Q.What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side $20\,\text{cm}$ oriented so that its faces are parallel to the coordinate planes?

The key idea here is Gauss's Law.

1. Gauss's Law states that the net electric flux ($\Phi_E$) through any closed surface is given by $\Phi_E = \frac{Q_{enc}}{\epsilon_0}$, where $Q_{enc}$ is the net electric charge enclosed within the surface.
2. A uniform electric field implies that the electric field lines are parallel and equally spaced. For such a field, any field line that enters a closed surface (like the cube) on one side must exit the surface on the opposite side. …

Electric flux is a measure of the "flow" of the electric field through a given surface. It quantifies how many electric field lines pass through that surface. For a uniform electric field $\vec{E}$ passing through a flat surface with area vector $\vec{A}$ (where the direction of $\vec{A}$ is normal to the surface), the electric flux $\Phi$ is given by the dot product:

The area vector $\vec{A}$ for a closed surface always points outwards.

The most fundamental principle governing electric flux through a closed surface is Gauss's Law.

Why This Approach Works

The problem specifies a "uniform electric field". This is the key piece of information. A uniform electric field means that the electric field lines are parallel, equally spaced, and point in the same direction everywhere in the region. Crucially, a uniform electric field is not generated by charges within the region where it is uniform. Instead, it's typically produced by charges located far away (e.g., an infinite sheet of charge or widely separated parallel plates).

Since the cube is placed in a region of uniform electric field, there are no electric charges inside the cube that would act as sources or sinks for the electric field lines. Therefore, the net electric charge enclosed by the cube, $Q_{enc}$, is zero.

According to Gauss's Law, if the net enclosed charge is zero, the net electric flux through the closed surface (the cube, in this case) must also be zero.

Step-by-Step Solution

We can solve this problem using two methods: directly applying Gauss's Law, or by calculating the flux through each face and summing them. Both methods lead to the same result and reinforce the underlying concept.

Method 1: Using Gauss's Law (The Most Direct Approach)

1. Identify the nature of the electric field: The problem states that the electric field is uniform.
2. Consider the charges enclosed by the cube: A uniform electric field implies that there are no net electric charges within the volume where the field is uniform. If there were charges inside the cube, the field lines would either originate from (positive charge) or terminate on (negative charge) these charges, making the field non-uniform in that region. Since the field is uniform, the net charge enclosed by the cube is zero.

3. Apply Gauss's Law: Substitute $Q_{enc} = 0$ into Gauss's Law:

Thus, the net flux through the cube is zero.

Method 2: Direct Calculation (Illustrating Cancellation)

This method involves calculating the flux through each of the six faces of the cube and then summing them. Let the side length of the cube be $L = 20\,\text{cm}$. The area of each face is $A = L^2$.

For simplicity and without loss of generality, let's assume the uniform electric field is directed along the positive x-axis:

We orient the cube such that its faces are parallel to the coordinate planes. Let one corner of the cube be at the origin $(0,0,0)$, and its edges extend along the positive x, y, and z axes.

1. Faces perpendicular to the x-axis:
   • Left face (at $x=0$): The area vector $\vec{A}_1$ points in the $-\hat{i}$ direction.

    The flux through this face is:

    (This is negative because field lines are entering the cube.)
*   **Right face (at $x=L$):** The area vector $\vec{A}_2$ points in the $+\hat{i}$ direction.

    The flux through this face is:

    (This is positive because field lines are exiting the cube.)

2. Faces perpendicular to the y-axis:

* Bottom face (at $y=0$): The area vector $\vec{A}_3$ points in the $-\hat{j}$ direction. …

Method: Gauss’s Law (Direct Flux Calculation)

Why this method?

Gauss’s Law relates the net electric flux through a closed surface to the charge enclosed. But here, we can also compute flux directly from the definition — because the field is uniform and the cube is aligned with coordinate axes.

Steps

1. Recall the definition of electric flux For a uniform field $\vec{E}$ through a flat surface of area $A$:

where $\theta$ is the angle between $\vec{E}$ and the area normal vector.

2. Identify the field direction From Exercise 1.14 (implied context), the uniform field is:

(Only the $x$-component is nonzero.)

3. Analyze each face of the cube The cube has side $a = 20\ \text{cm} = 0.2\ \text{m}$. Area of each face:

• Faces perpendicular to $x$-axis (left and right):
  • Right face: normal along $+\hat{\imath}$, so $\theta = 0^\circ$, $\Phi_{\text{right}} = E A \cos 0 = (3\times10^3)(0.04)(1) = 120\ \text{N m}^2/\text{C}$
  • Left face: normal along $-\hat{\imath}$, so $\theta = 180^\circ$, $\Phi_{\text{left}} = E A \cos 180^\circ = (3\times10^3)(0.04)(-1) = -120\ \text{N m}^2/\text{C}$ …

Common Mistakes with Gauss Law & Flux Through a Cube

Mistake 1: Thinking Flux Depends on Cube Size

The error: Students calculate the area of each face ($0.2 \times 0.2 = 0.04\ \text{m}^2$) and multiply by $E$, forgetting that flux through a closed surface in a uniform field is always zero.

Why it's wrong:

For a uniform electric field $\vec{E}$, the number of field lines entering one face exactly equals the number leaving the opposite face. The net flux through any closed surface is:

The cube's side length is irrelevant here — it cancels out.

How to avoid:

Always check: Is the field uniform? If yes, net flux through any closed surface = 0. Don't compute areas unless the field is non-uniform or the surface is open.

Mistake 2: Forgetting the Direction of Area Vectors

The error: Students take all area vectors pointing outward (correct) but then compute $\vec{E} \cdot \vec{A}$ for each face without considering sign.

Correct approach:

For a cube with faces parallel to coordinate planes:

• Face at $x = 0$: $d\vec{A} = -\hat{i}\,dA$
• Face at $x = L$: $d\vec{A} = +\hat{i}\,dA$
• Similarly for $y$ and $z$ faces.

If $\vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k}$, then:

• Flux through $x=0$ face: $-E_x A$
• Flux through $x=L$ face: $+E_x A$
• Sum = $0$

How to avoid:

Draw the cube, label outward normals explicitly, and compute dot products with sign.

Mistake 3: Confusing with Non-Uniform Fields

The error: Students recall Gauss Law: $\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$ and try to find enclosed charge — but here there's no charge mentioned inside the cube.

Why it's wrong:

Gauss Law still holds: if $q_{\text{enc}} = 0$, then $\Phi = 0$. But the direct flux calculation (Mistake 1) is simpler for uniform fields.

How to avoid:

• If field is uniform and no charge inside → flux = 0 …