UY1: Electric Potential Energy With Several Point Charges

Compute electric potential energy for test charges and multi-charge systems using superposition and pairwise energy sums.

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 Potential Energy With Several Point Charges. 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: two-charge energy idea (Electric Potential Energy), scalar superposition for potential (Electric Potential)
Outcomes: compute (1) potential at a point, (2) potential energy of a test charge, and (3) total system energy using pairwise sums
Key results: V = 1/4πε₀∑ᵢqᵢ/rᵢ, U = q₀V, U_sys = 1/4πε₀∑_(i < j)qᵢqⱼ/rᵢⱼ
Common trap: double-counting interaction pairs or mixing “test-charge energy” (U = q₀V) with “system energy” (U_sys)
Potential at a point from many source charges:

V = 1/4πε₀∑ᵢqᵢ/rᵢ

Potential energy of a test charge q₀ at that point:

U = q₀V = q₀/4πε₀∑ᵢqᵢ/rᵢ

Total energy of a system of charges:

U_sys = 1/4πε₀∑_(i < j)qᵢqⱼ/rᵢⱼ.

Motivation / intuition

For many charges, force-based work can get messy fast. Potential is scalar, so you can add contributions first, then get energies with one multiplication (test charge) or one pairwise sum (full system).

2) Setup

Assume electrostatics (charges fixed in place at the instant of evaluation).

Definitions:

rᵢ: distance from test charge q₀ to source charge qᵢ.
rᵢⱼ: separation between charges qᵢ and qⱼ.
Sign convention is automatic: like-charge pairs contribute positive energy, unlike-charge pairs negative energy.

3) Core derivation/explanation

Because electric potential is scalar, you can add contributions directly:

V = ∑ᵢ(1/4πε₀)qᵢ/rᵢ.

Then multiply by q₀:

U = q₀V.

For total assembly energy, imagine bringing charges in from infinity one by one. Each new charge interacts with previously placed charges. This leads to the pairwise sum:

U_sys = 1/4πε₀∑_(i < j)qᵢqⱼ/rᵢⱼ.

The condition i < j avoids double counting.

Units: [V] = J/C = V; [U] = q₀V = C · V = J.
Limits/signs: if all separations rᵢⱼ → ∞, then U_sys → 0 for the standard “assembled from infinity” reference; like-charge pairs contribute positive terms, unlike-charge pairs negative terms.

4) Worked example(s)

Three charges:

q₁ = +2.0 μC, q₂ = -3.0 μC, q₃ = +1.0 μC

with

r₁₂ = 0.20 m, r₁₃ = 0.30 m, r₂₃ = 0.25 m.

Then

U_sys = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃)

U_sys = 8.99 × 10⁹((-6 × 10⁻¹²)/0.20 + (2 × 10⁻¹²)/0.30 + (-3 × 10⁻¹²)/0.25) ≈ -0.318 J

Negative net energy means attractive interactions dominate.

5) Practice set (with hints + answers)

Two charges + q and + q are separated by r. Write system potential energy. Hint: one pair only. Answer: U = (1/4πε₀)q²/r.
At a point, source charges give potentials + 40 V and -15 V. Find U for q₀ = 3.0 μC. Hint: add potentials first. Answer: V = 25 V, so U = q₀V = 7.5 × 10⁻⁵ J.
Why does ∑_(i < j) appear instead of ∑_(i ≠ j)? Hint: pair counting. Answer: each interacting pair should be counted once, not twice.

6) Summary + next steps

Use superposition for potential, then multiply by test charge for potential energy.
For full systems, sum pairwise terms with i < j to avoid double counting.
Sign of each term is physically meaningful and comes from qᵢqⱼ directly.

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