---
title: "Flux through sphere, two line charges"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:flux-leaving-sphere-two-line-charges"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/flux-leaving-sphere-two-line-charges
site: BCT Engineering
---

# Flux through sphere, two line charges

Electromagnetism · Chapter 2: Electric field

- Asked in: 2076 Asoj, 6 marks
- Video: [Two uniform line charges, each 20 nC/m, are located at y = 1, z = ±1 m. Find the total flux le](https://www.youtube.com/watch?v=rirPFz_-6Z8) — Guy_Teaches_STEM
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/flux-leaving-sphere-two-line-charges
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Two uniform line charges, each 20 nC/m, are located at y = 1, z = ±1 m. Find the total electric flux leaving the surface of a sphere having a radius of 2 m, if it is centered at A(3, 1, 0).

## Short answer

By Gauss’s law, flux equals enclosed charge. Each line charge cuts the sphere; chord length L=2√(R²−d²)=2√3 m. Total enclosed charge = 2 × 20 nC/m × 2√3 m ≈ 138.6 nC.

## Answer

**Given:** Two uniform line charges, ρ_L = 20 nC/m each, at y = 1, z = ±1 m (running parallel to the x-axis). Sphere of radius 2 m centred at A(3, 1, 0). Find the total flux leaving the sphere.

**Method:** By Gauss's law, flux Ψ = charge enclosed by the sphere = ρ_L × (length of line inside the sphere), summed for both lines.

#### Distance of each line from the centre

The lines are parallel to the x-axis, so the perpendicular distance from the centre A(3, 1, 0) to the line at (y, z) = (1, ±1) is

```text
d = √( (1 − 1)² + (±1 − 0)² ) = 1 m
```

Since d = 1 < R = 2, each line cuts through the sphere.

#### Length of the line inside the sphere

The line passes at distance d from the centre, so the chord length is

```text
L = 2 √(R² − d²) = 2 √(4 − 1) = 2√3 = 3.464 m
```

```text
          ╭───────╮
         ╱    d    ╲        chord of length L
        │  ●────┼───│───    ← line charge
         ╲  A      ╱
          ╰───────╯
```

#### Charge enclosed and flux

```text
Q_one line = ρ_L L = 20 × 10⁻⁹ × 3.464 = 69.28 nC
Q_total = 2 × 69.28 = 138.6 nC
```

**Answer: Ψ = 138.6 nC ≈ 139 nC** (outward flux, since charges are positive).

## Check yourself

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

1. How is the total flux leaving the sphere found?
   A. Flux equals ρ_L times the sphere's surface area
   B. Flux equals the charge enclosed by the sphere (Gauss's law)
   C. Flux equals ρ_L times the radius
   D. Flux is zero since lines are infinite
   Answer: B. By Gauss's law, flux equals enclosed charge. (Key point: Gauss's law use)

2. What is the perpendicular distance from A(3, 1, 0) to each line (y = 1, z = ±1)?
   A. 1 m
   B. 2 m
   C. √2 m
   D. 0
   Answer: A. d = √((1-1)² + (±1-0)²) = 1 m, less than R = 2, so the lines cut the sphere. (Key point: Distance from centre)

3. The length of each line inside the sphere is:
   A. 2√(R² + d²)
   B. R - d = 1 m
   C. 2R = 4 m
   D. 2√(R² - d²) = 2√3 m
   Answer: D. A chord at distance d from the centre has length 2√(R² - d²) = 2√3. (Key point: Chord length)

4. The total flux leaving the sphere is:
   A. 69.3 nC
   B. 80 nC
   C. 138.6 nC
   D. 277 nC
   Answer: C. Each line holds 20 nC/m × 3.464 m = 69.28 nC; two lines give 138.6 nC. (Key point: Total flux)

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