A : We may have a Gaussian surface in which less number of field lines enter and more field lines come out.
R : The electric field E in the Gauss's law is only due to the enclosed charges.

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1).
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).
C. If Assertion is true statement but Reason is false, then mark (3).
D. If both Assertion and Reason are false statements, then mark (4).

Answer

To solve the assertion and reason question, we will analyze both statements step by step.

### Step 1: Understanding the Assertion
The assertion states that we may have a Gaussian surface in which fewer field lines enter and more field lines come out. This implies that the net electric flux through the Gaussian surface is positive.

**Hint:** Consider the concept of electric field lines and how they relate to electric flux.

### Step 2: Applying Gauss's Law
According to Gauss's Law, the electric flux (Φ) through a closed surface is given by the equation:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
where \(Q_{\text{enclosed}}\) is the charge enclosed by the Gaussian surface and \(\epsilon_0\) is the permittivity of free space.

**Hint:** Remember that electric flux is related to the number of electric field lines passing through a surface.

### Step 3: Analyzing the Electric Field Lines
If fewer field lines enter the Gaussian surface than leave it, this means that the net electric flux is positive. This indicates that the charge enclosed by the surface must also be positive (since positive flux corresponds to positive charge).

**Hint:** Think about how the direction of electric field lines indicates the nature of the charge (positive or negative).

### Step 4: Understanding the Reason
The reason states that the electric field \(E\) in Gauss's Law is only due to the enclosed charges. However, this is not entirely accurate. While the electric field inside the Gaussian surface is influenced by the enclosed charges, it can also be influenced by external charges.

**Hint:** Consider how electric fields from external charges can affect the net electric field at points inside a Gaussian surface.

### Step 5: Conclusion
- The assertion is **true** because it is indeed possible to have a Gaussian surface where fewer field lines enter than exit, resulting in a positive net flux.
- The reason is **false** because the electric field \(E\) in Gauss's Law is influenced by both enclosed charges and external charges.

**Final Answer:** The assertion is true, but the reason is false.