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. increase four times

Explanation

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

\[ \Phi_E = \frac{Q}{\epsilon_0} \]

where \( \epsilon_0 \) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the Charge and Gaussian Surface**:
- We have a charge \( Q \) enclosed by a Gaussian spherical surface of radius \( R \).

2. **Apply Gauss's Law**:
- According to Gauss's Law, the electric flux through the Gaussian surface is given by:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
- This equation shows that the electric flux depends only on the charge \( Q \) enclosed within the surface and not on the radius of the surface.

3. **Change the Radius of the Gaussian Surface**:
- Now, we double the radius of the Gaussian surface, making it \( 2R \).

4. **Reapply Gauss's Law**:
- Even with the new radius \( 2R \), the charge \( Q \) remains enclosed within the surface. Therefore, the electric flux through the new Gaussian surface is still:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]

5. **Conclusion**:
- Since the electric flux does not depend on the radius of the Gaussian surface, the outward electric flux remains the same when the radius is doubled.

### Final Answer:
The outward electric flux will remain the same.