Potential energy of a system of charges

Energy of a group of charges

Narrated video. Transcript:

Start with two charges, q₁ and q₂, far apart at infinity. How much work does it take to put them in place, a distance r₁₂ apart?

Bring q₁ in first. There is no field yet to push against, so this costs no work at all.

Now bring q₂ in from infinity. It moves in the potential of q₁, which at its final spot is kq₁/r₁₂. So the work is q₂ times that: kq₁q₂/r₁₂.

This work is stored as the potential energy of the pair: U = q₁q₂/4πε₀r₁₂. Bring them in the other order, and you get the same answer.

For like charges, q₁q₂ is positive, so U is positive: you had to push them together. For unlike charges, U is negative: they pull together, and you must do work to separate them.

A third charge q₃ feels the potential of both q₁ and q₂. The work to bring it in is q₃(kq₁/r₁₃ + kq₂/r₂₃).

Add it all up: U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃). There is one term for each pair of charges, and the order of assembly does not matter.

Worked example: 7 μC at x = −9 cm and −2 μC at x = +9 cm, so 18 cm apart. U = 9 × 10⁹ × 7 × 10⁻⁶ × (−2 × 10⁻⁶) ÷ 0.18. That is −0.7 joules.

The energy is negative, so the pair is bound. To pull them apart to infinity, an external agent must supply 0.7 joules of work.

Energy in an external field

Narrated video. Transcript:

Now place charges in an external field, made by sources outside our system. Let V(r) be the potential of that external field at the point r.

Bringing a single charge q from infinity to r against this field takes work qV(r). So its potential energy there is qV(r).

An electron moved through a potential difference of 1 volt gains or loses 1.6 × 10⁻¹⁹ joules. This handy unit of energy is the electron volt, eV.

For two charges in an external field, add three pieces: q₁V(r₁) and q₂V(r₂) for each charge in the field, plus their mutual energy, kq₁q₂/r₁₂.

Example: put our 7 μC and −2 μC pair in a field whose potential is V = A/r, with A = 9 × 10⁵ V m. Both charges are 9 cm from the origin, where V = 9 × 10⁵ ÷ 0.09 = 10⁷ volts.

Then q₁V = 7 × 10⁻⁶ × 10⁷ = 70 J, and q₂V = −20 J. Add the mutual energy, −0.7 J. The total is 49.3 joules.

Last, a dipole in a uniform field E. The forces +qE and −qE are equal and opposite, so there is no net force. But they form a couple, with torque τ = pE sin θ.

Turning the dipole from θ₀ to θ₁ against this torque takes work pE(cos θ₀ − cos θ₁). Choose the zero of energy at θ₀ = 90°. Then U(θ) = −pE cos θ = −p·E.

U is lowest, −pE, when p points along E: stable equilibrium. It is highest, +pE, when p points against E. Turning a dipole from 0° to 180° takes work 2pE.

Try it: assemble the charges

Interactive simulation. Things to try:

• The pair starts as the worked example: 7 μC and −2 μC, 18 cm apart, U = −0.70 J.
• Drag them closer. They are unlike charges, so U becomes more negative. Make both positive with the slider and try again.
• Add a third charge. The total has one term for each pair, and the “one by one” card gives the same total.
• Halve the distance between two charges: their pair energy doubles.

Try it: a dipole in a field

Interactive simulation. Things to try:

• Drag the + charge round. The point on the graph follows U = −pE cos θ.
• At θ = 0 the torque is zero and U is lowest: stable equilibrium. At 180° the torque is zero too, but the slightest nudge tips it over.
• Turn to θ = 0, press “Mark start θ₀”, then turn to 90° and to 180°. The work is pE and then 2pE.
• Where is the torque largest? Watch the τ readout as you turn.

Charges on a square

Charges +2 μC, −2 μC, +2 μC, −2 μC sit in turn at the corners of a square of side 10 cm. What is the potential energy of the system?

1. −0.93 J
2. −1.44 J
3. −1.95 J
4. +0.93 J

Turn the dipole round

A dipole with p = 4 × 10⁻⁸ C m lies along a uniform field of 5 × 10⁴ N/C (θ = 0). How much work is needed to turn it to θ = 180°?

1. 0, because the torque is zero at both ends
2. 2 × 10⁻³ J
3. 4 × 10⁻³ J
4. −4 × 10⁻³ J

Summary

• Two charges: U = q₁q₂/4πε₀r₁₂
• Three charges: U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃)
• Charge in an external field: U = qV(r)
• Dipole in a uniform field: U(θ) = −pE cos θ = −p·E

• The potential energy of a system is the work done to assemble it from infinity; there is one term for each pair, whatever the order.
• U is positive for like charges and negative for unlike charges.
• Two charges in an external field: U = q₁V(r₁) + q₂V(r₂) + kq₁q₂/r₁₂. 1 eV = 1.6 × 10⁻¹⁹ J.
• A dipole in a uniform field feels a torque pE sin θ but no net force; it is stable at θ = 0.