Potential energy of a system of charges and of a dipole
Assembling charges stores energy U = kq₁q₂/r for each pair. A dipole in a uniform field has U = −pE cos θ, lowest when p lines up with E.
Derivation: potential energy of two point charges
Build the system by bringing q₁ and then q₂ from infinity, slowly. The work done is stored as potential energy U.
- 1Bring q₁ to its place first. No other charge is present, so no work is needed.
- 2q₁ now sets up a potential V₁ = q₁/(4πε₀r₁₂) at the place where q₂ will go.
- 3Bring q₂ from infinity to that place. Work done = q₂ × V₁ = q₁q₂/(4πε₀r₁₂).
- 4★So, U = (1/4πε₀) q₁q₂/r₁₂. The force is conservative, so U does not depend on the order or the path.
- 5If q₁q₂ > 0 (like charges), U > 0: work is done to push them together. If q₁q₂ < 0, U < 0.
★ key step
Three or more charges
For more charges, add the energy of every pair, counting each pair once. For three charges: U = (1/4πε₀)(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃). For a single charge q placed in an external potential V, the potential energy is simply U = qV. A negative U means the system is bound: work must be supplied to pull the charges apart to infinity.
Derivation: potential energy of a dipole in a uniform field
A dipole of moment p is in a uniform field E. The torque on it is τ = pE sin θ.
- 1Turn the dipole slowly from angle θ₀ to θ₁ with an external torque equal and opposite to τ.
- 2Work done = ∫ pE sin θ dθ, with θ going from θ₀ to θ₁.
- 3Since ∫ sin θ dθ = −cos θ, W = pE(cos θ₀ − cos θ₁).
- 4This work is stored as the change in potential energy: U(θ₁) − U(θ₀) = pE(cos θ₀ − cos θ₁).
- 5Choose U = 0 when the dipole is at 90° to E (θ₀ = 90°, cos θ₀ = 0).
- 6★So, U(θ) = −pE cos θ = −p·E.
★ key step
Dipole energy at key angles
| Angle θ | U = −pE cos θ | Equilibrium |
|---|---|---|
| 0° (p along E) | −pE, the lowest | Stable |
| 90° | 0 | Not in equilibrium (torque is pE) |
| 180° (p against E) | +pE, the highest | Unstable |
Worked example: energy of two charges
Charges of +2 μC and −3 μC are 30 cm apart in air. (a) Find the potential energy of the pair. (b) How much work is needed to separate them completely?
- (a) U = kq₁q₂/r = 9 × 10⁹ × (2 × 10⁻⁶)(−3 × 10⁻⁶) ÷ 0.30.
- U = 9 × 10⁹ × (−6 × 10⁻¹²) ÷ 0.30 = −0.054 ÷ 0.30 = −0.18 J.
- (b) At infinite separation U = 0. Work needed = 0 − (−0.18 J) = +0.18 J.
(a) −0.18 J; (b) 0.18 J
Worked example: three charges at a triangle
Three charges of +3 μC each sit at the corners of an equilateral triangle of side 20 cm in air. Find the potential energy of the system. (1/4πε₀ = 9 × 10⁹ N m² C⁻²)
- There are three pairs, and each pair is 0.20 m apart.
- Energy of one pair: kq²/r = 9 × 10⁹ × (3 × 10⁻⁶)² ÷ 0.20 = 9 × 10⁹ × 9 × 10⁻¹² ÷ 0.20 = 0.081 ÷ 0.20 = 0.405 J.
- Total: U = 3 × 0.405 J = 1.215 J.
- U is positive: work had to be done to push the like charges together.
About 1.2 J
Every chemical bond stores energy as electric potential energy between charges. When LPG burns in a kitchen stove, the charges in the molecules settle into a lower energy, and the difference comes out as heat.
1 marks: “What is the sign of the potential energy of a proton and an electron held some distance apart?” Negative, because the charges are unlike: U = kq₁q₂/r with q₁q₂ less than zero.
2 marks: “A dipole of moment 2 × 10⁻⁸ C m lies along a uniform field of 5 × 10⁴ N C⁻¹. How much work is needed to turn it through 180°?” W = pE(cos 0° − cos 180°) = 2 × 10⁻⁸ × 5 × 10⁴ × (1 + 1) = 2 × 10⁻³ J.
3 marks: “Find the potential energy of a dipole in a uniform field E. In which position is it in stable equilibrium, and in which unstable?” Turning it against τ = pE sin θ needs W = pE(cos θ₀ − cos θ). With U = 0 at 90°, U = −pE cos θ = −p·E. Stable at θ = 0° (U = −pE, least); unstable at θ = 180° (U = +pE, greatest).
Like pair, plus energy; unlike pair, minus energy. For a dipole: lined up is lowest (−pE), turned back is highest (+pE).