Question

n electric dipoles are placed inside a closed surface. Total electric flux linked with the closed surface will be:

• A. \( \frac{q}{\epsilon_0} \)
• B. \( \frac{q}{n\epsilon_0} \)
• C. \( \frac{nq}{\epsilon_0} \)
• D. zero

Hint:

For electric dipoles, the net charge is zero, which leads to zero total electric flux linked with a closed surface.

Answer 1

The Correct Option is D

The total electric flux \( \Phi \) linked with a closed surface is given by Gauss's law:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \] Where:
- \( Q_{\text{enc}} \) is the total charge enclosed within the surface,
- \( \epsilon_0 \) is the permittivity of free space.
Now, for a dipole, the net charge enclosed is zero because a dipole consists of two equal and opposite charges. Therefore, the total charge enclosed by the surface is zero.
Hence, the total electric flux linked with the closed surface is:
\[ \Phi = \frac{0}{\epsilon_0} = 0 \] Thus, the correct answer is option (D) zero.