There is a closed three-dimensional surface and the electric flux associated with it is zero. Which of the following statements is correct?

Options

A. Electric field intensity everywhere on surface must be zero
B. Net enclosed charge within the surface in zero
C. No charged particle must be there inside surface
D. None

Explanation

To solve the question regarding the electric flux associated with a closed three-dimensional surface being zero, we can analyze the implications of this condition step by step.

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
Electric flux (Φ) through a closed surface is defined by Gauss's Law as:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
where \(Q_{\text{enclosed}}\) is the total charge enclosed within the surface and \(\epsilon_0\) is the permittivity of free space.

2. **Condition Given**:
The problem states that the electric flux associated with the closed surface is zero:
\[ \Phi = 0 \]

3. **Applying Gauss's Law**:
From Gauss's Law, if the electric flux is zero, this implies:
\[ \frac{Q_{\text{enclosed}}}{\epsilon_0} = 0 \]
Therefore, we conclude:
\[ Q_{\text{enclosed}} = 0 \]
This means that the net charge enclosed within the surface is zero.

4. **Electric Field Intensity**:
It is important to note that a zero electric flux does not necessarily mean that the electric field intensity (E) on the surface is zero. The electric field can exist, but the angle between the electric field vector and the area vector (dS) could be such that:
\[ E \cdot dS = E \cdot A \cdot \cos(\theta) = 0 \]
This can happen if \(\theta = 90^\circ\), meaning the electric field lines are parallel to the surface and do not penetrate it.

5. **Conclusion**:
Therefore, the correct statement is that the net enclosed charge within the surface is zero, but the electric field intensity on the surface can still be non-zero.

### Final Answer:
The correct statement is: **The net enclosed charge within the surface is zero.**

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