Question

A charge \( +q \) is at a distance \( L/2 \) above a square of side \( L \). Then what is the flux linked with the surface?

• A. \( \frac{q}{4 \epsilon_0} \)
• B. \( \frac{2q}{3 \epsilon_0} \)
• C. \( \frac{q}{3 \epsilon_0} \)
• D. \( \frac{6q}{\epsilon_0} \)

Hint:

Electric flux is given by \( \Phi = \frac{q_{\text{enc}}}{\epsilon_0} \), where \( q_{\text{enc}} \) is the charge enclosed by the surface.

Answer 1

The Correct Option is C

Step 1: Use Gauss’s law to calculate flux.
The electric flux \( \Phi \) through a surface is given by Gauss's law: \[ \Phi = \frac{q_{\text{enc}}}{\epsilon_0} \] where \( q_{\text{enc}} \) is the charge enclosed by the surface.
Step 2: Conclusion.
The flux linked with the surface is \( \frac{q}{3 \epsilon_0} \).
Final Answer: \[ \boxed{\frac{q}{3 \epsilon_0}} \]