Electric field \( E = \phi x \) N/C, with \( \phi = 1 \).
Cube side length is \( x = 1 \) cm, which is \( 0.01 \) m.
The cube is positioned at a distance of \( x = 1 \) cm (\( 0.01 \) m) from the origin.
The electric field is oriented along the positive \( x \)-axis.
Step 1: Determine the electric field at each relevant face
The cube possesses two faces that are perpendicular to the \( x \)-axis:
The left face is located at \( x = 0.01 \) m, as the cube starts 1 cm from the origin.
The right face is located at \( x = 0.02 \) m, due to the cube's 1 cm side length.
Calculated electric field values:
At the left face: \( E_{\text{left}} = \phi \times 0.01 = 1 \times 0.01 = 0.01 \) N/C.
At the 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
The formula for flux through a face is \( \Phi = E \cdot A \cdot \cos\theta \), where \( \theta \) represents the angle between the electric field and the face's normal vector.
For the left face:
The area \( A \) is \( (0.01)^2 = 1 \times 10^{-4} \) m².
The angle \( \theta \) is \( 180^\circ \) because the electric field points right, and the normal to the left face points left.
The flux \( \Phi_{\text{left}} \) is \( 0.01 \times 1 \times 10^{-4} \times \cos(180^\circ) = -1 \times 10^{-6} \) Wb.
For the right face:
The area \( A \) is \( (0.01)^2 = 1 \times 10^{-4} \) m².
The angle \( \theta \) is \( 0^\circ \) because both the electric field and the normal vector point right.
The flux \( \Phi_{\text{right}} \) is \( 0.02 \times 1 \times 10^{-4} \times \cos(0^\circ) = 2 \times 10^{-6} \) Wb.
For the remaining four faces, which are parallel to the \( x \)-axis, the flux is zero because the electric field is perpendicular to their respective normal vectors.
Step 3: Calculate the net flux
The net flux \( \Phi_{\text{net}} \) is the sum of the fluxes through the left and right faces: \( \Phi_{\text{net}} = \Phi_{\text{left}} + \Phi_{\text{right}} \).
