Q.An electric field is uniform, and in the positive $x$ direction for positive $x$, and uniform with the same magnitude but in the negative $x$ direction for negative $x$. It is given that $\mathbf{E} = 200\,\hat{\mathbf{i}}\ \text{N/C}$ for $x > 0$ and $\mathbf{E} = -200\,\hat{\mathbf{i}}\ \text{N/C}$ for $x < 0$. A right circular cylinder of length $20\,\text{cm}$ and radius $5\,\text{cm}$ has its centre at the origin and its axis along the $x$-axis so that one face is at $x = +10\,\text{cm}$ and the other is at $x = -10\,\text{cm}$ (Fig. 1.25).

The key idea is Gauss’s law: the net outward electric flux through a closed surface equals $\frac{q_{\text{enc}}}{\varepsilon_0}$. The field is symmetric about the origin, pointing outward from the $x=0$ plane on both sides.

(a) For the right face ($x = +10\ \text{cm}$), area vector $\Delta\mathbf{S}$ points along $+\hat{\mathbf{i}}$, and $\mathbf{E} = 200\,\hat{\mathbf{i}}\ \text{N/C}$. Flux is $\Phi_{\text{right}} = \mathbf{E} \cdot \Delta\mathbf{S} = (200)(\pi r^2) = 200 \times \pi (0.05)^2 = 200 \times 0.00785 \approx 1.57\ \text{N·m}^2/\text{C}$.

For the left face ($x = -10\ \text{cm}$), $\Delta\mathbf{S}$ points along $-\hat{\mathbf{i}}$ (outward normal), and $\mathbf{E} = -200\,\hat{\mathbf{i}}\ \text{N/C}$. So $\Phi_{\text{left}} = (-200\,\hat{\mathbf{i}}) \cdot (-\hat{\mathbf{i}}\,\pi r^2) = 200 \times \pi (0.05)^2 = 1.57\ \text{N·m}^2/\text{C}$.

(b) On the curved side, $\mathbf{E}$ is perpendicular to the area vector (which is radial), so $\Phi_{\text{side}} = 0$. …

The key insight here is that the electric field is piecewise uniform — it has the same magnitude everywhere, but flips direction at $x=0$. This is exactly the field you’d get from an infinite plane of charge at the origin (though we don’t need that detail yet). The cylinder is placed symmetrically about the origin, so the field at the left face points left (into the cylinder) and at the right face points right (out of the cylinder). That’s the whole story for the faces.

For the curved side, the field is everywhere perpendicular to the axis, and the side’s area vector is radial (perpendicular to the axis). So the field is parallel to the side surface — the dot product $\mathbf{E} \cdot d\mathbf{A}$ is zero everywhere on the side. That makes the side flux zero.

Let’s work through each part.

1. Flux through each flat face The area of each circular face is

For the right face at $x = +10\ \text{cm}$, the outward normal (pointing out of the cylinder) is $+\hat{\mathbf{i}}$. The field there is $\mathbf{E} = 200\,\hat{\mathbf{i}}\ \text{N/C}$. So the flux is

Numerically, $0.5\pi \approx 1.571\ \text{N·m}^2/\text{C}$.

For the left face at $x = -10\ \text{cm}$, the outward normal is $-\hat{\mathbf{i}}$ (pointing left, out of the cylinder). The field there is $\mathbf{E} = -200\,\hat{\mathbf{i}}\ \text{N/C}$. So

Same magnitude, same sign — both positive. That makes sense: at both faces, the field points outward from the cylinder.

2. Flux through the side of the cylinder …

Method: Gauss's Law — Flux Calculation via Surface Integration

We use Gauss's Law:

Step 1: Understand the field and geometry

• For $x > 0$: $\mathbf{E} = +200\,\hat{\mathbf{i}}\ \text{N/C}$ (points right)
• For $x < 0$: $\mathbf{E} = -200\,\hat{\mathbf{i}}\ \text{N/C}$ (points left)
• Cylinder: length $20\ \text{cm}$, radius $5\ \text{cm}$, axis along $x$-axis
• Left face at $x = -10\ \text{cm}$, right face at $x = +10\ \text{cm}$

Step 2: Flux through each flat face

Right face ($x = +10\ \text{cm}$):

