A charge `Q` is enclosed by a Gaussian spherical surface of radius `R`. If the radius is doubled, then the outward electric flux will

Options

A. be reduced to half
B. remain the same
C. be doubled
D. increases four times

Answer

To solve the problem, we will use Gauss's Law, which states that the electric flux (Φ) through a closed surface is proportional to the charge (Q) enclosed by that surface. The formula for Gauss's Law is given by:

\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]

where:
- \(\Phi\) is the electric flux,
- \(Q_{\text{enclosed}}\) is the total charge enclosed within the Gaussian surface,
- \(\epsilon_0\) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- A charge \(Q\) is enclosed by a Gaussian spherical surface of radius \(R\).
- The electric flux through this surface can be calculated using Gauss's Law.

2. **Calculate the initial electric flux:**
- According to Gauss's Law, the electric flux when the radius is \(R\) is:
\[ \Phi = \frac{Q}{\epsilon_0} \]

3. **Change the radius of the Gaussian surface:**
- Now, the radius of the Gaussian surface is doubled to \(2R\).
- However, the charge \(Q\) remains the same and is still enclosed by the surface.

4. **Calculate the new electric flux:**
- The electric flux through the new Gaussian surface (with radius \(2R\)) is still given by Gauss's Law:
\[ \Phi' = \frac{Q}{\epsilon_0} \]

5. **Compare the initial and new electric flux:**
- Since the charge \(Q\) enclosed by the Gaussian surface has not changed, the electric flux remains the same:
\[ \Phi' = \Phi = \frac{Q}{\epsilon_0} \]

### Conclusion:
The outward electric flux remains the same even when the radius of the Gaussian surface is doubled. Therefore, the answer is that the electric flux does not change.