Q.The electric field components in Fig. 1.24 are $E_x = \alpha x^{1/2}$, $E_y = E_z = 0$, in which $\alpha = 800\,\text{N/C m}^{1/2}$. Calculate

Only the two faces perpendicular to the $x$‑axis carry flux, since $E_y=E_z=0$ and $E_x=\alpha x^{1/2}$ is constant on each such face.

Left face ($x=a$, outward normal $-\hat x$): $\Phi_L=-\alpha a^{1/2}\,a^2=-\alpha a^{5/2}.$

Right face ($x=2a$, outward normal $+\hat x$): $\Phi_R=+\alpha(2a)^{1/2}\,a^2=\sqrt2\,\alpha a^{5/2}.$

1. Net flux:

   $\Phi=\Phi_L+\Phi_R=\alpha a^{5/2}(\sqrt2-1).$

   With $\alpha=800\,\text{N/C·m}^{1/2}$ and $a=0.1\,\text{m}$, $a^{5/2}=3.16\times10^{-3}$:

   $\Phi=800(3.16\times10^{-3})(0.414)\approx1.05\,\text{N·m}^2/\text{C}.$

2. Enclosed charge (Gauss's law):

   $q=\varepsilon_0\Phi=(8.854\times10^{-12})(1.05)\approx9.27\times10^{-12}\,\text{C}.$

   ✓Final answer

   Net flux $\Phi\approx1.05\,\text{N·m}^2/\text{C}$; enclosed charge $q\approx9.27\times10^{-12}\,\text{C}$.

The field points only along $x$, so flux passes only through faces whose normal has an $x$‑component. For the axis‑aligned cube, those are the left face at $x=a$ and the right face at $x=2a$; the four faces parallel to the $x$‑axis contribute nothing because $\vec E\cdot d\vec A=0$ there.

Left face ($x=a$). Here $E_x=\alpha a^{1/2}$ is uniform over the face of area $a^2$, and the outward normal points in $-\hat x$:

Right face ($x=2a$). Now $E_x=\alpha(2a)^{1/2}=\sqrt2\,\alpha a^{1/2}$, and the outward normal points in $+\hat x$:

1. Net flux.

   $\Phi=\Phi_L+\Phi_R=\alpha a^{5/2}(\sqrt2-1).$

   With $\alpha=800\,\text{N/C·m}^{1/2}$ and $a=0.1\,\text{m}$,

   $a^{5/2}=(0.1)^{5/2}=3.162\times10^{-3},\qquad \sqrt2-1=0.4142,$

   $\Phi=800\times3.162\times10^{-3}\times0.4142\approx1.05\,\text{N·m}^2/\text{C}.$

   Watch out

   Note $a^{5/2}=a^2\sqrt a$, not $a^{3/2}$ — the extra $a^2$ is the face area. Keep the minus sign on the left face, or the two contributions wrongly add.

2. Enclosed charge. Gauss's law $\Phi=q/\varepsilon_0$ gives

   $q=\varepsilon_0\Phi=(8.854\times10^{-12})(1.05)\approx9.27\times10^{-12}\,\text{C}.$

   ✓Final answer

   1. Net flux $\Phi\approx1.05\,\text{N·m}^2/\text{C}$.
   2. Enclosed charge $q\approx9.27\times10^{-12}\,\text{C}$.

Method: Gauss's Law Flux Calculation via Surface Integration

This problem uses Gauss's Law in integral form:

Step 1: Identify the non-zero field contribution

Given:

• $E_x = \alpha x^{1/2}$, where $\alpha = 800\ \text{N/C m}^{1/2}$
• $E_y = E_z = 0$
• Cube side length $a = 0.1\ \text{m}$, placed with one corner at origin

Since only $E_x$ is non-zero, flux only passes through faces perpendicular to the x-axis — the left face (at $x=0$) and the right face (at $x=a$).

