Key Idea: Topic 4.2 is about fields — the invisible regions of influence around charges, around charged plates and around currents. Three pictures run through it: the inverse-square force and field of a point charge, the uniform field between parallel plates, and the magnetic field of a current that lets two wires push or pull on each other. It is examined on Paper 1A (quick scaling MCQs — halve a charge or double a separation and find the new force or field; attract-or-repel for two currents) and on Paper 2 (substitute into F = k q₁q₂/r², E = kQ/r², E = V/d or F/L = μ₀ I₁ I₂/(2π r); find a resultant field by superposition; get the energy of a charge crossing plates in eV).
📐 Key formulas
Four of these are given in the data booklet — you do not memorise them, but you must know which one to reach for and how to rearrange it. Two are derived: E = kQ ÷ r² comes from Coulomb's law, and W = qV (the energy a charge gains crossing a voltage) must be remembered.
- electric force between the charges (N)
- Coulomb constant, 8.99 × 10⁹ N m² C⁻² (given)
- the two point charges (C)
- distance between the charges (m)
- electric field strength (N C⁻¹)
- force on the test charge (N)
- size of the small test charge (C)
- electric field strength (N C⁻¹)
- Coulomb constant, 8.99 × 10⁹ N m² C⁻²
- size of the charge making the field (C)
- distance from that charge to the point (m)
- uniform field strength between the plates (V m⁻¹, = N C⁻¹)
- potential difference across the plates (V)
- separation (gap) between the plates (m)
- work done / energy gained moving the charge (J)
- the charge moved (C)
- potential difference moved through (V)
- force per unit length on each wire (N m⁻¹)
- permeability of free space, 4π × 10⁻⁷ T m A⁻¹
- the currents in the two wires (A)
- separation between the two wires (m)
🔭 The three field pictures
Half of this topic is recognising which field you are looking at from its shape — and how its strength changes with distance.
| Field | Shape of the lines | Strength with distance |
|---|---|---|
| Point charge | radial — out of a + charge, in to a − charge | inverse-square: double r ⇒ field ÷ 4 (E = kQ ÷ r²) |
| Parallel plates | uniform — evenly-spaced parallel lines, + plate → − plate | the same everywhere in the gap (E = V ÷ d) |
| Straight current | concentric circles wrapping round the wire | weaker further out; lines spread apart |
🧲 The two direction rules
Each kind of interaction has a simple attract-or-repel rule — the formula gives the size, these give the direction.
| Situation | Attract | Repel |
|---|---|---|
| Two charges | unlike signs (+ and −) | like signs (+/+ or −/−) |
| Two magnet poles | unlike poles (N–S) | like poles (N–N or S–S) |
| Two parallel currents | same direction (parallel) | opposite directions (anti-parallel) |
Coulomb's law and the parallel-wires law are both products of factors, so for a 'what is the new force' question you rarely need k or μ₀. Apply each change as its own multiplying factor: halve a charge → × ½, double a separation → ÷ 2² = ÷ 4 for charges (or ÷ 2 for wires, since F/L ∝ 1/r), double one current → × 2. Multiply the original by the product of the factors.
➕ Superposition and the null point
The total electric field at a point is the vector sum of the field from each charge — work out each one with E = kQ ÷ r², then combine with directions. Same direction → add the sizes; opposite directions → subtract them. Between two like charges the fields oppose, so there is a null (zero-field) point where they are equal and opposite and cancel — at the midpoint for two equal charges, and closer to the smaller charge otherwise.
✍️ IB-style worked examples
Two small spheres carry charges q₁ = +4.0 × 10⁻⁶ C and q₂ = +6.0 × 10⁻⁶ C and sit 0.30 m apart. (a) Calculate the force between them and state whether it is attractive or repulsive. (b) The separation is then trebled to 0.90 m. State the new force. (k = 8.99 × 10⁹ N m² C⁻².)
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Two equal positive charges, each +5.0 × 10⁻⁹ C, are fixed 0.40 m apart. (a) Find the field strength each charge produces at the midpoint, 0.20 m from each. (b) State the resultant field at the midpoint, with a reason. (k = 8.99 × 10⁹ N m² C⁻².)
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Two parallel plates 0.050 m apart have a potential difference of 250 V across them. (a) Find the uniform field strength between the plates. (b) An alpha particle of charge +2e (e = 1.6 × 10⁻¹⁹ C) is released from rest at the positive plate and crosses the full 250 V. Find its kinetic energy in electronvolts and in joules.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Two long parallel wires 0.20 m apart carry currents of 5.0 A and 6.0 A in opposite directions. (a) Find the force per unit length on each wire and state whether they attract or repel. (b) The separation is then halved to 0.10 m (same currents). State the new force per unit length. (μ₀ = 4π × 10⁻⁷ T m A⁻¹.)
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
✅ Quick self-check
Tap each card to reveal the answer.
Double the separation between two point charges — what happens to the force? It is divided by 2² = 4 — Coulomb's law is inverse-square (F ∝ 1/r²). Same rule for a point-charge field E = kQ/r².
Which way do field lines point around a + charge and a − charge? Out of a positive charge, in to a negative charge — the way a small positive test charge would be pushed.
Where between two equal positive charges is the field zero? At the midpoint — the two equal fields point in opposite directions and cancel (the null point). Combine fields as vectors by superposition.
What does a uniform field look like, and what is its strength? Evenly-spaced parallel lines from the + plate to the − plate, the same strength everywhere. Strength E = V ÷ d; energy of a charge crossing it W = qV.
Two parallel currents the same way — attract or repel? And opposite ways? Same direction → attract; opposite directions → repel. Reversing one current flips it.
How does the force between two wires scale with separation r? F/L ∝ 1/r — double r → halve F/L. (Contrast Coulomb's law for charges, which is 1/r².)
🎯 Highest-yield exam reminders
Exam tips
- Coulomb's law and the point-charge field are inverse-SQUARE: multiply the separation by n and you divide F (or E) by n². Halving a charge only halves F. The most common lost mark is forgetting to square the distance factor.
- Put the SIZES of the charges into F = k q₁q₂/r² for the magnitude, then read the direction off the signs: like charges repel, unlike attract.
- Combine electric fields by VECTOR superposition: work out each field with E = kQ/r², then add (same direction) or subtract (opposite). Between two like charges the fields oppose, giving a null point — at the midpoint for equal charges, nearer the smaller charge otherwise.
- Between parallel plates the field is uniform (evenly-spaced parallel lines, + to −): E = V/d, the same everywhere, so the force F = qE on a charge does not change as it moves across the gap. Always put the gap d in metres first.
- For the energy of a charge crossing plates use W = qV (not F = qE) — the pd already bundles in the distance. A charge n×e crossing V volts gains n×V eV; multiply by 1.6 × 10⁻¹⁹ to get joules.
- Two parallel currents: same direction attract, opposite repel — and reversing one current flips it. For 'new force' questions scale by ratios (F/L ∝ I₁I₂/r), so μ₀/(2π) cancels and you rarely substitute it.
- Know the field shapes on sight: radial for a point charge (out of +, in to −), uniform parallel lines between plates, and concentric circles round a current-carrying wire (right-hand grip rule for the direction).