Potential energy of a system of charges and of a dipole in a field
Two charges store U = q₁q₂/(4πε₀r), the work done to bring them together from infinity. A dipole in a uniform field has U = −pE cos θ.
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: turning a dipole
A dipole of moment 2 × 10⁻⁸ C m lies along a uniform field of 5 × 10⁴ N C⁻¹. Find the work needed to turn it (a) through 90° and (b) through 180°.
- pE = 2 × 10⁻⁸ × 5 × 10⁴ = 1 × 10⁻³ J.
- (a) W = pE(cos 0° − cos 90°) = 1 × 10⁻³ × (1 − 0) = 1 × 10⁻³ J.
- (b) W = pE(cos 0° − cos 180°) = 1 × 10⁻³ × (1 + 1) = 2 × 10⁻³ J.
(a) 1 × 10⁻³ J; (b) 2 × 10⁻³ J
The energy in the food you eat and in the petrol of a scooter is really electrostatic potential energy, stored in how the charges in molecules are arranged. Digestion or burning moves these charges to a lower-energy arrangement and releases the difference.
3 marks: “Derive an expression for the potential energy of an electric dipole in a uniform electric field. Find the work done in turning it from the stable to the unstable position.” W = ∫ pE sin θ dθ = pE(cos θ₀ − cos θ₁); with U(90°) = 0, U = −pE cos θ. From 0° to 180°: W = 2pE.
2 marks: “Three charges of +1 μC each sit at the corners of an equilateral triangle of side 10 cm. Find the potential energy of the system.” Three equal pairs: U = 3kq²/a = 3 × 9 × 10⁹ × (10⁻⁶)² ÷ 0.10 = 3 × 0.09 = 0.27 J.
1 marks: “In which orientation is the potential energy of a dipole in a uniform field the least?” When p is along E (θ = 0°): U = −pE.
Count pairs, not charges: each pair once. Dipole energy is −pE cos θ: lowest when it lines up with the field.