Question

Given $ \lambda = 2 \, \text{nC/m} $ (linear charge density) for a wire passing through the body diagonal of a closed cube with side length $ \sqrt{3} \, \text{cm} $, find the flux through the cube.

• A. \( 1.44 \pi \)
• B. \( 0.72 \pi \)
• C. \( 2.16 \pi \)
• D. \( 6.84 \pi \)

Hint:

For electric flux problems involving linear charge density, use Gauss's law. The charge enclosed is determined by the length of the wire passing through the closed surface. Ensure to use the proper units for length and charge density to obtain the correct flux.

Answer 1

The Correct Option is C

Step 1: Total Length of the Wire Inside the Cube.
The total length of the wire inside the cube is the length of the body diagonal of the cube.
The body diagonal \( d \) of a cube with side length \( \ell \) is given by: \[ d = \sqrt{3} \ell \] Given that the side length \( \ell = \sqrt{3} \, \text{cm} = 3 \times 10^{-2} \, \text{m} \), we have: \[ d = 3 \times 10^{-2} \, \text{m} \]
Step 2: Calculate the Total Charge Enclosed.
The total charge \( Q_{\text{enc}} \) is the product of the linear charge density \( \lambda \) and the length of the wire inside the cube: \[ Q_{\text{enc}} = \lambda \times d = (2 \times 10^{-9} \, \text{C/m}) \times (3 \times 10^{-2} \, \text{m}) = 6 \times 10^{-11} \, \text{C} \]
Step 3: Apply Gauss's Law to Find the Flux.
Using Gauss's law, the flux \( \Phi_E \) through the cube is given by: \[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \] where \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2 / \text{N m}^2 \).
Now substitute the values: \[ \Phi_E = \frac{6 \times 10^{-11}}{8.854 \times 10^{-12}} = 6.8 \times 10^9 \, \text{N m}^2 / \text{C} \] Now, we multiply by \( 4\pi \) to get the flux: \[ \Phi_E = 2 \times 10^{-9} \times 3 \times 10^{-2} \times 4\pi \times 9 \times 10^9 = 2.16 \pi \]
Step 4: Final Answer.
The flux through the cube is \( 2.16 \pi \), corresponding to option (3).