Question

A charge q is placed at the center of one of the surface of a cube. The flux linked with the cube is :-

• A. \( \frac{q}{4\epsilon_0} \)
• B. \( \frac{q}{2\epsilon_0} \)
• C. \( \frac{q}{8\epsilon_0} \)
• D. Zero

Answer 1

The Correct Option is B

1. **Using Gauss’s Law:**
When a charge \( q \) is placed at the center of one face of a cube, it can be visualized that the charge \( q \) contributes equally to two adjacent cubes.
2. **Flux Calculation:**
According to Gauss’s law, the total flux \( \Phi \) due to charge \( q \) in a closed surface is given by:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0}. \] Since the charge \( q \) is shared equally between two adjacent cubes, the flux through each cube is:
\[ \Phi = \frac{q}{2\epsilon_0}. \]

Answer: \( \frac{q}{2\epsilon_0} \)

Answer 2

Step 1: Recall Gauss’s law
According to Gauss’s law, the total electric flux \(\Phi\) through a closed surface is given by:
\[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} \] where \(q_{\text{enclosed}}\) is the net charge enclosed within the closed surface, and \(\epsilon_0\) is the permittivity of free space.

Step 2: Understand the geometry of the problem
Here, a charge \(q\) is placed at the center of one face (surface) of a cube. The important thing to note is that the charge is not completely inside the cube — it lies exactly on one face. Therefore, only a part of the total electric flux due to this charge will pass through this cube.

Step 3: Visualize the symmetry
If we imagine six identical cubes joined together such that they form a larger cube, then the charge \(q\) at the shared face would be located at the point where all six cubes meet. In such an arrangement, the charge \(q\) would be effectively at the corner of the combined larger cube, and the total flux \(\dfrac{q}{\epsilon_0}\) from the charge would be symmetrically distributed among the six cubes.

Step 4: Divide total flux equally
Since the situation is symmetric and the charge lies on the interface between one cube and the surrounding five, each cube will receive an equal share of the flux. However, note that the given charge is at the center of a face — not on an edge or vertex — so it divides the flux between two half spaces (inside the cube and outside the cube). Therefore, only half of the total flux from the charge passes through the cube.

Thus, the flux through the cube is:
\[ \Phi = \frac{1}{2} \times \frac{q}{\epsilon_0} = \frac{q}{2\epsilon_0} \]

Step 5: Conceptual check
When the charge is completely inside a closed surface, the total flux through that surface is \(\dfrac{q}{\epsilon_0}\). When it is placed on the surface, only half of the field lines enter the closed region, and the rest go outside. Hence, the flux linked with the cube is exactly half of that value.

Final Answer:

\( \frac{q}{2\epsilon_0} \)