If `oint_(s) E.ds = 0` Over a surface, then

Options

A. the electric field inside the surface and on it is zero.
B. the electric field inside the surface is necessarily uniform.
C. all charges must necessarily be outside the surface.
D. all of these.

Explanation

To solve the question "If \(\oint_{s} \mathbf{E} \cdot d\mathbf{s} = 0\) over a surface, then...", we will analyze the implications of this integral being equal to zero.

### Step-by-Step Solution:

1. **Understanding the Integral**:
The integral \(\oint_{s} \mathbf{E} \cdot d\mathbf{s}\) represents the electric flux \(\Phi\) through a closed surface \(s\). According to Gauss's law, this is related to the charge enclosed by the surface.

2. **Applying Gauss's Law**:
Gauss's law states that:
\[ \Phi = \oint_{s} \mathbf{E} \cdot d\mathbf{s} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \(Q_{\text{enc}}\) is the total charge enclosed by the surface and \(\epsilon_0\) is the permittivity of free space.

3. **Setting the Integral to Zero**:
Given that \(\oint_{s} \mathbf{E} \cdot d\mathbf{s} = 0\), we can set the right-hand side of Gauss's law to zero:
\[ \frac{Q_{\text{enc}}}{\epsilon_0} = 0 \]

4. **Conclusion about Enclosed Charge**:
From the equation above, it follows that:
\[ Q_{\text{enc}} = 0 \]
This means that there is no net charge enclosed within the surface.

5. **Implications for the Electric Field**:
While the electric flux is zero, it does not necessarily imply that the electric field \(\mathbf{E}\) is zero everywhere on the surface or inside it. The electric field could be present but could also cancel out over the surface.

6. **Final Conclusion**:
Thus, we can conclude that:
- The electric field inside the surface is not necessarily zero.
- The electric field can be non-uniform.
- The only certain conclusion is that there is no net charge inside the surface.

### Answer:
The correct conclusion is that the total charge enclosed by the surface is zero, which implies that all charge must be outside the surface.