Question

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.

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.

Answer 1

Given an electric field \( \vec{E} = (10x + 5) \hat{i} \) N/C, where \( x \) denotes the position along the x-axis. The electric flux \( \Phi_E \) through a closed surface is calculated using Gauss's law as \( \Phi_E = \oint \vec{E} \cdot d\vec{A} \), which equates the net flux to the enclosed charge divided by \( \epsilon_0 \). Here, \( d\vec{A} \) is the vector area element of the cube's surface. Step 1: Electric Field Representation The electric field is defined as \( \vec{E} = (10x + 5) \hat{i} \). This indicates the field has an x-component that varies with the x-coordinate. Step 2: Flux Calculation per Face A cube has six faces. The flux through each face is determined by the dot product of the electric field and the face's area vector \( \vec{E} \cdot d\vec{A} \). The area vector is normal to the surface. The flux through a face is given by \( \Phi_{\text{face}} = \vec{E} \cdot A \), where \( A \) is the face area. Since \( \vec{E} \) is purely in the x-direction, only the faces perpendicular to the x-axis (at \( x = 0 \) and \( x = L \)) contribute to the net flux. Step 3: Flux at \( x = 0 \) and \( x = L \) 1. Face at \( x = 0 \): At \( x = 0 \), \( \vec{E} = 5 \hat{i} \) N/C. The face area is \( A = L^2 \). The area vector points in the negative x-direction. Thus, the flux is: \[\Phi_1 = E_x \cdot A = 5 \times L^2 = 5L^2\] 2. Face at \( x = L \): At \( x = L \), \( \vec{E} = (10L + 5) \hat{i} \) N/C. The area vector points in the positive x-direction. Thus, the flux is: \[\Phi_2 = E_x \cdot A = (10L + 5) \times L^2 = (10L + 5) L^2\] Step 4: Total Flux The total flux is the difference between the flux through the face at \( x = L \) and the face at \( x = 0 \): \[\Phi_{\text{total}} = \Phi_2 - \Phi_1 = (10L + 5) L^2 - 5L^2\] \[\Phi_{\text{total}} = 10L^3\] The net electric flux through the cube is \( \Phi_{\text{total}} = 10L^3 \, \text{Nm}^2/\text{C} \).