• Area vector $d\mathbf{A}$ points outward → along $+\hat{\mathbf{i}}$
• $\mathbf{E}$ is also $+\hat{\mathbf{i}}$, so $\mathbf{E} \cdot d\mathbf{A} = E \cdot dA$
• Area $A = \pi r^2 = \pi (0.05)^2 = 0.0025\pi\ \text{m}^2$
• Flux = $E \cdot A = 200 \times 0.0025\pi = 0.5\pi\ \text{N·m}^2/\text{C}$

Left face ($x = -10\ \text{cm}$):

• Area vector points outward → along $-\hat{\mathbf{i}}$
• $\mathbf{E}$ is $-\hat{\mathbf{i}}$, so $\mathbf{E} \cdot d\mathbf{A} = (-200)(-1)\,dA = +200\,dA$
• Same area $A = 0.0025\pi\ \text{m}^2$
• Flux = $200 \times 0.0025\pi = 0.5\pi\ \text{N·m}^2/\text{C}$

Answer (a): Flux through each flat face = $\boxed{0.5\pi\ \text{N·m}^2/\text{C}}$

Step 3: Flux through the curved side

• On the side, $d\mathbf{A}$ is radial (perpendicular to $x$-axis)
• $\mathbf{E}$ is purely along $x$-axis
• So $\mathbf{E} \cdot d\mathbf{A} = 0$ everywhere on the side

Answer (b): Flux through side = $\boxed{0}$

Step 4: Net outward flux through the cylinder …

Common Mistakes Students Make with This Gauss Law Problem

Here are the most frequent errors, along with how to avoid each one.

1. Sign Confusion on the Left Face

The Mistake: Students often compute the flux through the left face as positive, or they use the wrong area vector direction.

Why It Happens: The electric field points left ($-200\,\hat{\mathbf{i}}$) for $x < 0$, but the outward normal for the left face also points left ($-\hat{\mathbf{i}}$). Students forget to check the direction of $\hat{\mathbf{n}}$ relative to the field.

How to Avoid:

• Always draw the outward normal for each surface.
• For the left face (at $x = -10\,\text{cm}$), the outward normal is $-\hat{\mathbf{i}}$ (pointing away from the interior).
• Flux is $\Phi = \mathbf{E} \cdot \mathbf{A} = \mathbf{E} \cdot (A\,\hat{\mathbf{n}})$.
• Here: $\mathbf{E} = -200\,\hat{\mathbf{i}}$ and $\hat{\mathbf{n}} = -\hat{\mathbf{i}}$, so:

• Key result: Both faces give positive flux because the field points outward from the cylinder on both sides.

2. Forgetting the Area Vector Direction for the Curved Surface

The Mistake: Students assume the flux through the side is zero without checking, or they incorrectly compute it.

Why It Happens: The side surface is curved, and students don't systematically apply $\Phi = \mathbf{E} \cdot \mathbf{A}$.

How to Avoid:

• For the curved surface, the outward normal at every point is radially outward (perpendicular to the axis).
• The electric field is purely along $\hat{\mathbf{i}}$ (the x-axis).
• Since $\mathbf{E}$ is perpendicular to $\hat{\mathbf{n}}$ at every point on the side:

• Therefore, flux through the side is zero.

3. Using Wrong Area for the Flat Faces

The Mistake: Using the curved surface area ($2\pi r L$) instead of the flat face area ($\pi r^2$).

How to Avoid:

• Flat faces are circles of radius $r = 5\,\text{cm}$.
• Area of each flat face: $A = \pi r^2 = \pi (0.05)^2 = 0.0025\pi\ \text{m}^2$
• The curved surface area is only used if you're computing flux through the side (which is zero here anyway).

4. Unit Conversion Errors

The Mistake: Using cm instead of m in area calculations.

How to Avoid:

• Always convert to SI units before plugging into formulas.
• $5\,\text{cm} = 0.05\,\text{m}$, so $A = \pi (0.05)^2 = 0.0025\pi\ \text{m}^2$
• $10\,\text{cm} = 0.1\,\text{m}$ (this is the distance from centre to face, not needed for area).

5. Adding Fluxes Incorrectly for Net Flux

The Mistake: Subtracting fluxes instead of adding them, or forgetting that both faces give positive flux.

How to Avoid: …