Force superposition and equilibrium

Coulomb's law tells you the force between two charges and nothing more. What lets you handle three is a separate claim, and it was tested before it was assumed: bring a third charge up and the force between the first two does not change. Each pair acts as though alone, using only its own two charges and its own separation, and the total on any one charge is the vector sum of those independent contributions. Nothing screens — a charge parked directly between two others does not weaken the force they exert on each other. This is linearity, and it is why N charges never become an N-body tangle: they become N − 1 routine calculations and one addition. The linearity does fail eventually, but only in fields of order 10¹⁸ V m⁻¹, where the vacuum itself starts producing pairs. Nothing in this course comes within twelve orders of magnitude of that.

Draw the picture, choose axes, and write them down. Most superposition errors are sign errors that a declared frame would have caught. Then for each source: get the magnitude from Coulomb's law using |q₀qᵢ|, set the direction from the signs — away from a like charge, toward an unlike one — and resolve onto the axes. Take Q = +2.0 nC at the origin, with +5.0 nC at (3.0 cm, 0) and −5.0 nC at (0, 4.0 cm). The first is a repulsion of k(5.0 nC)(2.0 nC)/(0.030 m)² = 99.9 μN pointing along −x. The second is an attraction of the same charge product over (0.040 m)², which is 56.2 μN pointing along +y. The columns read Fₓ = −99.9 μN and F_y = +56.2 μN, so |F| = √(99.9² + 56.2²) = 114.6 μN at atan2(56.2, −99.9) = 150.6° from the +x axis. Never add the magnitudes: 99.9 + 56.2 = 156.1 μN is not a force that exists anywhere in this problem.

Look for mirror pairs before reaching for the calculator. Put equal charges q at (−a, 0) and (+a, 0), and a test charge Q at (0, y) on their perpendicular bisector. Both sources are the same distance r = √(a² + y²) away, so the two magnitudes are equal; their x components point opposite ways and cancel exactly, while their y components are identical and add. The result is F = 2kqQy/(a² + y²)³⁄² along y — one term where the brute-force route carried four. The same habit kills the entire sum for n equal charges spaced evenly round a circle, but say why carefully. For even n each charge does have a partner directly opposite, and the pairs cancel one by one. For odd n — three at the corners of an equilateral triangle, say — no charge has an opposite partner at all, so use the rotational argument instead: turning the whole ring by 360°/n leaves the arrangement identical, so it must leave the force at the centre identical, and the only vector a turn leaves unchanged is the zero vector. Either way the central force vanishes for every n ≥ 2. Symmetry also hands you free checks. The bisector force must vanish at y = 0, and must approach 2kqQ/y² when y ≫ a because the pair then looks like a single charge 2q, so it has to peak in between — at y = a/√2, as differentiating confirms.

Two point charges have at most one place on their line where a test charge feels nothing, and that place is fixed before you decide what to put there, because the test charge's own value cancels from both sides of the equation. Setting magnitudes equal, k|q₁|/r₁² = k|q₂|/r₂², gives r₁/r₂ = √(|q₁/q₂|), so the point always sits further from the larger charge. Where it sits depends on the signs. With like signs the two pushes oppose each other only between the charges, so the null point is there, at x = d/(1 + √(q₂/q₁)) from q₁. With opposite signs both forces point the same way everywhere between them, so the cancellation has to happen outside the pair, beyond the weaker charge, at x = d/(√(|q₁/q₂|) − 1) from it. The magnitude equation knows none of this and returns two roots. At one of them the forces reinforce instead of opposing, so check the directions at a root before you quote it.

A harder question asks that every charge, not just a test charge, feel zero force. Take q₁ and q₂ = βq₁, both positive, a distance d apart, and ask where a third charge q₃ must go and what it must be. Putting q₃ at the null point of the other two costs nothing, since it is in equilibrium wherever they cancel: x = d/(1 + √β) from q₁. Now demand equilibrium for q₁ itself. The repulsion from q₂ is kq₁q₂/d², so q₃ has to pull q₁ the other way, which makes it negative, with a magnitude fixed by kq₁|q₃|/x² = kq₁q₂/d². That gives |q₃| = βq₁/(1 + √β)². For β = 4 it means q₃ = −4q₁/9 placed a third of the way from q₁, and checking q₂ shows it balances too — the third condition comes free. The arrangement is complete, and, as the next section shows, still useless as a way of storing charge.

Equilibrium is a statement about a first derivative; stability is a statement about a second. Test it. Put equal positive charges q at (±a, 0) and a positive Q at the origin, where the net force is zero. Displace Q a small distance x along the line: the near charge now pushes harder than the far one, and expanding 1/(a ± x)² to first order gives Fₓ ≈ −4kqQx/a³, a restoring force. Displace it by y instead, perpendicular to the line, and both charges push outward along y, giving F_y ≈ +2kqQy/a³, which drives it away. The same happens along z. So the three gradients ∂Fₓ/∂x, ∂F_y/∂y and ∂F_z/∂z are −4, +2 and +2 in units of kqQ/a³ — one negative and restoring, two positive and unstable — and they sum to zero. That is no accident of this arrangement: at any point of empty space ∂Fₓ/∂x + ∂F_y/∂y + ∂F_z/∂z = 0, so the three can never all be negative. Earnshaw's theorem follows, which is why real traps use oscillating fields, active feedback, or a mechanical constraint such as a bead threaded on a wire.