Field superposition and null points

Superposition is a physical claim, not a definition. Bring a second charge into a region and the field of the first is unchanged; the two fields simply coexist, and a test charge placed there responds to their vector sum. This works because Coulomb's law is linear in the source charge: double q and you double its own contribution without touching anyone else's. A problem with N charges is therefore N separate one-charge problems plus one vector addition, and the order of that addition never matters. What can fail is never the linearity — it is the list of charges you feed into it. Push a charge up to a metal sphere and the sphere's own charge rearranges; superposition still gives the exact field, but only once you sum over the new distribution, and that distribution is now something you have to solve for. For the fixed point charges assumed here the list never changes, so the sum is immediate.

Write every contribution on the same pair of axes before adding anything. The compact form Eₓ = Σ k qᵢ (x − xᵢ)/rᵢ³ does the sign bookkeeping for you: the factor (x − xᵢ) is negative when the field point lies to the left of the source, and a negative qᵢ flips the whole vector, so no separate direction argument is needed. Add the x-components, add the y-components, and only then recombine with |E| = √(Eₓ² + E_y²) and θ = tan⁻¹(E_y/Eₓ). Skipping that discipline is the standard route to a wrong answer. If one charge contributes 300 N C⁻¹ along +x and another 400 N C⁻¹ along +y, the resultant is √(300² + 400²) = 500 N C⁻¹ at 53° above the x-axis, not the 700 N C⁻¹ that adding magnitudes suggests — 40% too large. Contributions add arithmetically only when they are parallel.

A mirror symmetry is worth more than a line of algebra. Put equal charges +q at (−a, 0) and (+a, 0) and ask for the field at (0, y) on the perpendicular bisector. Each contributes the same magnitude kq/(y² + a²), because the distances are equal, and reflecting the picture in the y-axis swaps the two charges while reversing x — so the two x-components must cancel exactly. Only the y-components survive, each carrying the factor cos θ = y/√(y² + a²), which gives E = 2kqy/(y² + a²)³⁄² along +y. Test it at both ends. For y ≫ a it becomes 2kq/y², the field of a single charge 2q, which is what a distant observer should see. For y → 0 it falls to zero, because at the midpoint the two fields are equal and opposite — the null point of a like pair.

For two charges a null must lie on the line through them. Anywhere off that line the two field vectors point along different directions — each lies along its own line from source to field point — so they can never be antiparallel and can never cancel. That leaves three regions on the line. If the charges share a sign, their fields point the same way everywhere outside the pair and oppose only between them, so the null is between. If the signs differ, then between the charges both fields point from the positive charge toward the negative one and add rather than cancel; they oppose only outside the pair, and the magnitudes can balance only where the weaker charge is the closer one, so the null sits beyond the weaker charge. Equal and opposite charges have no finite null at all.

Setting k|q₁|/r₁² = k|q₂|/r₂² fixes only a ratio of distances, r₂/r₁ = √(|q₂|/|q₁|), and two separate regions of the line can meet that ratio, so there are two candidate points and only one of them is a null. Take +9.0 nC at x = 0 and +4.0 nC at x = 0.50 m. The ratio is √(4/9) = 2/3, and the same-sign formula gives x₀ = 0.50/(1 + 2/3) = 0.30 m — between the charges and nearer the weaker one, as it must be. The other root, x = 0.50/(1 − 2/3) = 1.50 m, is honest arithmetic: there the two fields really do have equal magnitudes, 36 N C⁻¹ each. But at 1.50 m both of them point along +x, so rather than cancelling they add to 72 N C⁻¹. Equal magnitudes are necessary for a null, never sufficient; check the directions in the region you landed in.

E is a vector sum while V = Σ kqᵢ/rᵢ is a signed scalar sum, so the two vanish under different conditions and almost never at the same place. Put +6.0 nC and −6.0 nC at x = ±0.20 m. Every point on the perpendicular bisector is equidistant from both charges, so the signed scalars cancel and V = 0 all along it — yet at the midpoint the two fields both point from the positive charge toward the negative one and add to 2.7 × 10³ N C⁻¹. Now make both charges +6.0 nC. At the midpoint the fields are equal and opposite, so E = 0, but the two potentials are both positive and add to 5.4 × 10² V. A null point is a place where a charge feels no force, not a place where it has no energy.