A point charge q is placed at one corner of a cube of edge a. The flux through each of the cube faces is

Options

A. `q/(epsilon_(0))`
B. `q/(16 epsilon_(0))`
C. `q/(24 epsilon_(0))`
D. `q/(48 epsilon_(0))`

Answer

To solve the problem of finding the electric flux through each face of a cube with a point charge \( q \) placed at one corner, we can follow these steps:

### Step 1: Understand the Setup
We have a cube with edge length \( a \) and a point charge \( q \) located at one of its corners. The cube has 6 faces, and we need to determine the electric flux through each face.

### Step 2: Use Gauss's Law
According to Gauss's Law, the electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the total charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Enclosed Charge Calculation
Since the charge \( q \) is located at a corner of the cube, it is shared by multiple cubes. If we consider a larger cube that encompasses the smaller cube with the charge at the corner, we can visualize that the charge \( q \) is effectively shared among 8 smaller cubes (since each corner of a cube is shared by 8 adjacent cubes).

Thus, the charge enclosed by the larger cube is:
\[ Q_{\text{enc}} = q \]

### Step 4: Total Flux Through the Larger Cube
Using Gauss's Law for the larger cube, the total electric flux through the entire surface of the larger cube is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

### Step 5: Flux Distribution
Since the charge is at the corner, the total flux is distributed equally among the 8 smaller cubes. Therefore, the flux through one cube is:
\[ \Phi_{\text{cube}} = \frac{q}{8\epsilon_0} \]

### Step 6: Flux Through Each Face
Now, we need to find the flux through each face of the cube. The cube has 3 faces that are directly in contact with the charge. The flux through these faces is zero because the electric field lines from the charge do not penetrate these surfaces (the electric field lines are directed outward).

The remaining flux through the other 3 faces (which are not in contact with the charge) must account for the total flux through the cube. Since the flux is uniformly distributed, we can calculate the flux through each of these 3 faces as follows:
\[ \Phi_{\text{each face}} = \frac{\Phi_{\text{cube}}}{3} = \frac{q}{8\epsilon_0 \cdot 3} = \frac{q}{24\epsilon_0} \]

### Final Answer
Thus, the electric flux through each face of the cube is:
\[ \Phi = \frac{q}{24\epsilon_0} \]