---
title: "Sheet and line charge, E at origin"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:sheet-line-charge-e-origin"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/sheet-line-charge-e-origin
site: BCT Engineering
---

# Sheet and line charge, E at origin

Electromagnetism · Chapter 2: Electric field

- Asked in: 2074 Asoj, 4 marks
- Video: [Example of electric field intensity calculation due to infinite line charge by Prof. Niraj KumarVITC](https://www.youtube.com/watch?v=BIzqGbqiyTQ) — RF Design Basics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/sheet-line-charge-e-origin
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Let a uniform line charge density of 3 nC/m be at y = 3 and a uniform surface charge density of 0.2 nC/m² be at x = 2. Find E at the origin.

## Short answer

Superpose fields at origin. Line charge at y=3 gives E_L = (ρ_L/(2πε₀·3))(−a_y) = −17.98 a_y V/m. Sheet at x=2 gives E_S = (ρ_s/(2ε₀))(−a_x) = −11.29 a_x V/m. Total E = −11.3 a_x − 18.0 a_y V/m.

## Answer

Assumptions: the line charge at y = 3 is an infinite line parallel to the z-axis (x = 0, y = 3), and the sheet at x = 2 is the infinite plane x = 2. Superpose the two fields at the origin. Take ε₀ = 8.854×10⁻¹² F/m.

#### Field of the line charge
E_L = ρ_L/(2πε₀ρ) a_ρ. The distance from the line to the origin is ρ = 3 m, and the field points away from the line, from (0,3) towards the origin, which is the −a_y direction.

```text
|E_L| = (3×10⁻⁹)/(2π × 8.854×10⁻¹² × 3)
      = (3×10⁻⁹)(17.98×10⁹)/3
      = 17.98 V/m
E_L = −17.98 a_y V/m
```

#### Field of the sheet
E_S = (ρ_s/2ε₀) a_n. The origin lies at x < 2, so the normal pointing from the sheet to the origin is −a_x.

```text
|E_S| = 0.2×10⁻⁹/(2 × 8.854×10⁻¹²) = 11.29 V/m
E_S = −11.29 a_x V/m
```

#### Total
```text
E = E_S + E_L = −11.29 a_x − 17.98 a_y V/m
|E| = √(11.29² + 17.98²) = 21.2 V/m
```

**Answer: E = −11.3 a_x − 18.0 a_y V/m.**

## Check yourself

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

1. The field of the line at y = 3 at the origin points along:
   A. +a_y
   B. −a_y, away from the line
   C. −a_x
   D. +a_x
   Answer: B. The origin lies at smaller y, so the field points in the −a_y direction. (Key point: Line direction)

2. The line charge field magnitude at ρ = 3 m is:
   A. 17.98 V/m
   B. 11.29 V/m
   C. 53.9 V/m
   D. 5.99 V/m
   Answer: A. E = ρ_L/(2πε₀ρ) = 3×10⁻⁹ × 17.98×10⁹ / 3 = 17.98 V/m. (Key point: Line magnitude)

3. The sheet at x = 2 gives at the origin:
   A. +11.29 a_x V/m
   B. −22.6 a_x V/m
   C. −11.29 a_y V/m
   D. −11.29 a_x V/m
   Answer: D. ρ_s/2ε₀ = 11.29 V/m, pointing away from the sheet toward x < 2. (Key point: Sheet field)

4. The total E is:
   A. 11.3 a_x + 18.0 a_y V/m
   B. −18.0 a_x − 11.3 a_y V/m
   C. −11.3 a_x − 18.0 a_y V/m
   D. −29.3 a_x V/m
   Answer: C. Adding the two contributions. (Key point: Total)

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