Q.Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{-22}\,\text{C/m}^2$. What is E:

Concept: Gauss’s law for parallel plate capacitors — fields from each plate superpose.

Reasoning:

1. Each large plate alone produces a field of magnitude $\frac{\sigma}{2\varepsilon_0}$ on either side, directed away from positive charge and toward negative charge.

2. For two plates with equal and opposite surface charge densities $\sigma = 17.0 \times 10^{-22}\,\text{C/m}^2$, the fields add in the region between the plates and cancel in the outer regions.

3. Using $\varepsilon_0 = 8.85 \times 10^{-12}\,\text{C}^2/\text{N·m}^2$:

   • Outer region of first plate: fields from both plates are equal and opposite → net $E = 0$.
   • Outer region of second plate: same cancellation → net $E = 0$. …

Field of the pair. Each plate produces a field $\sigma/2\varepsilon_0$. Outside the pair the two fields are opposite and cancel; between the plates they add.

1. Outer region of first plate: $E=0.$
2. Outer region of second plate: $E=0.$
3. Between the plates: the fields add, …

Method: Gauss's Law for Parallel Plate Capacitors

This problem uses Gauss's Law applied to the superposition principle for infinite charged sheets.

Key Concept

For a single infinite thin sheet with surface charge density $\sigma$, the electric field is:

directed away from the sheet if $\sigma > 0$, and toward the sheet if $\sigma < 0$.

Given Data

• Plate 1 inner face: $+\sigma = +17.0 \times 10^{-22} \, \text{C/m}^2$
• Plate 2 inner face: $-\sigma = -17.0 \times 10^{-22} \, \text{C/m}^2$
• $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2$

Steps

Step 1: Identify the three regions

• Region (a): Outer side of plate 1 (left of plate 1)
• Region (b): Outer side of plate 2 (right of plate 2)
• Region (c): Between the plates

Step 2: Apply superposition

Each plate produces its own field. The net field is the vector sum of fields from both plates.

Step 3: Calculate field magnitude

Step 4: Determine direction and net field in each region

• (a) Outer region of first plate: Fields from both plates point left (away from + plate, toward – plate). They cancel. $\boxed{E_a = 0}$ …

Here are the common mistakes students make when solving this classic Gauss’s law problem, along with how to avoid each.

Mistake 1: Forgetting that the field due to a single plate is $\frac{\sigma}{2\varepsilon_0}$, not $\frac{\sigma}{\varepsilon_0}$

Why it happens

Students often memorise the formula for an infinite sheet as $\frac{\sigma}{\varepsilon_0}$, but that’s the total field on one side when using a Gaussian pillbox that encloses both surfaces. For a single thin conducting plate, the field on one side is half that: $\frac{\sigma}{2\varepsilon_0}$.

How to avoid

• Draw a Gaussian pillbox that cuts through the plate. The flux goes out both sides, so $2EA = \frac{\sigma A}{\varepsilon_0}$ → $E = \frac{\sigma}{2\varepsilon_0}$.
• Remember: one plate → half the flux per side.

Mistake 2: Adding fields without considering direction (sign convention)

Why it happens

Students treat all fields as positive magnitudes and add them algebraically, ignoring that fields from opposite charges point in opposite directions.

How to avoid

• Always draw a diagram. Mark the direction of $\vec{E}$ from each plate (away from positive, toward negative).
• Use a sign convention (e.g., right = positive). Then add vectors, not magnitudes.
• For this problem:
  • Outer region of plate 1: fields from both plates point away from plate 1 → they add.
  • Between plates: fields point in opposite directions → they subtract.

Mistake 3: Using the wrong value of $\sigma$ for each plate

Why it happens

The problem gives one magnitude $17.0 \times 10^{-22}\,\text{C/m}^2$, but the plates have opposite signs. Students sometimes use $+\sigma$ for both.

How to avoid

• Label plate 1 with $+\sigma$ and plate 2 with $-\sigma$ (or vice versa).
• The magnitude is the same, but the sign matters for direction.
• In the formula $E = \frac{|\sigma|}{2\varepsilon_0}$, the sign only tells you the direction — keep the magnitude positive, then assign direction from the diagram.

Mistake 4: Forgetting that the field is zero inside a conductor in electrostatic equilibrium

Why it happens

Students try to compute the field inside the metal plates using the same formulas, not realising that charges rearrange to cancel the internal field.

How to avoid

• Recall: Inside a conductor in equilibrium, $\vec{E} = 0$.
• The given surface charges are on the inner faces only. The field inside the metal is zero — that’s why all charge sits on the surfaces.

Mistake 5: Not using superposition correctly for the three regions

Why it happens …