A point charge causes an electric flux of ` -1.0 xx10^3 Nm^(-2)C^(-1)` to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. If the radius of the Gaussian surface were three times, how much flux would pass through the surface

Options

A. `3.0 xx10^3 Nm^2//C`
B. `-1.0 xx10^3 Nm^2//C`
C. `-3.0 xx10^3 Nm^2//C`
D. `-2.0 xx10^3 Nm^2//C`

Explanation

To solve the problem, we need to understand the concept of electric flux and how it relates to a point charge and a Gaussian surface. The electric flux (Φ) through a closed surface is given by Gauss's law, which states:

\[ Φ = \frac{Q_{enc}}{ε_0} \]

where \(Q_{enc}\) is the charge enclosed by the surface and \(ε_0\) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The electric flux through a spherical Gaussian surface of radius 10.0 cm is given as \(Φ = -1.0 \times 10^3 \, \text{Nm}^{-2}\text{C}^{-1}\).
- The radius of the Gaussian surface is increased to three times its original size, which means the new radius is \(3 \times 10.0 \, \text{cm} = 30.0 \, \text{cm}\).

2. **Understand Gauss's Law:**
- According to Gauss's law, the electric flux through a closed surface depends only on the charge enclosed within that surface.
- Since the charge is at the center of the Gaussian surface, changing the radius of the surface does not change the amount of charge enclosed.

3. **Determine the Charge Enclosed:**
- Since the flux through the original surface is \(Φ = -1.0 \times 10^3 \, \text{Nm}^{-2}\text{C}^{-1}\), this indicates that the charge enclosed by the surface is constant regardless of the radius of the surface.

4. **Calculate the Electric Flux for the New Surface:**
- When the radius of the Gaussian surface is increased to 30.0 cm, the charge enclosed remains the same. Therefore, the electric flux through the new surface will also remain the same.
- Thus, the electric flux through the new Gaussian surface of radius 30.0 cm is still \(Φ = -1.0 \times 10^3 \, \text{Nm}^{-2}\text{C}^{-1}\).

### Final Answer:
The electric flux that would pass through the surface with a radius three times larger is still:
\[ Φ = -1.0 \times 10^3 \, \text{Nm}^{-2}\text{C}^{-1} \]