Q.Consider a uniform electric field $\mathbf{E} = 3 \times 10^{3}\,\hat{\mathbf{i}}\ \text{N/C}$.

Concept: Electric flux through a surface is given by $\Phi = \mathbf{E} \cdot \mathbf{A} = EA\cos\theta$, where $\theta$ is the angle between the field and the area normal vector.

Step 1: The square has side $10\ \text{cm} = 0.1\ \text{m}$, so its area is $A = (0.1)^2 = 0.01\ \text{m}^2$.

Step 2: (a) When the plane is parallel to the $yz$ plane, the normal is along $\hat{\mathbf{i}}$, so $\theta = 0^\circ$. Flux is $\Phi = EA\cos 0 = (3\times10^3)(0.01)(1) = 30\ \text{N m}^2/\text{C}$. …

The idea behind electric flux is beautifully simple: it measures how much of the electric field “flows through” a given surface. For a uniform field $\mathbf{E}$ and a flat surface of area $A$, the flux is

where $\theta$ is the angle between the field direction and the normal (the outward-pointing perpendicular) to the surface. This is the core of Gauss’s law — but here we’re just applying the definition directly.

Let’s work through each part.

1. Set up the area vector. The square has side $10\ \text{cm} = 0.1\ \text{m}$, so its area is

The area vector $\mathbf{A}$ is defined as having magnitude $A$ and direction along the normal to the surface.

2. Part (a): Plane parallel to the $yz$-plane. If the square’s plane is parallel to the $yz$-plane, then its normal points along the $x$-axis. Since the field $\mathbf{E} = 3\times10^3\,\hat{\mathbf{i}}$ is also along the $x$-axis, the angle between them is $\theta = 0^\circ$. Hence $\Phi_a = E A \cos 0^\circ = (3\times10^3)(0.01)(1) = 30\ \text{N m}^2/\text{C}.$ …

Method: Electric Flux Calculation Using Dot Product Definition

This problem uses the definition of electric flux through a surface:

where:

• $\mathbf{E}$ = uniform electric field
• $\mathbf{A}$ = area vector (magnitude = area, direction = normal to surface)
• $\theta$ = angle between $\mathbf{E}$ and the normal to the surface

Step 1: Identify given data

• $\mathbf{E} = 3 \times 10^{3}\,\hat{\mathbf{i}}\ \text{N/C}$ (field along $+x$ direction)
• Side of square = $10\ \text{cm} = 0.1\ \text{m}$
• Area of square: $A = (0.1)^2 = 1 \times 10^{-2}\ \text{m}^2$

Step 2: Part (a) — Square parallel to $yz$ plane

• Plane parallel to $yz$ → normal is along $x$-axis → $\mathbf{A} = A\,\hat{\mathbf{i}}$
• $\theta = 0^\circ$ (field and normal are parallel)

Step 3: Part (b) — Normal makes $60^\circ$ with $x$-axis …

Mistake 1: Forgetting to convert cm to m

The error:

Students use side length $10\ \text{cm}$ directly in the area formula, getting $A = 10 \times 10 = 100\ \text{cm}^2$ and then plugging it into flux formula without converting to $\text{m}^2$.

Why it’s wrong:

Electric field is in N/C (SI units), so area must be in $\text{m}^2$ for flux to be in $\text{N·m}^2/\text{C}$.

How to avoid:

Always convert cm → m before any calculation.

$10\ \text{cm} = 0.1\ \text{m}$

So area:

Mistake 2: Using the wrong angle in the flux formula

The error:

In part (b), students take $\theta = 60^\circ$ directly as the angle between $\mathbf{E}$ and the area vector $\mathbf{A}$.

Why it’s wrong:

The flux formula is

where $\phi$ is the angle between $\mathbf{E}$ and the normal to the surface.

Here, the normal makes $60^\circ$ with the $x$-axis, and $\mathbf{E}$ is along $+x$. So the angle between them is $60^\circ$ — but many students mistakenly use the angle between the surface and the field instead.

How to avoid:

• Draw a diagram: surface normal vs field direction.
• If the problem says “normal makes angle $\theta$ with the $x$-axis” and $\mathbf{E}$ is along $x$, then $\phi = \theta$.
• If it said “plane makes angle $\theta$ with the $x$-axis”, then $\phi = 90^\circ - \theta$.

Mistake 3: Forgetting the direction of the area vector

The error:

In part (a), students compute flux as $E \times A$ without considering that the area vector is perpendicular to the surface.

Why it’s wrong:

For a plane parallel to the $yz$-plane, the normal is along $\pm \hat{\mathbf{i}}$. Since $\mathbf{E}$ is along $+\hat{\mathbf{i}}$, the dot product is $E A \cos 0^\circ = EA$. But if they choose the opposite normal, they’d get $-EA$.

How to avoid:

• Always define the area vector $\mathbf{A} = A\,\hat{\mathbf{n}}$ where $\hat{\mathbf{n}}$ is the outward normal.
• In part (a), the natural choice is $\hat{\mathbf{n}} = \hat{\mathbf{i}}$ (parallel to $\mathbf{E}$), so $\Phi = EA$.

Mistake 4: Using the wrong formula for flux

The error:

Some students write $\Phi = \oint \mathbf{E} \cdot d\mathbf{A}$ even for a flat surface, then get confused.

Why it’s wrong:

For a uniform field and flat surface, the flux simplifies to

No integral needed.

How to avoid:

• Recognize when the field is uniform and surface is flat — use the dot product formula directly.
• Save the integral for non-uniform fields or curved surfaces.

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