Electric potential energy

Start from the definition W = ∫ F ⋅ dl and split each step dl into a piece along the line joining the charges and a piece perpendicular to it. The Coulomb force is purely radial, so the perpendicular piece contributes nothing, and the whole integral collapses to ∫ Fᵣ dr taken between the starting and finishing separations. With Fᵣ = k q₁q₂ / r², that integral is elementary: W = k q₁q₂ (1/rₐ − 1/r_b). Take q₁ = +5.0 nC and q₂ = +8.0 nC, and let the separation grow from 4.0 cm to 10.0 cm. Then k q₁q₂ = 3.596 × 10⁻⁷ J m and W = 3.596 × 10⁻⁷ (25 − 10) = 5.39 × 10⁻⁶ J. Positive, as it must be: the charges repel and they are moving apart. Nothing in the answer records how they got there. The force does the same work on a charge dragged out along a spiral as on one carried straight out, because every sideways step is perpendicular to it.

Because the work depends only on the endpoints, you can attach a number to each separation and read the work off as a difference. Define it so that the stored energy falls by exactly the work the force does: ΔU = −W. Then U(r_b) − U(rₐ) = k q₁q₂ (1/r_b − 1/rₐ), which fixes U to within one additive constant. Choosing U = 0 at infinite separation kills the constant and leaves U(r) = k q₁q₂ / r. Return to the pair above: U(4.0 cm) = 3.596 × 10⁻⁷ / 0.040 = 8.99 μJ and U(10.0 cm) = 3.60 μJ, a drop of 5.39 μJ — exactly the work found by integrating. The zero is a convention, not a measurement. Shift it by 3 J and both numbers shift by 3 J, while every difference, and therefore every speed you will ever predict, is untouched.

Enter both charges signed and U arrives with its own sign. Like charges make q₁q₂ positive, so U > 0 at every separation and grows without bound as r → 0: assembling such a pair costs work, and the pair sits loaded like a compressed spring, ready to convert U into kinetic energy the moment it is released. Unlike charges make q₁q₂ negative, so U < 0 and falls towards −∞ as r → 0: the pair is bound, and an outside agent must supply the whole of |U| to carry them to infinity. In a hydrogen atom the electron sits about 0.0529 nm from the proton, giving U = −k e²/r = −4.36 × 10⁻¹⁸ J = −27.2 eV. That minus sign is not a statement that the energy is small; it is a debt that must be paid before the two can be separated.

Build the arrangement one charge at a time from infinity. The first arrives free, because there is nothing for it to interact with. The second costs k q₁q₂ / r₁₂. The third costs k q₃ (q₁/r₁₃ + q₂/r₂₃), one term for each charge already in place. Add them and the total is a sum over distinct pairs — and it comes out the same whatever order you use, which is what lets you call it a property of the configuration. Put +6.0 nC, +6.0 nC and −6.0 nC at the corners of an equilateral triangle of side 5.0 cm: the like pair contributes +6.47 μJ, each unlike pair contributes −6.47 μJ, and the total is −6.47 μJ. Negative means net bound, so 6.47 μJ must be supplied to scatter all three to infinity. Three charges give three terms, four give six, and no charge is ever paired with itself.

Once U is known everywhere, release the charge and impose K + U = constant. Fire a proton from far away straight at a bead holding +2.0 nC fixed, with initial speed 3.0 × 10⁵ m s⁻¹. Far away U ≈ 0, so the total energy is K₀ = ½(1.673 × 10⁻²⁷ kg)(3.0 × 10⁵ m s⁻¹)² = 7.53 × 10⁻¹⁷ J, which is 470 eV. At the closest approach the proton has stopped, so K = 0 and all of that energy has gone into U: k(2.0 × 10⁻⁹)(1.602 × 10⁻¹⁹) / rₘᵢₙ = 7.53 × 10⁻¹⁷ J, giving rₘᵢₙ = 3.8 × 10⁻² m. The structure of that result is worth as much as the number. Since U goes as 1/r, rₘᵢₙ goes as 1/K₀ and therefore as 1/v₀², so doubling the launch speed cuts the closest approach to a quarter.

Three limits come with the model. It is electrostatic: charges are assumed to be moved slowly, so nothing radiates and no energy leaks into the electromagnetic field on the way. It treats the charges as points, so U diverges as r → 0 and the energy stored inside each particle is excluded by construction — that exclusion is precisely why a configuration of N point charges carries N(N − 1)/2 terms and not one more. And it counts only the electric interaction: the insulating rod, the clamp or the hand that holds a charge in place carries energy of its own that never appears here. One habit follows from all three. U is not the energy of a charge. It takes two charges to have a pair energy, and moving either one of them changes it by the same amount.