Gauss theorem:
Gauss theorem is mainly used to find out the electric flux linked to a closed surface. It does not depend upon the shape or size of surface. According to this theorem, the electric flux linked to a closed surface is equal to `((1)/(epsilon_(0)))` times the charge enclosed by the surface.
Let we have a charge q, now if we want to find out the net flux linked to a closed surface around it them,
Electric flux `phi = oint_(s) vecE. vecds = (q)/(epsilon_(0))`
If we increase the charge enclosed by the surface then electric flux will:

Options

A. Increases
B. Decreases
C. Remain same
D. Both (a) and (b)

Explanation

To solve the problem regarding the effect of increasing the charge enclosed by a closed surface on the electric flux linked to that surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Gauss's Law**:
Gauss's Law states that the electric flux (Φ) through a closed surface is directly proportional to the total charge (q) enclosed within that surface. Mathematically, it can be expressed as:
\[ \Phi = \frac{q}{\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Identify the Variables**:
- Let \( q \) be the charge enclosed by the surface.
- \( \epsilon_0 \) is a constant (approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)).
- The electric flux \( \Phi \) is what we want to analyze as we change \( q \).

3. **Consider the Effect of Increasing Charge**:
If we increase the charge \( q \) enclosed by the surface, we denote the new charge as \( q' \) where \( q' > q \). According to Gauss's Law:
\[ \Phi' = \frac{q'}{\epsilon_0} \]

4. **Establish the Relationship**:
Since \( \epsilon_0 \) is a constant, if \( q' \) increases, the electric flux \( \Phi' \) will also increase. This can be expressed as:
\[ \Phi' = \frac{q'}{\epsilon_0} > \frac{q}{\epsilon_0} = \Phi \]
This shows that the electric flux is directly proportional to the charge enclosed.

5. **Conclusion**:
Therefore, if we increase the charge enclosed by the surface, the electric flux linked to that surface will also increase.

### Final Answer:
The electric flux will **increase** if we increase the charge enclosed by the surface.

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