The electric flux through a chlosed surface area S enclosing charge Q is `phi`. If the surface area is doubled, then the flux is

Options

A. `2phi`
B. `phi//2`
C. `phi//4`
D. `phi`

Explanation

To solve the problem, we need to understand the relationship between electric flux, charge, and the area of the surface. We will use Gauss's law, which states that the electric flux (Φ) through a closed surface is directly proportional to the charge (Q) enclosed by that surface.

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
The electric flux (Φ) through a closed surface is given by Gauss's law:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
where \(Q_{\text{enclosed}}\) is the charge enclosed by the surface and \(\epsilon_0\) is the permittivity of free space.

2. **Initial Conditions**:
We are given that the electric flux through a closed surface area \(S\) enclosing charge \(Q\) is \(\Phi\). Thus, we can write:
\[ \Phi = \frac{Q}{\epsilon_0} \]

3. **Doubling the Surface Area**:
Now, we are told that the surface area is doubled. However, it is important to note that the electric flux depends only on the charge enclosed and not on the area of the surface itself.

4. **Charge Enclosed**:
Since the charge \(Q\) remains the same when we double the surface area, the enclosed charge does not change. Therefore, we still have:
\[ Q_{\text{enclosed}} = Q \]

5. **Calculating New Flux**:
The electric flux through the new surface area (which is double the original) can still be calculated using Gauss's law:
\[ \Phi' = \frac{Q_{\text{enclosed}}}{\epsilon_0} = \frac{Q}{\epsilon_0} \]
Since \(Q_{\text{enclosed}} = Q\), we find that:
\[ \Phi' = \Phi \]

6. **Conclusion**:
The electric flux remains the same, even after doubling the surface area. Therefore, the new flux is:
\[ \Phi' = \Phi \]

### Final Answer:
The electric flux through the closed surface after doubling the surface area remains \(\Phi\).

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