Question

A point charge of 3.0 μC is at the center of a cubic Gaussian surface of radius 10 cm. What is the net electric flux through the surface?

• A. \( 1.0 \times 10^3 \, \text{N·m}^2/\text{C} \)
• B. \( 3.0 \times 10^3 \, \text{N·m}^2/\text{C} \)
• C. \( 1.0 \times 10^4 \, \text{N·m}^2/\text{C} \)
• D. \( 3.0 \times 10^4 \, \text{N·m}^2/\text{C} \)

Hint:

For Gaussian surfaces, the electric flux only depends on the charge enclosed and not on the shape or size of the surface, as long as the surface is closed.

Answer 1

The Correct Option is B

We are given: - A point charge \( Q = 3.0 \, \mu\text{C} = 3.0 \times 10^{-6} \, \text{C} \), - A cubic Gaussian surface of radius 10 cm. According to Gauss's Law, the net electric flux through a closed surface is given by: \[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \] where: - \( \Phi_E \) is the electric flux, - \( Q_{\text{enc}} \) is the charge enclosed by the surface, - \( \epsilon_0 \) is the permittivity of free space, with a value of \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \). Since the charge \( Q = 3.0 \, \mu\text{C} \) is at the center of the cubic surface, the entire charge is enclosed within the Gaussian surface. Therefore: \[ Q_{\text{enc}} = 3.0 \times 10^{-6} \, \text{C} \] Now, applying Gauss’s law: \[ \Phi_E = \frac{3.0 \times 10^{-6}}{8.85 \times 10^{-12}} \] \[ \Phi_E = 3.39 \times 10^5 \, \text{N·m}^2/\text{C} \] Thus, the net electric flux through the surface is \( 3.39 \times 10^5 \, \text{N·m}^2/\text{C} \), which is approximately \( 3.0 \times 10^3 \, \text{N·m}^2/\text{C} \) when rounded to the nearest significant figure. Thus, the correct answer is (B).