---
title: "Potential from point and line charge"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:potential-point-and-line-charge"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/potential-point-and-line-charge
site: BCT Engineering
---

# Potential from point and line charge

Electromagnetism · Chapter 2: Electric field

- Asked in: 2067 Mangsir, 6 marks
- Video: [KTU Electromagnetics Module-2 |  Electric Potential Due to Uniformly Charged Infinite Line](https://www.youtube.com/watch?v=ZX_ph6YAQ40) — 100 Million Thoughts
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/potential-point-and-line-charge
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Find the potential at a point P(2, 3, 3) due to a 1 nC charge located at Q(3, 4, 4) and a 1 nC/m uniform line charge located at x = 2, y = 1, if V = 0 V at (3, 4, 5).

## Short answer

Point charge contribution (Q/4πε₀)(1/R_P−1/R_ref)=−3.80 V. Line charge contribution −(ρ_L/2πε₀) ln(ρ_P/ρ_ref)=+8.24 V. Total V_P = 4.44 V.

## Answer

Use ε₀ = 8.854×10⁻¹² F/m. The reference point is V = 0 at (3, 4, 5). The line (x = 2, y = 1) is parallel to the z-axis. Add the potentials of the two sources, each measured relative to the reference point.

#### Point charge Q = 1 nC at (3, 4, 4)
V = (Q/4πε₀)(1/R_P − 1/R₀).

```text
R_P : from (3,4,4) to P(2,3,3):  (−1,−1,−1)  →  √3 = 1.732 m
R₀  : from (3,4,4) to (3,4,5):   (0,0,1)      →  1 m

Q/(4πε₀) = (10⁻⁹)(8.988×10⁹) = 8.988 V·m

V₁ = 8.988 (1/1.732 − 1/1) = 8.988 (0.5774 − 1) = −3.80 V
```

#### Line charge ρ_L = 1 nC/m at x = 2, y = 1
V = −(ρ_L/2πε₀) ln(ρ_P/ρ₀) relative to the reference radius.

```text
ρ_P : from line to P(2,3):   (0, 2)   → 2 m
ρ₀  : from line to (3,4):    (1, 3)   → √10 = 3.162 m

ρ_L/(2πε₀) = (10⁻⁹)(17.975×10⁹) = 17.975 V

V₂ = −17.975 ln(2/3.162) = −17.975 (−0.4581) = +8.24 V
```

#### Total
```text
V_P = V₁ + V₂ = −3.80 + 8.24 = 4.44 V
```

**Answer: V_P ≈ 4.44 V.**

## Check yourself

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

1. With V = 0 at (3,4,5), how is the point-charge potential at P written?
   A. V = Q/(4πε0 R_P)
   B. V = (Q/4πε0)(1/R_P − 1/R0)
   C. V = (Q/4πε0)(R_P − R0)
   D. V = (Q/4πε0)(1/R0 − 1/R_P)
   Answer: B. Potential is relative to the reference point, so V = (Q/4πε0)(1/R_P − 1/R0), where R0 is the reference distance from the charge. (Key point: Potential with a finite reference)

2. For Q at (3,4,4), what are R_P (to P(2,3,3)) and R0 (to (3,4,5))?
   A. √3 m and 1 m
   B. 3 m and 1 m
   C. √3 m and 2 m
   D. 1 m and √3 m
   Answer: A. P − Q = (−1,−1,−1) gives √3 m; reference − Q = (0,0,1) gives 1 m. (Key point: Distances for point charge)

3. For the line at x = 2, y = 1, which distances enter the log term?
   A. ρ_P = 2 m and ρ0 = √10 m
   B. ρ_P = 3 m and ρ0 = 5 m
   C. ρ_P = √10 m and ρ0 = 2 m
   D. ρ_P = 2 m and ρ0 = 1 m
   Answer: A. Only x and y matter: P gives (0,2) so 2 m; reference (3,4) gives (1,3) so √10 m. (Key point: Line charge radial distances)

4. How is the total potential at P obtained, and what is it?
   A. Subtract: −3.80 − 8.24 = −12.04 V
   B. Add scalars: −3.80 + 8.24 = 4.44 V
   C. Multiply the two potentials
   D. Use only the line charge: 8.24 V
   Answer: B. Potentials are scalars and add: V1 = −3.80 V (point charge) and V2 = +8.24 V (line), total 4.44 V. (Key point: Superposition)

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