---
title: "Parallel plate sheets, field in regions"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:parallel-plate-sheets-three-regions"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/parallel-plate-sheets-three-regions
site: BCT Engineering
---

# Parallel plate sheets, field in regions

Electromagnetism · Chapter 2: Electric field

- Asked in: 2074 Asoj, 4 marks
- Video: [1.17-Electric Field due to Parallel infinite charge sheets-Electrostatics](https://www.youtube.com/watch?v=W8JexXrgCwI) — Jaya physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/parallel-plate-sheets-three-regions
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Find the electric field intensity in all three regions due to an infinite sheet parallel plate capacitor having surface charge density ρ_s C/m² and −ρ_s C/m² and placed at y = 0 and y = b respectively.

## Short answer

By superposition, the electric fields of the sheets cancel outside the plates and add inside. Hence E = 0 for y<0 and y>b. Between the plates (0<y<b) E = (ρ_s/ε₀) a_y, uniform and directed from the positive to the negative plate.

## Answer

The field of one infinite sheet of charge ρ_s is E = (ρ_s/2ε₀) a_n, where a_n is the unit normal pointing away from the sheet (for positive ρ_s). Superpose the fields of the two sheets.

```text
   Region I      |   Region II     |   Region III
   (y < 0)       |  (0 < y < b)    |   (y > b)
                 |                 |
               +ρ_s              −ρ_s
              (y = 0)           (y = b)
                 |   ──→ a_y       |
```

#### Field of each sheet
- Sheet +ρ_s at y = 0: E₁ = +(ρ_s/2ε₀) a_y for y > 0, and −(ρ_s/2ε₀) a_y for y < 0.
- Sheet −ρ_s at y = b: the field points towards a negative sheet. So E₂ = +(ρ_s/2ε₀) a_y for y < b, and E₂ = −(ρ_s/2ε₀) a_y for y > b.

#### Total field
| Region | E₁ | E₂ | E = E₁ + E₂ |
|---|---|---|---|
| y < 0 | −(ρ_s/2ε₀) a_y | +(ρ_s/2ε₀) a_y | 0 |
| 0 < y < b | +(ρ_s/2ε₀) a_y | +(ρ_s/2ε₀) a_y | (ρ_s/ε₀) a_y |
| y > b | +(ρ_s/2ε₀) a_y | −(ρ_s/2ε₀) a_y | 0 |

**Answer:** E = 0 for y < 0, E = (ρ_s/ε₀) a_y V/m for 0 < y < b (from the positive to the negative plate), and E = 0 for y > b. The field is confined between the plates and is uniform there.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The field of one infinite sheet of charge ρ_s is:
   A. ρ_s / ε₀ a_n
   B. ρ_s / (4πε₀ r²)
   C. ρ_s / (2πε₀ r)
   D. (ρ_s / 2ε₀) a_n, pointing away from a positive sheet
   Answer: D. An infinite sheet gives uniform field ρ_s/2ε₀ directed away from positive charge. (Key point: Single sheet field)

2. In 0 < y < b, E equals:
   A. (ρ_s/2ε₀) a_y
   B. 0
   C. (ρ_s/ε₀) a_y
   D. −(ρ_s/ε₀) a_y
   Answer: C. Both sheets' fields point in +a_y and add. (Key point: Field between plates)

3. For y < 0 and y > b, E is:
   A. ρ_s/ε₀
   B. 0, because the two fields cancel
   C. ρ_s/2ε₀
   D. ρ_s/4ε₀
   Answer: B. Equal and opposite sheet fields cancel outside. (Key point: Outside the plates)

4. The field between the plates points:
   A. From the positive to the negative plate
   B. From the negative to the positive plate
   C. Parallel to the plates
   D. Randomly
   Answer: A. Field lines go from +ρ_s at y = 0 to −ρ_s at y = b. (Key point: Direction)

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