Question

A charge $Q\mu\text{C}$ is placed at the centre of a cube. The flux through two opposite faces of the cube is ($\varepsilon_0 = $ permittivity of free space)}

• A. $\frac{Q}{6\varepsilon_0}$
• B. $\frac{Q}{3\varepsilon_0}$
• C. $\frac{Q}{\varepsilon_0}$
• D. $\frac{Q}{2\varepsilon_0}$

Hint:

Flux through $n$ faces of a symmetric $N$-faced polyhedral with central charge $= \frac{n}{N} \frac{Q}{\varepsilon_0}$.

Answer 1

The Correct Option is B

Step 1: Concept
According to Gauss's Law, the total electric flux $\Phi_{total}$ through a closed surface is $\frac{Q_{enclosed}}{\varepsilon_0}$.
Step 2: Analysis
For a cube with a charge $Q$ at its center, the total flux is $\frac{Q}{\varepsilon_0}$. Since a cube has 6 identical faces, the flux through one face is $\Phi_{face} = \frac{1}{6} \frac{Q}{\varepsilon_0}$.
Step 3: Calculation
The question asks for the flux through two opposite faces. $\Phi_{2 faces} = 2 \times \Phi_{face} = 2 \times \frac{Q}{6\varepsilon_0} = \frac{Q}{3\varepsilon_0}$.
Step 4: Conclusion
Hence, the correct answer is $\frac{Q}{3\varepsilon_0}$.
Final Answer: (B)