Question

Two point charges 2q and q are placed at vertex A and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :
\includegraphics[width=0.3\linewidth]{Screenshot 2026-02-04 151639.png}

• A. 3q/ε₀
• B. 3q/(4ε₀)
• C. q/ε₀
• D. 3q/(2ε₀)

Hint:

To remember sharing fractions: a vertex is shared by 8 cubes, an edge by 4, and a face by 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
According to Gauss's Law, the total electric flux through a closed surface is equal to the net charge enclosed divided by $\varepsilon_0$. When charges are on the boundary (vertex or face), we use symmetry to determine what fraction of the charge is "inside" the volume.

Step 2: Key Formula or Approach:
1. Flux due to charge at vertex: $\phi_v = \frac{q_{enclosed}}{\varepsilon_0} = \frac{1}{8} \frac{Q}{\varepsilon_0}$.
2. Flux due to charge at face center: $\phi_f = \frac{q_{enclosed}}{\varepsilon_0} = \frac{1}{2} \frac{Q}{\varepsilon_0}$.
Step 3: Detailed Explanation:
For the charge $2q$ at vertex A: This charge is shared by 8 identical cubes meeting at that vertex. Thus, the contribution to one cube is $\frac{2q}{8\varepsilon_0} = \frac{q}{4\varepsilon_0}$.
For the charge $q$ at the center of face CDEF: This charge is shared by 2 identical cubes meeting at that face. Thus, the contribution to one cube is $\frac{q}{2\varepsilon_0}$.
Total flux $\Phi = \frac{q}{4\varepsilon_0} + \frac{q}{2\varepsilon_0} = \frac{q + 2q}{4\varepsilon_0} = \frac{3q}{4\varepsilon_0}$.
Step 4: Final Answer:
The electric flux passing through the cube is 3q/(4ε₀).