Question

For the given diagram, E=ϕxN/C, the net flux through the cube of side x=1 cm, placed at x=1 cm from the origin is:

• A. \(2 \times 10^{-6} \, \text{Wb} \\\)
• B. \(1 \times 10^{-6} \, \text{Wb} \\\)
• C. \(3 \times 10^{-6} \, \text{Wb} \\\)
• D. \(4 \times 10^{-6} \, \text{Wb} \\\)

Answer 1

The Correct Option is B

Given:

• Electric field \( E = \phi x \) N/C (where \( \phi = 1 \))
• Cube side length \( x = 1 \) cm \( = 0.01 \) m
• Cube is placed at \( x = 1 \) cm \( = 0.01 \) m from the origin
• Electric field is directed along the positive \( x \)-axis

Step 1: Determine the electric field at each face

The cube has two faces perpendicular to the \( x \)-axis:

• Left face at \( x = 0.01 \) m (since the cube is placed 1 cm from the origin)
• Right face at \( x = 0.02 \) m (since the cube has side length 1 cm)

Electric field values:

• At left face: \( E_{\text{left}} = \phi \times 0.01 = 1 \times 0.01 = 0.01 \) N/C
• At right face: \( E_{\text{right}} = \phi \times 0.02 = 1 \times 0.02 = 0.02 \) N/C

Step 2: Calculate the flux through each face

Flux through a face is given by \( \Phi = E \cdot A \cdot \cos\theta \), where \( \theta \) is the angle between \( E \) and the normal to the face.

For the left face:

• Area \( A = (0.01)^2 = 1 \times 10^{-4} \) m²
• \( \theta = 180^\circ \) (since \( E \) points right and the normal to the left face points left)
• \( \Phi_{\text{left}} = 0.01 \times 1 \times 10^{-4} \times \cos(180^\circ) = -1 \times 10^{-6} \) Wb

For the right face:

• Area \( A = (0.01)^2 = 1 \times 10^{-4} \) m²
• \( \theta = 0^\circ \) (since \( E \) and the normal both point right)
• \( \Phi_{\text{right}} = 0.02 \times 1 \times 10^{-4} \times \cos(0^\circ) = 2 \times 10^{-6} \) Wb

For the other four faces (parallel to the \( x \)-axis), the flux is zero because \( E \) is perpendicular to their normals.

Step 3: Calculate the net flux

Net flux \( \Phi_{\text{net}} = \Phi_{\text{left}} + \Phi_{\text{right}} \)

\( \Phi_{\text{net}} = -1 \times 10^{-6} + 2 \times 10^{-6} = 1 \times 10^{-6} \) Wb

Answer 2

\(\textbf{Solution:} \text{ The electric field } E = \phi x \, \text{N/C}, \text{ where } \phi = 1. \text{ The flux through the cube is calculated using:}\)
\(\Phi = E \times A\)
\(\text{Given that the area } A \text{ is the surface area of the cube and the electric field changes with position, the net flux is computed as:}\)
\(\Phi = \phi \times 1 \times 10^{-6} = 1 \times 10^{-6} \, \text{Wb}\)
\(\text{Thus, the correct answer is } 1 \times 10^{-6} \, \text{Wb}.\)