Step 2: Calculate flux through each x-face

Right face ($x = a = 0.1\ \text{m}$):

• Area vector: $d\vec{A} = \hat{i}\, dy\, dz$ (outward normal is $+\hat{i}$)
• Field at this face: $E_x = \alpha a^{1/2}$ (constant over the face)
• Flux:

Left face ($x = 0$):

• Area vector: $d\vec{A} = -\hat{i}\, dy\, dz$ (outward normal is $-\hat{i}$)
• Field at this face: $E_x = \alpha (0)^{1/2} = 0$
• Flux: $\Phi_{\text{left}} = 0$

Step 3: Total flux through the cube

Substitute values:

Since $(0.1)^{1/2} = \sqrt{0.1} \approx 0.3162$:

Step 4: Find enclosed charge using Gauss's Law

Final Answer Summary

Common Mistakes Students Make with This Gauss Law Problem

Mistake 1: Forgetting That Flux Depends Only on the Perpendicular Component

The error: Students often try to integrate $E_x$ over all six faces of the cube, including faces where the field is parallel to the surface.

Why it's wrong: Flux through a surface is $\Phi = \int \vec{E} \cdot d\vec{A}$. Since $E_y = E_z = 0$, only the two faces perpendicular to the x-axis contribute. The four side faces (parallel to the x-axis) have zero flux because $\vec{E} \cdot d\vec{A} = 0$ there.

How to avoid: Always check which field components are non-zero. If $E_y = E_z = 0$, only faces with normals along $\hat{i}$ matter. Draw the cube and label each face's normal vector.

Mistake 2: Using the Same $x$ Value for Both Faces

The error: Plugging $x = a$ into $E_x = \alpha x^{1/2}$ for both the left and right faces.

Why it's wrong: The left face is at $x = 0$, the right face is at $x = a$. The field strength is different at each location:

• Left face: $E_x(0) = \alpha \cdot 0^{1/2} = 0$
• Right face: $E_x(a) = \alpha a^{1/2}$

How to avoid: Write the coordinates explicitly:

• Left face: $x = 0$, area vector $d\vec{A} = -dA\,\hat{i}$
• Right face: $x = a$, area vector $d\vec{A} = +dA\,\hat{i}$

Then compute each flux separately.

Mistake 3: Ignoring the Direction of the Area Vector

The error: Treating both faces as having $+dA\,\hat{i}$ and getting zero net flux.

Why it's wrong: By convention, the area vector points outward from the closed surface:

• Left face: outward normal is $-\hat{i}$, so $d\vec{A} = -dA\,\hat{i}$
• Right face: outward normal is $+\hat{i}$, so $d\vec{A} = +dA\,\hat{i}$

The flux through the left face is:

How to avoid: Always draw outward normals on each face before computing dot products.

Mistake 4: Forgetting That $E_x$ Varies Over the Face

The error: Treating $E_x$ as constant over the entire right face and writing $\Phi = E_x \cdot A$.

Why it's wrong: $E_x = \alpha x^{1/2}$ depends on $x$. On the right face, $x = a$ is constant, so this actually works here — but only because the face is perpendicular to the x-axis. Students often carry this habit to problems where the field varies across the face.

How to avoid: Check if the field component is constant over the surface. Here, since the right face is at fixed $x = a$, $E_x$ is uniform across it. But be cautious — this is a special case.

Mistake 5: Incorrectly Computing the Net Flux

The error: Adding magnitudes without signs, e.g., $\Phi_{\text{net}} = \alpha a^{1/2} \cdot a^2 + 0 = \alpha a^{5/2}$.

Why it's wrong: The left face contributes negative flux because $\vec{E}$ points inward there (field enters the cube). The correct calculation:

The left face has $E_x = 0$, so its flux is zero. The net flux is just from the right face: $\Phi_{\text{net}} = \alpha a^{5/2}$.

How to avoid: Compute each face's flux with its correct sign, then sum. Don't shortcut.

Mistake 6: Using the Wrong Formula for Charge from Flux

The error: Writing $q = \Phi \cdot \varepsilon_0$ instead of $q = \Phi \varepsilon_0$.

Why it's wrong: Gauss's law states:

How to avoid: Memorize the exact form: flux = charge enclosed divided by epsilon-zero. Rearrange carefully.

Mistake 7: Unit Errors in the Final Answer

The error: Reporting flux in N/C or charge in C without checking dimensions.

Why it's wrong: Flux has units $\text{N·m}^2/\text{C}$. With $\alpha = 800\,\text{N/C·m}^{1/2}$ and $a = 0.1\,\text{m}$:

Then $q = \Phi \varepsilon_0 = 2.53 \times 8.85 \times 10^{-12} = 2.24 \times 10^{-11}\,\text{C}$.

How to avoid: Track units at every step. Write the unit of each quantity before substituting numbers.

Quick Checklist to Avoid These Mistakes