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.
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.
Two point charges +q and −q are held at (a, 0) and (−a, 0) in x-y plane. Obtain an expression for the net electric field due to the charges at a point (0, y). Hence, find electric field at a far off point (y ≫ a).