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