Question

Consider an imaginary cube with different vertices of the cube marked as shown. A point charge is placed at the vertex H and flux through the face ABCD is $\phi$ in this case. Now, the charge is shifted to centroid of the cube. Flux through the face AEHD in this case is $k\phi$. What is the value of $k$?

Answer 1

4

Answer 2

The problem involves calculating electric flux using Gauss's Law and principles of symmetry.

Case 1: Point charge 'q' is placed at vertex H.

1. Total flux through a cube due to a charge at its vertex: A point charge placed at a vertex of a cube is effectively shared by 8 identical cubes that meet at that vertex. Therefore, the total electric flux passing through the surface of one such cube is one-eighth of the total flux produced by the charge. $\Phi_{\text{cube, H}} = \frac{1}{8} \frac{q}{\epsilon_0}$

2. Flux through faces meeting at the vertex: The three faces that meet at vertex H (AEHD, DHGC, and EFGH) have the charge lying on their surface. For a point charge on a surface, the electric field lines are tangential to the surface, and thus the electric flux through these faces is considered to be zero.

3. Flux through faces not meeting at the vertex: The remaining three faces (ABCD, AEFB, and BFGC) are symmetric with respect to the charge at H. Therefore, the total flux $\Phi_{\text{cube, H}}$ passes equally through these three faces. Flux through face ABCD, $\phi = \frac{1}{3} \Phi_{\text{cube, H}} = \frac{1}{3} \left(\frac{1}{8} \frac{q}{\epsilon_0}\right) = \frac{q}{24\epsilon_0}$.

Case 2: Point charge 'q' is shifted to the centroid of the cube.

1. Total flux through the cube: When the charge is at the centroid, it is entirely enclosed within the cube. According to Gauss's Law, the total electric flux through the closed surface of the cube is: $\Phi_{\text{cube, centroid}} = \frac{q}{\epsilon_0}$

2. Flux through a single face: A cube has 6 identical faces. Since the charge is at the centroid, it is symmetrically located with respect to all six faces. Therefore, the total flux is equally distributed among these six faces. Flux through face AEHD = $\frac{1}{6} \Phi_{\text{cube, centroid}} = \frac{1}{6} \frac{q}{\epsilon_0}$.

3. Relating the fluxes: The problem states that the flux through face AEHD in this case is $k\phi$. So, $k\phi = \frac{q}{6\epsilon_0}$.

Now, substitute the value of $\phi$ from Case 1 into this equation: $k \left(\frac{q}{24\epsilon_0}\right) = \frac{q}{6\epsilon_0}$

To find $k$, we can cancel out $\frac{q}{\epsilon_0}$ from both sides: $\frac{k}{24} = \frac{1}{6}$ $k = \frac{24}{6}$ $k = 4$

The value of $k$ is 4.