Question

In a cuboid of dimension \(2L \times 2L \times L\), a charge q is placed at the center of the surface ‘S’ having area of 4 L². The flux through the opposite surface to ‘S’ is:

• A. \( \frac{q}{2\epsilon_0} \)
• B. \( \frac{q}{6\epsilon_0} \)
• C. \( \frac{q}{12\epsilon_0} \)
• D. \( \frac{q}{3\epsilon_0} \)

Hint:

For flux calculations:

• Use Gauss’s Law \( \Phi = \frac{q}{\epsilon_0} \) for symmetric surfaces.
• Divide the flux equally among faces if the charge is symmetrically placed.

Answer 1

The Correct Option is B

• The total flux through the cuboid is: \[ \Phi_{\text{total}} = \frac{q}{\epsilon_0}. \]
• As the cuboid has 6 faces, the flux through each face is: \[ \Phi = \frac{q}{6\epsilon_0}. \]