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

# Potential due to sheet and line charge

Electromagnetism · Chapter 2: Electric field

- Asked in: 2068 Baisakh, 8 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/sheet-and-line-charge-potential
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

A uniform sheet of charge ρ_s = 40ε0 C/m² is located in the plane x = 0 in free space. A uniform line charge ρ_L = 0.6 nC/m lies along the line x = 9, y = 4 in free space. Find the potential at point P(6, 8, −3) if V = 10 V at A(2, 9, 3).

## Short answer

Sheet field is 20 a_x V/m, contribution −20(6−2)=−80 V. Line contribution uses −(ρ_L/2πε₀) ln(ρ_P/ρ_A)=+5.85 V. Total V_P = 10 − 74.15 = −64.2 V.

## Answer

Use ε₀ = 8.854×10⁻¹² F/m. Treat the sheet as the infinite plane x = 0 and the line as an infinite line parallel to the z-axis through (x, y) = (9, 4). Both A(2, 9, 3) and P(6, 8, −3) lie on the side x > 0 of the sheet.

#### Potential difference due to the sheet
E_S = (ρ_s/2ε₀) a_x = (40ε₀/2ε₀) a_x = 20 a_x V/m for x > 0.

```text
V_P − V_A (sheet) = −∫(A→P) E·dl = −20 (x_P − x_A) = −20 (6 − 2) = −80 V
```

#### Potential difference due to the line
E_L = ρ_L/(2πε₀ρ) a_ρ, so moving from ρ_A to ρ_P changes the potential by −ρ_L/(2πε₀) ln(ρ_P/ρ_A).

```text
Distance to line:
 A: (2−9, 9−4) = (−7, 5)   ρ_A = √74 = 8.602 m
 P: (6−9, 8−4) = (−3, 4)   ρ_P = 5 m

ρ_L/(2πε₀) = (0.6×10⁻⁹)(17.975×10⁹) = 10.785 V

V_P − V_A (line) = −10.785 ln(5/8.602)
                 = −10.785 (−0.5426)
                 = +5.852 V
```

#### Total
```text
V_P − V_A = −80 + 5.85 = −74.15 V
V_P = V_A − 74.15 = 10 − 74.15 = −64.15 V
```

**Answer: V_P ≈ −64.1 V.**

## Check yourself

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

1. With ρ_s = 40ε0, what is the sheet's field for x > 0?
   A. 40 a_x V/m
   B. 20 a_x V/m
   C. 10 a_x V/m
   D. −20 a_x V/m
   Answer: B. E = ρ_s/(2ε0) a_x = 40ε0/(2ε0) = 20 a_x V/m on the x > 0 side, uniform and independent of distance. (Key point: Field of an infinite sheet)

2. Since the sheet's field is uniform, what is V_P − V_A due to the sheet (x_A = 2, x_P = 6)?
   A. −80 V
   B. +80 V
   C. −20 V
   D. −160 V
   Answer: A. V_P − V_A = −∫E·dl = −20(6 − 2) = −80 V; moving along the field direction lowers the potential. (Key point: Potential difference due to the sheet)

3. Why is the line charge contribution to V_P − V_A positive (+5.85 V)?
   A. The line charge is negative
   B. P (5 m) is closer to the positive line than A (8.6 m), so potential is higher at P
   C. The sheet and line cancel each other
   D. ln(5/8.602) is positive
   Answer: B. V_P − V_A = −(ρ_L/2πε0) ln(ρ_P/ρ_A). Since ρ_P < ρ_A, the logarithm is negative and the result positive: potential rises nearer a positive line. (Key point: Line charge potential change)

4. Using V_P − V_A = −80 + 5.85 and V_A = 10 V, what is V_P?
   A. −64.1 V
   B. −74.1 V
   C. 84.1 V
   D. −70.0 V
   Answer: A. Superposition gives V_P − V_A = −74.15 V, so V_P = 10 − 74.15 ≈ −64.1 V. (Key point: Combining to get V_P)

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