---
title: "Potential on axis of charged disk"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:disc-axial-potential-field"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/disc-axial-potential-field
site: BCT Engineering
---

# Potential on axis of charged disk

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2067 Asar, 5 marks
- Video: [Electric Potential due to Disc of Charge](https://www.youtube.com/watch?v=Ahae5cqpmTg) — Andrey K
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/disc-axial-potential-field
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

Derive an expression for the electric potential at any point on the axis of a uniformly charged disk. Extend your result to calculate electric field.

## Short answer

By integrating ring contributions, the axial potential is V = (σ/(2ε₀))[√(z²+R²) − z]. The electric field from E = −dV/dz is E = (σ/(2ε₀))[1 − z/√(z²+R²)] directed away from the disk.

## Answer

Disc of radius R with uniform surface charge density σ. Find V and E at point P on the axis, distance z from the centre.

```text
            P
            .
           /|
        s / | z
         /  |
   -----o---O-----   disc
      ring  R
      (radius w, width dw)
```

#### Potential
1. Divide the disc into thin rings of radius w and width dw. Charge on a ring: dq = σ (2πw dw).
2. Every point on the ring is at distance s = √(w² + z²) from P, so its potential is

dV = (1/4πε0) dq/s = (1/4πε0) σ 2πw dw / √(w² + z²) = (σ/2ε0) w dw / √(w² + z²)

3. Integrate w from 0 to R:

V = (σ/2ε0) ∫₀ᴿ w dw/√(w² + z²) = (σ/2ε0) [√(w² + z²)]₀ᴿ

**V = (σ/2ε0) [√(z² + R²) − z]**

#### Electric field
Since E = −∇V and by symmetry E is along the axis:

E = −dV/dz = −(σ/2ε0) [ z/√(z² + R²) − 1 ]

**E = (σ/2ε0) [1 − z/√(z² + R²)]**, directed along the axis away from the disc (σ > 0).

#### Limiting cases
- R → ∞ (infinite sheet): E = σ/(2ε0), uniform.
- z ≫ R: expanding gives V → Q/(4πε0 z) and E → Q/(4πε0 z²) with Q = σπR², i.e. a point charge.

## Check yourself

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

1. What is the charge on a ring of radius w and width dw?
   A. σ π w^2
   B. σ 2πw dw
   C. σ w dw
   D. σ π dw
   Answer: B. Ring area is 2πw dw, so dq = σ(2πw dw). (Key point: Ring element)

2. The axial potential of the disc is
   A. (σ/2ε0)[sqrt(z^2+R^2) − z]
   B. (σ/2ε0)[sqrt(z^2+R^2) + z]
   C. (σ/2ε0) z/sqrt(z^2+R^2)
   D. σR^2/(4ε0 z)
   Answer: A. Integrating w dw/sqrt(w^2+z^2) from 0 to R gives sqrt(R^2+z^2) − z. (Key point: Potential)

3. How is the axial field obtained from V?
   A. E = V/z
   B. E = −dV/dz
   C. E = ∫V dz
   D. E = dV/dz^2
   Answer: B. By symmetry E lies along the axis and equals −dV/dz. (Key point: Field from V)

4. As R → ∞, the disc field becomes
   A. Zero
   B. σ/ε0
   C. σ/(2ε0), uniform
   D. Q/(4πε0 z^2)
   Answer: C. z/sqrt(z^2+R^2) → 0, leaving E = σ/(2ε0). (Key point: Infinite sheet)

## Related in Engineering Physics

- [Field on axis of charged ring](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aring-axial-field-maximum) — Engineering Physics, chapter 7, asked 4×
- [Field of charged rod along its axis](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acharged-rod-field-along-axis) — Engineering Physics, chapter 7
- [Field on axis of charged disc](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adisc-axial-field) — Engineering Physics, chapter 7, asked 3×
- [Field inside uniformly charged cylinder](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acylinder-uniform-charge-field-inside) — Engineering Physics, chapter 7
- [Ring on axis: potential then field](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aring-axial-potential-then-field) — Engineering Physics, chapter 7, asked 2×
- [Field inside uniformly charged sphere](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asphere-uniform-charge-field-inside) — Engineering Physics, chapter 7

## Related in other subjects

- [Electric dipole potential and field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Adipole-potential-and-field-derive) — Electromagnetism, chapter 2, asked 4×
- [Potential gradient gives electric field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Apotential-gradient-gives-electric-field) — Electromagnetism, chapter 2
- [pn junction 10¹⁸/10¹⁶: potential, width, field](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-1e18-1e16-max-field) — Electrical Engineering Material, chapter 6, asked 2×
- [Derive PET expression by Penman's method](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aderive-penman-pet) — Engineering Hydrology, chapter 3