UY1: Electric Field Of Uniformly Charged Disk

Derive the on-axis field of a uniformly charged disk by summing ring contributions, with limits to infinite-sheet and point-charge cases.

University Physics Year 1

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Learning objectives

Construct electric-field and potential models for discrete and continuous charge distributions.

This page gives the UY1 working model/result for Electric Field Of Uniformly Charged Disk. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Module path: Electromagnetism (UY1)
Practice: UY1 Electromagnetism Quiz
Full routing: UY1 Assessment Map
Math toolkit: Mathematics for Undergraduate Physics

1) At a glance

Prerequisites: ring field idea (stacking rings) (Field of a Ring of Charge), basic integration (Integration Techniques)
Outcomes: derive the on-axis disk field and verify the infinite-sheet and point-charge limits
Key result (on axis):

vector E(x) = (σ/2ε₀)(1-x/(square root of (x² + R²)))x hat

Common trap: mixing up σ (C/m²) with total charge Q = σπ R², or forgetting the direction flips on the opposite side of the disk
Near-disk limit (R → ∞): E → σ/(2ε₀) (infinite sheet).

Motivation / intuition

The disk result is a “calculus upgrade” of the ring: every thin ring contributes a known on-axis field, and integrating rings from 0 → R builds the full disk. The limiting cases then connect three big models: disk, sheet, and point charge.

2) Setup

Disk radius: R, surface charge density: σ (uniform).
Field point is on axis through disk center.
Symmetry removes transverse components; only axial component survives.
Use differential ring of radius r and thickness dr.

3) Core derivation/explanation

Ring element area:

dA = 2π r dr

Charge on ring:

dq = σ dA = 2πσ r dr

Axial field from ring element:

dEₓ = (1/4πε₀)(x dq)/((x² + r²)^(3/2))

Substitute dq:

dEₓ = ((σ x)/2ε₀)(r dr)/((x² + r²)^(3/2))

Integrate from r = 0 to R:

Eₓ = (σ x)/2ε₀∫₀^R(r dr)/((x² + r²)^(3/2))

Using

∫(r dr)/((x² + r²)^(3/2)) = -1/(square root of (x² + r²))

we obtain:

Eₓ = (σ/2ε₀)(1-x/(square root of (x² + R²)))

Direction is away from disk if σ > 0.

Checks:

x≫ R: behaves like point charge Q = σπ R².
R≫ x: approaches infinite-sheet field σ/(2ε₀).

This is the on-axis field of a thin, uniformly charged disk. Off-axis fields are more complicated.
The “infinite sheet” limit is an approximation that becomes accurate when R≫ x (you are close compared to the disk radius).

Units: [σ/ε₀] = (C/m²)/(C²/(N m²)) = N/C.
Limits/signs: R → ∞ ⇒ E → σ/(2ε₀); x≫ R ⇒ E ≈ kQ/x²; changing the sign of σ flips the direction.

4) Worked example(s)

Given σ = 3.0 × 10⁻⁶ C m⁻², R = 0.20 m, and x = 0.10 m:

E = (σ/2ε₀)(1-0.10/(square root of (0.10² + 0.20²)))

σ/2ε₀ = (3.0 × 10⁻⁶)/(2(8.854 × 10⁻¹²)) ≈ 1.69 × 10⁵ N C⁻¹

Bracket term:

1-0.10/(square root of 0.05) ≈ 1-0.447 = 0.553

So:

E ≈ 9.37 × 10⁴ N C⁻¹

Direction: + x hat for positive disk and x > 0.

5) Practice set (with hints + answers)

What is E at the center of a uniformly charged disk along its axis from this formula? Hint: set x = 0. Answer: E = σ/(2ε₀) on the immediate side of the disk (direction normal to surface).
If x is doubled while R and σ are fixed, does E increase or decrease? Hint: inspect the bracket term. Answer: decreases.
In the limit R → ∞, what does E become? Hint: x/square root of (x² + R²) → 0. Answer: E = σ/(2ε₀).

6) Summary + next steps

Disk field is a ring-integration result with clean symmetry.
Limiting cases check physics: infinite sheet nearby, point charge far away.
Sign and direction follow the sign of σ and side of the disk.

Next: Electric Field Of Two Oppositely Charged Infinite Sheets Previous: Electric Field Of A Line Of Charge Back To Electromagnetism