← Back to course

Electric Potential Energy

You’ll be able to

The Coulomb force is conservative

The work done by the electric force on a charge moving between two points is the same for every path connecting them — the Coulomb force is conservative, just like gravity. That single fact lets us define an electric potential energy U for a configuration of charges: as the charges move, the field’s work shows up as a change in U, with ΔU = −W_field. When the field does positive work (pushing charges the way they “want” to go), the stored energy drops.

Potential energy from work
ΔU = U_b − U_a = −W_field = −∫ₐᵇ →F·d→l
The line integral is path-independent, so U depends only on the configuration, not on how the charges got there.
Worked example

Derive the potential energy of two point charges q₁ and q₂ separated by distance r, taking U = 0 at infinite separation.

  1. 1.Hold q₁ fixed and let q₂ move radially from separation r out to infinity. The Coulomb force on q₂ is F = k·q₁q₂/r′² along the outward radial direction.
  2. 2.Work done by the field during that trip: W = ∫ᵣ^∞ (k·q₁q₂ / r′²) dr′.
  3. 3.Integrate: W = k·q₁q₂·[−1/r′] from r to ∞ = k·q₁q₂·(0 − (−1/r)) = k·q₁q₂/r.
  4. 4.Use ΔU = −W: U(∞) − U(r) = −k·q₁q₂/r. With U(∞) = 0, this gives U(r) = k·q₁q₂/r.
  5. 5.Sign check: for like charges (q₁q₂ > 0), U > 0 — the field does positive work flinging them apart, spending stored energy. For opposite charges U < 0: the pair is bound.
Answer: U(r) = k·q₁q₂ / r, with signs of the charges included and U(∞) = 0
Potential energy of two point charges
U = k·q₁q₂ / r, k = 1/(4πε₀) ≈ 8.99 × 10⁹ N·m²/C²
Plug the charges in *with their signs*. Note the single power of r — energy goes as 1/r, force as 1/r².

U belongs to the system, and it adds as a scalar

Potential energy is a property of the pair, not of either charge alone — it is the work an external agent must do to assemble the configuration from infinity. For several charges, add the energy of every distinct pair: U_total = Σ k·qᵢqⱼ/rᵢⱼ over all pairs. These are ordinary signed numbers, not vectors — no components, no directions. A three-charge system has exactly three pair terms.

Energy conservation with charges

Because the force is conservative, mechanical energy is conserved for a charge moving freely in an electrostatic field: K + U = constant. A released charge converts potential energy into kinetic energy exactly as a dropped mass does. This is usually the fastest route to a speed — no kinematics, no force components, just energy bookkeeping between the initial and final configurations.

Worked example

A +2.0 μC charge is bolted down. A 10 g bead carrying +3.0 μC is released from rest 0.30 m away. How fast is the bead moving when it is very far from the fixed charge?

  1. 1.Initial energy: U_i = k·q₁q₂/r = (8.99 × 10⁹)(2.0 × 10⁻⁶)(3.0 × 10⁻⁶)/0.30.
  2. 2.Numerator: (8.99 × 10⁹)(6.0 × 10⁻¹²) = 5.39 × 10⁻² J·m; dividing by 0.30 m gives U_i = 0.18 J. The bead starts at rest, so K_i = 0.
  3. 3.Far away, U_f → 0 (r → ∞). Conservation: K_f = U_i = 0.18 J.
  4. 4.Solve ½mv² = 0.18 J with m = 0.010 kg: v² = 2(0.18)/0.010 = 36 m²/s².
  5. 5.v = 6.0 m/s.
Answer: ≈ 6.0 m/s, directed radially away — all 0.18 J of stored energy becomes kinetic energy
Tip

Always carry the signs of the charges into U = kq₁q₂/r and let the algebra keep track. A negative U is meaningful — it flags a bound pair — and a decrease in U (more negative counts!) always means the field did positive work.

Checkpoint

Two positive point charges are pushed closer together. The electric potential energy of the pair:

Checkpoint

A charge is carried around a closed loop through the electrostatic field of several fixed charges, returning to its start. The net work done on it by the electric force is:

Answer the 2 checkpoints as you read.

Sign in to save your progress