If we increase the charge enclosed by the surface then electric flux will :

Options

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

Answer

To solve the question, we need to understand the relationship between the charge enclosed by a surface and the electric flux through that surface, as described by Gauss's Law.

### Step-by-Step Solution:

1. **Understanding Electric Flux**: Electric flux (Φ) through a closed surface is defined as the total electric field (E) passing through that surface. Mathematically, it is given by:
\[ \Phi = \int \vec{E} \cdot d\vec{A} \]
where \(d\vec{A}\) is an infinitesimal area vector on the surface.

2. **Gauss's Law**: According to Gauss's Law, the electric flux through a closed surface is proportional to the net charge (Q) enclosed within that surface:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \(Q_{\text{enc}}\) is the total charge enclosed by the surface and \(\epsilon_0\) is the permittivity of free space.

3. **Increasing the Charge**: If we increase the charge enclosed by the surface, say from \(Q\) to \(Q + \Delta Q\) (where \(\Delta Q\) is the additional charge), the new flux will be:
\[ \Phi' = \frac{Q + \Delta Q}{\epsilon_0} \]

4. **Comparing Initial and New Flux**: The initial flux was:
\[ \Phi = \frac{Q}{\epsilon_0} \]
The new flux after increasing the charge is:
\[ \Phi' = \frac{Q + \Delta Q}{\epsilon_0} \]
Since \(\Delta Q > 0\) (we are increasing the charge), it follows that:
\[ \Phi' > \Phi \]

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

### Final Answer:
If we increase the charge enclosed by the surface, then the electric flux will **increase**.

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