An electric field \( \vec{E} = (10x + 5) \hat{i} \, \text{N/C} \) exists in a region in which a cube of side \( L \) is kept as shown in the figure. Here \( x \) and \( L \) are in metres. Calculate the net flux through the cube.
Show Hint
When calculating the electric flux through a cube in a non-uniform electric field, consider the contribution of the flux from the faces perpendicular to the electric field direction. Only the faces parallel to the direction of the electric field contribute to the net flux.
We are given an electric field \( \vec{E} = (10x + 5) \hat{i} \) N/C, where \( x \) is the position along the x-axis. The flux through the cube can be calculated using Gauss's law, which states that the net electric flux \( \Phi_E \) through a closed surface is equal to the charge enclosed divided by the permittivity of free space \( \epsilon_0 \). Mathematically:
\[
\Phi_E = \oint \vec{E} \cdot d\vec{A}
\]
where \( d\vec{A} \) is the vector area element on the surface of the cube.
Step 1: Electric Field Expression
The electric field is given as \( \vec{E} = (10x + 5) \hat{i} \). This means the electric field has a component along the x-axis, and it depends on the x-coordinate.
Step 2: Flux Through Each Face
The cube has six faces, and the flux through each face depends on the electric field and the orientation of the face. The area vector for each face is perpendicular to the surface, and the flux through each face is calculated by the dot product \( \vec{E} \cdot d\vec{A} \).
For a face of the cube, the electric flux is given by:
\[
\Phi_{\text{face}} = \vec{E} \cdot A
\]
where \( A \) is the area of the face of the cube. Since the electric field is only along the x-axis, the flux through the faces that are parallel to the yz-plane (i.e., the faces at \( x = 0 \) and \( x = L \)) will contribute to the total flux.
Step 3: Calculate the Flux Through the Faces at \( x = 0 \) and \( x = L \)
1. Face at \( x = 0 \):
The electric field at \( x = 0 \) is \( \vec{E} = 5 \hat{i} \) N/C. The area of the face at \( x = 0 \) is \( A = L^2 \), and the area vector is in the negative x-direction. Therefore, the flux through this face is:
\[
\Phi_1 = E_x \cdot A = 5 \times L^2 = 5L^2
\]
2. Face at \( x = L \):
The electric field at \( x = L \) is \( \vec{E} = (10L + 5) \hat{i} \) N/C. The area vector is in the positive x-direction, so the flux through this face is:
\[
\Phi_2 = E_x \cdot A = (10L + 5) \times L^2 = (10L + 5) L^2
\]
Step 4: Total Flux
The total flux through the cube is the sum of the flux through the two faces (at \( x = 0 \) and \( x = L \)):
\[
\Phi_{\text{total}} = \Phi_2 - \Phi_1 = (10L + 5) L^2 - 5L^2
\]
\[
\Phi_{\text{total}} = 10L^3
\]
Thus, the net electric flux through the cube is \( \Phi_{\text{total}} = 10L^3 \, \text{Nm}^2/\text{C} \).
