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

# Line charge and point charge, find E

Electromagnetism · Chapter 2: Electric field

- Asked in: 2075 Asoj, 4 marks
- Video: [Electric Field due to Point, Line & Surface charge | EMFT lecture](https://www.youtube.com/watch?v=xghSPIgqBmU) — CADDD Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/line-and-point-charge-find-e
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

An infinite uniform line charge ρ_L = 2 nC/m lies along the x-axis in free space, while the point charge of 8 nC is located at (0, 0, 1). Find E at (2, 3, −4).

## Short answer

Total E is the vector sum of line and point charge fields. Line: using ρ_L/(2πε₀ρ) with ρ=5 m gives 4.31 a_y − 5.75 a_z V/m. Point: using Q/(4πε₀R²) with R from (0,0,1) to (2,3,−4) gives 0.614 a_x + 0.921 a_y − 1.535 a_z V/m. Result: E = 0.61 a_x + 5.23 a_y − 7.29 a_z V/m.

## Answer

The total field is the vector sum of the field of the infinite line charge and the field of the point charge (superposition). Use ε₀ = 8.854×10⁻¹² F/m, so 1/(4πε₀) = 8.99×10⁹ and 1/(2πε₀) = 17.98×10⁹.

#### Field of the line charge (along the x-axis)
For a line charge, E = ρ_L/(2πε₀ρ) a_ρ, where ρ is the perpendicular distance from the line, measured from the axis to the field point (2, 3, −4).

```text
ρ vector = 0 a_x + 3 a_y − 4 a_z ,  ρ = √(9+16) = 5 m
a_ρ = (3 a_y − 4 a_z)/5 = 0.6 a_y − 0.8 a_z
|E_L| = (2×10⁻⁹)(17.98×10⁹)/5 = 7.19 V/m
E_L = 7.19 (0.6 a_y − 0.8 a_z) = 4.31 a_y − 5.75 a_z V/m
```

#### Field of the point charge at (0, 0, 1)
```text
R = (2,3,−4) − (0,0,1) = 2 a_x + 3 a_y − 5 a_z
|R| = √(4+9+25) = √38 = 6.164 m
|E_P| = Q/(4πε₀R²) = (8×10⁻⁹)(8.99×10⁹)/38 = 1.892 V/m
a_R = (2 a_x + 3 a_y − 5 a_z)/6.164
E_P = 1.892 (0.324 a_x + 0.487 a_y − 0.811 a_z)
    = 0.614 a_x + 0.921 a_y − 1.535 a_z V/m
```

#### Total field
```text
E = E_L + E_P
  = (0 + 0.614) a_x + (4.314 + 0.921) a_y + (−5.752 − 1.535) a_z
  = 0.61 a_x + 5.23 a_y − 7.29 a_z V/m
```

**Answer: E = 0.61 a_x + 5.23 a_y − 7.29 a_z V/m** (magnitude about 9.0 V/m).

## Check yourself

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

1. How is the total E found?
   A. Sum of magnitudes
   B. Only the line charge matters
   C. Product of the fields
   D. Vector sum of the line charge field and the point charge field
   Answer: D. Fields from separate sources add vectorially. (Key point: Superposition)

2. The line charge field at (2, 3, -4) has magnitude:
   A. 17.98 V/m
   B. 1.89 V/m
   C. 7.19 V/m, directed along 0.6 a_y − 0.8 a_z
   D. 35.9 V/m
   Answer: C. E = ρ_L/(2πε₀ρ) = (2×10⁻⁹)(17.98×10⁹)/5 = 7.19 V/m. (Key point: Line charge field)

3. The distance from (0, 0, 1) to (2, 3, -4) is:
   A. 5 m
   B. √38 = 6.164 m
   C. √29
   D. √13
   Answer: B. R = (2, 3, -5), |R| = √(4 + 9 + 25) = √38. (Key point: Point charge distance)

4. The total E is:
   A. 0.61 a_x + 5.23 a_y − 7.29 a_z V/m
   B. 0.61 a_x − 5.23 a_y + 7.29 a_z
   C. 5.23 a_x + 0.61 a_y − 7.29 a_z
   D. 0 a_x + 4.31 a_y − 5.75 a_z
   Answer: A. Adding E_L and E_P gives 0.61 a_x + 5.23 a_y - 7.29 a_z V/m. (Key point: Final answer)

## Close variants set in other exams

- [Line and point charges, find D](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aline-and-point-charges-find-d) — Electromagnetism, chapter 2, asked 2×

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- [Line charge and sheet charge, find D](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aline-charge-and-sheet-find-d) — Electromagnetism, chapter 2

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