POTENTIAL ENERGY IN AN EXTERNAL FIELD

This section finds the potential energy of charges placed in an external electric field, one produced by charges outside the system, in three steps: a single charge, a system of two charges, and an electric dipole.

An external field is produced by sources that are not part of the system being studied and are assumed to stay undisturbed when the system is brought in. Its potential V(r) gives the work needed, per unit charge, to bring a charge from infinity to the point r. The three sub-lessons build on this idea: a single charge q has potential energy qV(r); two charges have q₁V(r₁) + q₂V(r₂) + q₁q₂/(4πε₀r₁₂), which adds their mutual interaction; and a dipole in a uniform field feels a torque τ = p × E and has potential energy U(θ) = −pE cos θ = −p·E, lowest when it lines up with the field.

Potential Energy in an External Field: What the Section Covers

The earlier section on the potential energy of a system of charges looked at the energy stored by charges because of their effect on one another. Here the charges sit in an external field E, produced by some other set of charges, for example the charged plates of a capacitor. Those sources are taken as fixed, so that bringing the system in does not change the field. The field is described by its potential V(r), with V = 0 at infinity, and V(r) is the work done by an external agent, per unit charge, in bringing a charge from infinity to r.

2.7.1 Potential energy of a single charge

• Bringing a charge q from infinity to the point r in the external field takes work qV(r). This work is stored as the charge's potential energy: U = qV(r).
• Because the electrostatic force is conservative, the result does not depend on the path taken.
• This section also introduces a convenient unit of energy, the electron volt: the energy gained by an electron accelerated through a potential difference of 1 V, so 1 eV = 1.6 × 10⁻¹⁹ J.

2.7.2 Potential energy of a system of two charges in an external field

• Bringing q₁ to r₁ takes work q₁V(r₁). Bringing q₂ to r₂ takes work against the external field, q₂V(r₂), and against the field of q₁, which is q₁q₂/(4πε₀r₁₂).
• Adding these gives U = q₁V(r₁) + q₂V(r₂) + q₁q₂/(4πε₀r₁₂), where r₁₂ is the distance between the charges.
• Example: charges of 7 μC and −2 μC placed 18 cm apart have a mutual energy of (9 × 10⁹)(7 × 10⁻⁶)(−2 × 10⁻⁶)/0.18 = −0.7 J, so 0.7 J of work would separate them completely. If both sit 9 cm from the origin in an external field whose potential is V(r) = A/r with A = 9 × 10⁵ V m, then V = 10⁷ V at each charge, and the total energy is 70 J − 20 J − 0.7 J = 49.3 J.

2.7.3 Potential energy of a dipole in an external field

• A dipole of charges +q and −q has dipole moment p. In a uniform field the net force on it is zero, but it feels a torque τ = p × E, of magnitude pE sin θ, which tries to turn p along E.
• Rotating the dipole slowly from angle θ₀ to θ₁ against this torque takes work W = pE(cos θ₀ − cos θ₁), which is stored as potential energy.
• Choosing the zero of energy at θ₀ = π/2 gives U(θ) = −pE cos θ = −p·E. The energy is lowest, −pE, at θ = 0 (stable equilibrium) and highest, +pE, at θ = π (unstable equilibrium). The same result follows from applying the single-charge formula to each charge of the dipole.
• Example: a mole of molecules, each with dipole moment 10⁻²⁹ C m, fully aligned in a field of 10⁶ V m⁻¹ has a total moment of 6 × 10⁻⁶ C m and energy −6 J. If the field direction suddenly turns through 60°, the energy becomes −6 cos 60° = −3 J, so when the dipoles realign with the new direction the substance releases 3 J as heat.

Together, the three sub-lessons show how one idea, the work needed to assemble charges against a field, gives the energy of any arrangement of charges in an external field.