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Electric Potential · 12.1

Electric Potential Energy

Electric potential energy belongs to a system of charges. Its change tells us exactly how much work the electric field does.

01

Build the model

Choose a reference, then keep every sign.

Choose a reference before assigning potential energy. For two point charges, U = 0 at infinite separation is the usual choice. Like charges then have positive U and unlike charges negative U; the electric field does work whenever the system moves toward lower U.

Simple definition
Electric potential energy is energy a system has because of the positions of its charges relative to a chosen reference.
Example
When a positive and negative charge move closer after release, their electric potential energy decreases while kinetic energy increases.
Two point chargesU = kQq/r

The stored energy of two charges held a distance apart — positive for like charges (they want to fly apart), negative for opposites.

The signs of Q and q belong in the calculation

Energy changeΔU = Uᵦ − Uₐ

Only changes in stored energy matter: final minus initial, once you fix a reference point.

Final minus initial, after fixing one reference

Work by the fieldWfield = −ΔU

When the field does positive work (charges move the way they're pulled), stored energy drops by the same amount.

Positive field work means the system loses potential energy

01

Energy belongs to the pair

U describes the configuration of both interacting charges. It is not stored inside one charge by itself.

02

Signs predict the tendency

Like charges have U > 0 and naturally separate. Unlike charges have U < 0 and naturally move closer when released.

03

Keep the work agent clear

Work by the electric field is −ΔU. Slow external work with negligible kinetic-energy change is +ΔU.

02

Change one quantity at a time

Make potential visible.

The model uses U = 0 at infinite separation. Move either charge slider through zero to see why the interaction disappears.

Initial U−0.067 J

Final U−0.18 J

Change ΔU−0.112 J

Work by field+0.112 J

Natural tendencyattraction lowers U by decreasing r

03

Catch the common trap

Predict the sign before calculating.

A positive charge moves freely in the direction of an electric field. What happens?

Choose an answer, then explain the sign before revealing the feedback.

04

Worked examples

Reference, relationship, sign, sense-check.

EasyTwo +1.0 μC charges are 0.10 m apart. Find their potential energy.
  1. U = kq₁q₂/r = 8.99 × 10⁹ × 10⁻¹² ÷ 0.10.
  2. U ≈ +0.090 J — positive, they'd fly apart.

AnswerU ≈ +0.090 J

MediumA +3.0 μC charge and a −2.0 μC charge are 0.20 m apart. Take U = 0 at infinity. Find their electric potential energy.
  1. Use U = kQq/r and retain both charge signs.
  2. U = (8.99 × 10⁹)(+3.0 × 10⁻⁶)(−2.0 × 10⁻⁶)/(0.20).
  3. The negative result matches an attractive, bound configuration relative to infinity.

AnswerU = −0.270 J

HardThree +2.0 μC charges occupy the corners of an equilateral triangle of side 0.30 m. Find the total stored energy of the arrangement.
  1. Three equal pairs: U(pair) = 8.99 × 10⁹ × 4.0 × 10⁻¹² ÷ 0.30 ≈ 0.12 J.
  2. U(total) = 3 × 0.12.
  3. ≈ 0.36 J — the work needed to assemble the triangle from infinity.

AnswerU ≈ 0.36 J

ChallengingAn electron is released from rest 0.53 × 10⁻¹⁰ m from a proton (hydrogen's radius). Using energy methods, find its speed when it has fallen to half that distance.
  1. U = −ke²/r: initially −4.35 × 10⁻¹⁸ J; at r/2 it is −8.70 × 10⁻¹⁸ J.
  2. K gained = 4.35 × 10⁻¹⁸ J.
  3. v = √(2K/mₑ) = √(9.55 × 10¹²) ≈ 3.1 × 10⁶ m/s — 1% of light speed.

Answerv ≈ 3.1 × 10⁶ m/s