What will be the total electric flux through the faces of the cube as given in the figure with side of length 'a' if a charge Q is placed at B, midpoint of an edge of the cube (see figure)?
Show Hint
Remember these symmetry rules for a charge \( Q \) and a cube:
- At the center: \( \Phi = \frac{Q}{\epsilon_0} \)
- On a face center: \( \Phi = \frac{Q}{2\epsilon_0} \) per cube.
- On an edge midpoint: \( \Phi = \frac{Q}{4\epsilon_0} \) per cube.
- On a vertex: \( \Phi = \frac{Q}{8\epsilon_0} \) per cube.
Step 1: Understanding the Question:
The question asks for the net electric flux through a single cube when a point charge is placed exactly at the midpoint of one of its edges. We can use Gauss's Law and symmetry to solve this. Step 2: Key Formula or Approach:
Gauss's Law states that the total flux through a closed surface is \( \Phi = \frac{Q_{enclosed}}{\epsilon_0} \).
If a charge is on the boundary of a surface, we can construct a larger symmetrical surface consisting of identical units to enclose the charge completely. Step 3: Detailed Explanation:
1. The charge \( Q \) is placed at the midpoint of an edge.
2. To enclose this point charge completely and symmetrically, we need 4 such identical cubes sharing that same edge.
3. According to Gauss's Law, the total flux through this large composite surface (consisting of 4 cubes) is \( \Phi_{total} = \frac{Q}{\epsilon_0} \).
4. Since the charge is placed symmetrically with respect to all 4 cubes, the flux through each individual cube will be one-fourth of the total flux.
\[ \Phi_{cube} = \frac{1}{4} \Phi_{total} = \frac{Q}{4\epsilon_0} \]
Step 4: Final Answer:
The total electric flux through the faces of the cube is \(\frac{Q}{4\epsilon_0}\).
