IB Physics HLTopic 4 — Force FieldsPaper 1 & 2F = kq₁q₂/r²~17 min read
Coulomb’s Law
We have been leaning on one idea for two pages without ever writing it down: that the electric force fades with distance. The comb must be held close. The near side of the paper wins. Charles-Augustin de Coulomb pinned that fading down in 1785, and what he found should stop you in your tracks — it is the same equation Newton had written for gravity a century earlier. Different force. Different universe of causes. Identical shape.
📘 What you need to know
All charged particles generate an electric field, which exerts a force on other charges nearby
Coulomb’s law: the force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation
F = kq1q2 / r2, where r is measured centre to centre
The Coulomb constant in a vacuum is k = 8.99 × 109 N m2 C−2
k = 1 / 4πε0, where ε0 = 8.85 × 10−12 C2 N−1 m−2 is the permittivity of free space
A positive product q1q2 means repulsion; a negative product means attraction
Relative permittivityεr = ε/ε0 (the dielectric constant) has no units. In a material, k = 1/4πε and the force is reduced
It applies only to point charges, or spheres much smaller than their separation. Never to irregular shapes
Unlike gravity, the electric force can be attractive or repulsive — and it is vastly stronger
The law itself
Two charges, sitting in space, pushing or pulling on each other. Coulomb measured how hard, and found only two things mattered: how much charge, and how far apart.
Coulomb’s law — statementthe electric force between two point charges is directly proportional to the product of the charges, and inversely proportional to the square of their separation
Coulomb’s law — equationF = kq1q2 / r2F = force (N) • q1, q2 = charges (C) • r = separation of centres (m)k = 8.99 × 109 N m2 C−2, the Coulomb constant
Whichever charge is bigger, both feel the same size of force. And r is always measured from centre to centre, never surface to surface.
The small print: point charges only
Coulomb’s law is written for point charges — charges with no size at all. In practice you may apply it to charged spheres, but only when the spheres are much smaller than their separation. That is the point charge approximation, and it comes with two warnings:
r must be taken from the centres of the spheres
You cannot use it for charge spread over an irregularly-shaped object
Which way does the force point?
Feed the charges in with their signs and the equation tells you the answer itself:
Two charges of the same type: the product q1q2 is positive → F is positive → repulsion
Two opposite charges: the product is negative → F is negative → attraction
In an exam I would not bother. Put the magnitudes into the equation, get a positive number for the size of the force, and then look at the two signs and write “attractive” or “repulsive” underneath. Nobody has ever lost a mark for a force of 8.4 × 10−5 N labelled attractive. Plenty have lost marks by carrying a minus sign into a square root later on.
The inverse square law
The r2 underneath is doing enormous work. Move the charges twice as far apart and the force does not halve — it drops to a quarter. Three times as far, and it is a ninth.
Halving the distance quadruples the force. This is why the near edge of the paper wins, and why a spark leaps a small gap and not a large one.
For comparison questions you rarely need k at all. Just use the proportionality:
Proportional reasoningF ∝ q1q2 and F ∝ 1/r2multiply the charge factors, divide by the square of the distance factor
Coulomb’s constant and permittivity
That k is not a fundamental constant. It is a bundle, and unpacking it tells you something about the space between the charges.
Where k comes fromk = 1 / 4πε0ε0 = 8.85 × 10−12 C2 N−1 m−2, the permittivity of free space
Put the numbers in and you get 8.99 × 109, exactly the k in the data booklet. Permittivity ε measures how easily an electric field can be set up inside a material — or, turned around, the resistance a material offers to having a field created within it.
Charges sitting in a vacuum use ε0. Put a material between them instead, and you use its permittivity:
With a material between the chargesk = 1 / 4πε where ε = εrε0every material has ε > ε0, so k gets smaller and the force gets weaker
Relative permittivity (the dielectric constant)εr = ε / ε0a ratio of two like quantities, so it has no units
Material
Relative permittivity εr
Free space (vacuum)
1
Air
1.0005
Paraffin
2.3
Polystyrene
3
Paper
4
Pure water
80
Ceramic
100 – 15 000
Look at air: εr = 1.0005. That is 1, to any accuracy you will ever need. So charges in air are treated exactly as charges in a vacuum, and unless the question names a material, assume vacuum. Now look at pure water: εr = 80. Drop two charges into water and the force between them collapses to one-eightieth of its value in air. That is not a small correction. That is why salt dissolves.
Because k and ε sit on opposite sides of a fraction, the shortcut for any medium is pleasingly simple:
Force in a mediumFmedium = Fvacuum / εr
Coulomb versus Newton
Set the two laws side by side. The resemblance is not a coincidence, and it is worth a mark or two whenever a question asks you to compare them.
Two constants, two pairs of “somethings”, one r2. The only structural difference is that charge comes in two flavours, and mass in one.
Work out the ratio for two protons and you get about 1.2 × 1036. Notice that the r2 cancels, so the answer is the same at any separation. Gravity is pitifully, absurdly weak. The only reason it runs the solar system is that charge comes in two signs and cancels itself out, while mass only ever adds up.
