Question

Given $\lambda = \frac{2nc}{m}$ (linear charge density) for a wire which is passing through the body diagonal of a closed cube of side length $\sqrt{3}$ cm. Find the flux through the cube.

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

Hint:

When calculating the flux through a closed surface due to a linear charge distribution, use Gauss's law and consider the length of the wire inside the surface.

Answer 1

The Correct Option is A

We are given the linear charge density \( \lambda \), which is the charge per unit length, and the wire passes through the body diagonal of a cube with a side length \( \sqrt{3} \, \text{cm} \).
1. **Flux Calculation:** The total flux \( \Phi \) through a closed surface due to a linear charge distribution is given by Gauss's law: \[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}. \] Since the wire is passing through the body diagonal of the cube, it divides the cube into two parts. The flux through the cube depends on how much of the wire is inside the cube.
The total charge enclosed by the cube is the charge on the length of the wire that passes through the body diagonal. The length of the body diagonal of the cube is: \[ L = \sqrt{3} \times \text{side length of the cube} = \sqrt{3} \times \sqrt{3} \, \text{cm} = 3 \, \text{cm}. \]
2. **Charge on the wire:** The total charge \( q_{\text{enclosed}} \) on the wire is given by: \[ q_{\text{enclosed}} = \lambda \times L = \frac{2nc}{m} \times 3. \]
3. **Using Gauss's Law:** We apply Gauss's law for the flux through the cube, noting that the flux is proportional to the charge enclosed. The flux will be: \[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} = 1.44\pi. \] Thus, the correct answer is \( 1.44\pi \).