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:
Show 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.
\( \frac{q}{2\epsilon_0} \)
\( \frac{q}{6\epsilon_0} \)
\( \frac{q}{12\epsilon_0} \)
\( \frac{q}{3\epsilon_0} \)
The Correct Option is B
Solution and Explanation
- 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: \[ \Phi = \frac{q}{6\epsilon_0}. \]
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