Select the correct statement.

Options

A. Gauss's law is valid for symmetric charge distributions only
B. Electric field intensity calculated by Gauss's law is due to charges enclosed in the Gaussian surface only
C. Charge should not lie on a Gaussian surface.
D. Electric flux associated with a Gaussian surface is not affected by the charges outside the surface

Explanation

To solve the question, we need to analyze the statements given and determine which one is correct based on the principles of Gauss's Law in electrostatics.

### Step-by-Step Solution:

1. **Understanding Gauss's Law**:
Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. Mathematically, it is expressed as:
\[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
where \(\Phi_E\) is the electric flux, \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is the differential area vector, \(Q_{\text{enclosed}}\) is the charge enclosed by the surface, and \(\epsilon_0\) is the permittivity of free space.

2. **Evaluating Option A**:
The statement claims that Gauss's Law is valid only for symmetric charge distributions. This is incorrect. Gauss's Law can be applied to any closed surface regardless of the symmetry of charge distribution. Symmetry simplifies calculations but is not a requirement for the law to hold.

3. **Evaluating Option B**:
This option states that the electric field intensity calculated by Gauss's Law is due to the charge enclosed in the Gaussian surface only. This is also incorrect. The electric field at a point on the Gaussian surface is influenced by all charges, both inside and outside the surface. However, only the charge inside contributes to the net electric flux through the surface.

4. **Evaluating Option C**:
This option suggests that a charge should not lie on the Gaussian surface. This statement is misleading. While it is true that charges on the surface can complicate calculations, Gauss's Law still holds. The electric flux can still be calculated, but one must consider the contribution of the charge on the surface appropriately.

5. **Evaluating Option D**:
This option states that the electric flux associated with a Gaussian surface is not affected by a charge outside the surface. This is correct. Charges outside the Gaussian surface do not contribute to the net electric flux through that surface, as per Gauss's Law.

### Conclusion:
The correct statement is **Option D**: "Electric flux associated with a Gaussian surface is not affected by a charge outside the surface."