What is the flux link through a cube of side l if a charge 2q is placed at one end of cube ?

Options

A. `(q)/(epsilon_(0))`
B. `(q)/(8epsilon_(0))`
C. `(2q)/(epsilon_(0))`
D. `(q)/(4epsilon_(0))`

Explanation

To solve the problem of finding the electric flux through a cube of side \( l \) when a charge \( 2q \) is placed at one end of the cube, 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 \( Q \) enclosed by that surface divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q}{\epsilon_0} \]

2. **Identify the Charge and Geometry**:
In this case, we have a charge \( 2q \) placed at one end of the cube. Since the charge is not enclosed within the cube, we need to consider how much of the total flux from this charge actually passes through the cube.

3. **Determine the Total Flux from the Charge**:
The total electric flux produced by the charge \( 2q \) can be calculated using Gauss's Law:
\[ \Phi_{\text{total}} = \frac{2q}{\epsilon_0} \]

4. **Calculate the Fraction of Flux through the Cube**:
Since the charge \( 2q \) is located at one corner of the cube, we can visualize that the total flux emanating from the charge is distributed equally among the eight surrounding cubes (since a charge at a corner of a cube is shared by eight cubes). Therefore, the flux through the cube is:
\[ \Phi_{\text{cube}} = \frac{\Phi_{\text{total}}}{8} = \frac{2q}{8\epsilon_0} = \frac{q}{4\epsilon_0} \]

5. **Final Result**:
Thus, the electric flux through the cube is:
\[ \Phi = \frac{q}{4\epsilon_0} \]

### Conclusion:
The electric flux through the cube when a charge \( 2q \) is placed at one end of the cube is \( \frac{q}{4\epsilon_0} \).