Mini Physics
Part of Mini EducationSwitch subject

UY1: Electric Potential Of A Ring Of Charge

Derive the on-axis electric potential of a uniformly charged ring and verify it against the electric field relation.

  • University Physics Year 1
On this page

Learning objectives

  • Construct electric-field and potential models for discrete and continuous charge distributions.
Why this matters + quick links

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

1) At a glance

  • Prerequisites: potential definition (Electric Potential) and symmetry/integration (Integration Techniques)
  • Outcomes: derive V(x) on the ring axis and verify it by differentiating to recover Eₓ = -dV/dx
  • Key result (on axis, V(∞) = 0):
V(x) = (1/4πε₀)Q/(square root of (a² + x²))
  • Common trap: mixing potential (scalar) with field (vector) or forgetting to state the reference V(∞) = 0
  • For a thin ring of radius a carrying total charge Q, the potential at an axis point x from the center is:
V(x) = (1/4πε₀)Q/(square root of (a² + x²))
  • This result assumes V(∞) = 0.
  • Since potential is a scalar, all charge elements add directly (no vector components needed).

Motivation / intuition

This integral is unusually simple because every charge element on the ring is the same distance from an on-axis point. That’s why r = square root of (a² + x²) is constant and the integration is basically “sum dq to get Q”.

2) Setup

Assume:

  • Ring is thin and uniformly charged.
  • Total charge is Q and radius is a.
  • Observation point P is on the symmetry axis at coordinate x.
  • Reference potential is zero at infinity.

Define linear charge density:

λ = Q/(2π a)

Small element of charge on arc a dθ:

dq = λ a dθ

Distance from every element to P:

r = square root of (a² + x²)

3) Core derivation/explanation

Potential contribution from dq:

dV = (1/4πε₀)dq/r = (1/4πε₀)(λ a dθ)/(square root of (a² + x²))

Integrate around the full ring (θ:0 → 2π):

V(x) = ∫₀^2π dV; = (1/4πε₀)(λ a)/(square root of (a² + x²))∫₀^2πdθ; = (1/4πε₀)Q/(square root of (a² + x²))

Sign convention: if Q > 0, then V > 0; if Q < 0, then V < 0.

Check against field relation on axis:

Eₓ = -dV/dx = (1/4πε₀)Qx/((a² + x²)^(3/2))

which matches the standard ring-field result.

Quick checks (units + limits/sign)
  • Units: [kQ/square root of (a² + x²)] = (N m² C⁻²)C/m = N m/C = V.
  • Limits/signs: x → ∞ ⇒ V → 0 (consistent with V(∞) = 0); x = 0 ⇒ V(0) = kQ/a; the sign of Q sets the sign of V.

4) Worked example(s)

Given Q = 6.0 μC, a = 0.30 m, x = 0.40 m:

square root of (a² + x²) = square root of (0.30² + 0.40²) = 0.50 m
V = (1/4πε₀)Q/(square root of (a² + x²)) = (8.99 × 10⁹)(6.0 × 10⁻⁶)/0.50 ≈ 1.08 × 10⁵ V

5) Practice set (with hints + answers)

  1. A ring has Q = 2.0 μC and a = 0.20 m. Find V at the center (x = 0). Hint: center distance to every element is a. Answer: V = (1/4πε₀)Q/a ≈ 9.0 × 10⁴ V.

  2. For fixed Q and a, what happens to V(x) as x → ∞? Hint: compare with point-charge form. Answer: V → 0, consistent with the chosen reference V(∞) = 0.

  3. If Q is doubled and x is unchanged, how does V change? Hint: inspect linearity in Q. Answer: V doubles.

6) Summary + next steps

  • Symmetry makes the ring-potential integration straightforward because r is constant around the ring.
  • The final form V(x) = (1/4πε₀)Q/(square root of (a² + x²)) is physically consistent with Eₓ = -dV/dx.
  • Keep sign conventions explicit: charge sign sets potential sign.

Next: Electric Potential Of A Line Of Charge Previous: Potential Gradient Back To Electromagnetism (UY1)