Question

The line $ag$ in figure is a diagonal of a cube. A point charge $q$ is located at the vertex $a$ of the cube. Determine the electric flux through each of the sides of the cube which meet at the point $a$.

ILLUSTRATION 53

• A. $\frac{q}{8\epsilon_0}$
• B. $\frac{q}{6\epsilon_0}$
• C. $\frac{q}{24\epsilon_0}$
• D. $\frac{q}{4\pi\epsilon_0}$

Answer 1

Answer: (C)

$\frac{q}{24\epsilon_0}$

Answer 2

The problem asks for the electric flux through each of the three faces of a cube that meet at a vertex where a point charge $q$ is located.

We can solve this problem using the concept of solid angles. The electric flux $\Phi$ through a surface is given by the formula: $\Phi = \frac{q}{4\pi\epsilon_0} \Omega$ where $\Omega$ is the solid angle subtended by the surface at the location of the charge $q$.

Let the charge $q$ be placed at vertex $a$ of the cube. There are three faces of the cube that meet at vertex $a$. Let these faces be $F_1, F_2,$ and $F_3$. These three faces are mutually perpendicular, forming a corner of the cube.

The solid angle subtended by the entire cube at vertex $a$ is $\frac{1}{8}$ of the total solid angle of a sphere ($4\pi$ steradians). Therefore, the solid angle of the cube at vertex $a$ is: $\Omega_{cube} = \frac{4\pi}{8} = \frac{\pi}{2} \text{ steradians}$

This solid angle $\Omega_{cube}$ is formed by the three faces $F_1, F_2,$ and $F_3$ meeting at vertex $a$. By symmetry, the solid angle subtended by each of these three faces at vertex $a$ is equal. Thus, the solid angle subtended by each face is: $\Omega_{face} = \frac{\Omega_{cube}}{3} = \frac{\pi/2}{3} = \frac{\pi}{6} \text{ steradians}$

Now, we can calculate the electric flux through each of these three faces using the flux formula: $\Phi_{face} = \frac{q}{4\pi\epsilon_0} \Omega_{face} = \frac{q}{4\pi\epsilon_0} \left(\frac{\pi}{6}\right) = \frac{q}{24\epsilon_0}$

Therefore, the electric flux through each of the sides of the cube which meet at the point $a$ is $\frac{q}{24\epsilon_0}$.