Question

A point charge 'q' is placed at the corner of a cube of side 'a' as shown in the figure. What is the electric flux through the face ABCD?

• A. 0
• B. $\frac{q}{ 24 \epsilon_{0} }$
• C. $\frac{q}{ 6 \epsilon_{0} }$
• D. $\frac{q}{ 72 \epsilon_{0} }$

Answer 1

The Correct Option is A

We are given a point charge $q$ placed at the corner of a cube. The problem asks for the electric flux through the face ABCD of the cube. To solve this, we will use Gauss's law, which states that the electric flux through a closed surface is proportional to the charge enclosed by that surface: $$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$$ However, the point charge $q$ is placed at the corner of the cube, which is shared by 8 adjacent cubes. Therefore, the charge $q$ is equally divided among these 8 cubes. Now, we consider the fact that a cube has 6 faces, and the flux is distributed evenly across all faces if the charge is symmetrically placed within the cube. Since the point charge is placed at a corner, it does not contribute any net flux through any single face. Thus, the flux through each face of the cube is: $$\Phi_{\text{face}} = \frac{1}{8} \times \frac{q}{\epsilon_0 \times 6} = 0$$ Hence, the electric flux through the face ABCD is: $${0}$$

Thus, the correct answer is (A) 0.

Answer 2

In this problem, a point charge $q$ is placed at the corner of a cube. To calculate the electric flux through the face ABCD, we use Gauss's Law: $$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$$ where $\Phi$ is the electric flux through a closed surface and $q_{\text{enclosed}}$ is the charge enclosed by that surface. Since the point charge $q$ is placed at the corner of the cube, only a fraction of the total charge contributes to the flux through the face ABCD. A cube has 8 corners, and the charge placed at one corner is shared equally by 8 neighboring cubes. Thus, the charge enclosed by the face ABCD is: $$q_{\text{enclosed}} = \frac{q}{8}$$ The total flux through the entire cube is: $$\Phi_{\text{total}} = \frac{q}{\epsilon_0}$$ Since the cube has 6 faces, and the flux is uniformly distributed, the flux through each face is: $$\Phi_{\text{each face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6 \epsilon_0}$$ However, since the point charge is at the corner, the flux through the face ABCD will be zero, as it is shared among the 8 adjacent faces. Therefore, the electric flux through the face ABCD is: $${0}$$

Thus, the correct answer is (A) 0.