UY1: Electric Potential

Define electric potential and potential difference, connect them to work and field, and compute potentials for point-charge systems.

University Physics Year 1

On this page

Learning objectives

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

This page gives the UY1 working model/result for Electric Potential. 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: work/energy sign conventions (Electric Potential Energy), dot product + line integrals (Vector Calculus)
Outcomes: define V and Δ V, compute potentials for point-charge systems, and use Δ V = -∫ vector E · d vector l with correct sign/reference
Key results: V = U/q₀, V_b-Vₐ = -∫ₐ^b vector E · d vector l, V(r) = (1/4πε₀)q/r (V(∞) = 0)
Common trap: dropping the minus sign in Δ V = -∫ vector E · d vector l or switching reference points mid-solution
Electric potential is potential energy per unit charge:

V = U/q₀

Potential difference and work relation:

V_b-Vₐ = -∫ₐ^b vector E · d vector l

Point charge potential:

V(r) = (1/4πε₀)q/r

Unit: 1 V = 1 J C⁻¹.

Motivation / intuition

Electric potential turns a vector-field problem into a scalar-field problem. That’s why it is so useful for multiple charges: you add potentials directly, then convert back to forces/fields only when you need directions.

2) Setup

Pick reference point for potential (often V(∞) = 0).
Potential is scalar, so you add values algebraically (not vectorially).
Sign rule in uniform field: moving along vector E reduces V.

3) Core derivation/explanation

Start from work-energy relation for a test charge q₀:

W_(a → b) = -(U_b-Uₐ)

Divide by q₀:

(W_(a → b))/q₀ = Vₐ-V_b

Why this step matters

Dividing by q₀ removes dependence on the test charge and defines a property of the field configuration itself. That is why potential is reusable for any charge placed at that point.

Using vector F = q₀ vector E,

(W_(a → b))/q₀ = ∫ₐ^b vector E · d vector l

So:

V_b-Vₐ = -∫ₐ^b vector E · d vector l

For a point charge with V(∞) = 0:

V(r) = (1/4πε₀)q/r

For multiple point charges:

Vₙₑₜ = ∑ᵢ (1/4πε₀)qᵢ/rᵢ

Dropping the minus sign in V_b-Vₐ = -∫ₐ^b vector E · d vector l.
Mixing up potential (scalar) with field (vector): V adds algebraically, vector E adds by components.
Using inconsistent reference points (for example, switching between finite reference and V(∞) = 0 mid-solution).

Units: [V] = [U/q₀] = J/C = V and [∫ vector E · d vector l] = (N/C)m = V.
Limits/signs: for a point charge with V(∞) = 0, you must have V → 0 as r → ∞; moving in the direction of vector E makes V decrease (because of the minus sign).

4) Worked example(s)

Two charges create potential at point P:

q₁ = +4.0 nC, r₁ = 0.20 m; q₂ = -2.0 nC, r₂ = 0.30 m

V(P) = (1/4πε₀)(q₁/r₁ + q₂/r₂) = (8.99 × 10⁹)((4.0 × 10⁻⁹)/0.20-(2.0 × 10⁻⁹)/0.30)

V(P) ≈ 120 V

Quick sign check: positive contribution dominates, so net V is positive.

5) Practice set (with hints + answers)

Find potential at r = 0.50 m from q = +3.0 nC. Hint: V = kq/r. Answer: V ≈ 54 V.
A particle moves in a uniform field from a to b such that ∫ₐ^b vector E · d vector l = 25 V. Find V_b-Vₐ. Hint: include minus sign. Answer: V_b-Vₐ = -25 V.
Why is potential easier than field for multiple point charges? Hint: scalar vs vector addition. Answer: potentials add algebraically; no component decomposition needed.

6) Summary + next steps

Potential translates field-work ideas into a scalar quantity.
The negative sign in Δ V = -∫ vector E · d vector l encodes direction of decreasing potential.
Point-charge and superposition formulas are the base for gradient and equipotential topics.

Next: Potential Gradient Previous: Electric Potential Energy With Several Point Charges Back To Electromagnetism