]

### Summary of the Solution
The electric flux through the cube due to the point charge \( Q \) located at one of its vertices is given by:
\[ \Phi = \frac{Q}{8\epsilon_0} \]

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