Two charges, separation r
multiply the charges
q1q2
divide by r², scale by k
F = kq1q2/r2
🧲 Working a Coulomb’s law question
Convert everything. nC and µC into coulombs; cm and mm into metres. Always.
Is r centre to centre? Halve any diameter. Add any radii if you are given a surface gap.
Use magnitudes.F = kq1q2/r2, then state attractive or repulsive from the signs.
Square the r, never the q. The commonest slip in the topic.
Comparing two situations? Skip k entirely. Use F ∝ q1q2/r2.
A material named? Divide the vacuum answer by εr. If none is named, assume vacuum.
WE 1
Two small charged spheres carry charges of +3.0 nC and −5.0 nC. Their centres are 4.0 cm apart in a vacuum. Calculate the magnitude of the electric force between them, and state whether it is attractive or repulsive. (k = 8.99 × 109 N m2 C−2)
Step 1 — convert to SI unitsq₁ = 3.0 × 10⁻⁹ C | q₂ = 5.0 × 10⁻⁹ C | r = 0.040 mStep 2 — work out the top and bottom separatelyq₁q₂ = 1.5 × 10⁻¹⁷ C²r² = (0.040)² = 1.6 × 10⁻³ m²Step 3 — put it togetherF = (8.99 × 10⁹) × (1.5 × 10⁻¹⁷) / (1.6 × 10⁻³)F = 8.4 × 10⁻⁵ NStep 4 — direction, from the signs
The charges are opposite, so the product is negative.
attractiveThat is 84 microNewtons — roughly the weight of a grain of sand. Note how the nano prefixes crushed the answer: forget to convert 3.0 nC and you are out by a factor of 10¹⁸.
WE 2
Two point charges separated by a distance r exert a force F on each other. Determine the new force, in terms of F, if: (a) both charges are doubled and the separation is doubled; (b) one charge is trebled and the separation is halved; (c) the separation alone is trebled.
Step 1 — write down what mattersF ∝ q₁q₂ / r²
Multiply the charge factors; divide by the square of the distance factor.
(a) charges ×2 and ×2, separation ×2factor = (2 × 2) / 2² = 4 / 4the force is unchanged, F(b) one charge ×3, separation ×½factor = 3 / (½)² = 3 / 0.2512F(c) separation ×3factor = 1 / 3²F/9Part (a) is the one that catches people. Doubling both charges quadruples the top; doubling the distance quadruples the bottom. They cancel exactly, and no calculator was needed anywhere on this question.
WE 3
Two charges in a vacuum experience a force of 6.0 mN. They are then completely immersed in paraffin, which has a relative permittivity of 2.3. (a) Calculate the permittivity of paraffin. (b) Determine the new force between the charges. (c) Explain, in terms of permittivity, why the force changed. (ε0 = 8.85 × 10−12 C2 N−1 m−2)
(a) Step 1 — rearrange the definition of relative permittivityεr = ε / ε₀ → ε = εrε₀ε = 2.3 × (8.85 × 10⁻¹²)ε = 2.0 × 10⁻¹¹ C² N⁻¹ m⁻²(b) Step 2 — k = 1/4πε, so a bigger ε means a smaller kkparaffin = kvacuum / εrFparaffin = 6.0 / 2.3F = 2.6 mN(c) Step 3 — say why
Paraffin has a higher permittivity than free space.
It offers more resistance to an electric field forming inside it.
so k falls, and the force is weakerEvery material has ε > ε₀, so filling the gap with anything can only ever reduce the force. Never increase it. If your answer came out bigger than 6.0 mN, you divided the wrong way round.
💡 Top tips
k and ε0 are both in the data booklet. Don’t memorise them — memorise that k = 1/4πε0.
If no material is mentioned, assume a vacuum. Air is close enough anyway.
Square the distance, not the charges. Write r2 on the bottom before you touch the calculator.
r runs from centre to centre. Watch for diameters, radii and surface-to-surface gaps.
Put in magnitudes, get the size, then read the signs and write “attractive” or “repulsive”.
Ratio questions need no constants. F ∝ q1q2/r2 does the whole job.
In a medium, F is divided by εr. It always gets weaker, never stronger.
⚠ Common mistakes
Forgetting to square r, or squaring the charges instead
Leaving r in cm or mm, and charges in nC or µC
Using a diameter, or a surface-to-surface gap, as r
Applying Coulomb’s law to irregular objects, or to spheres that are large compared with their separation
Carrying the minus sign of an attractive force into later working, where it becomes nonsense
Multiplying by εr instead of dividing. A material always weakens the force
Confusing ε (permittivity, has units) with εr (relative permittivity, no units)
Thinking gravity and Coulomb are unrelated. Both are inverse square laws — that’s a marked comparison
Quick recap: The force between two point charges is F = kq1q2/r2, with r measured centre to centre and k = 1/4πε0 = 8.99 × 109 N m2 C−2. A positive product means repulsion, a negative product means attraction. It is an inverse square law: double r and the force falls to a quarter. Fill the gap with a material and k = 1/4πε with ε = εrε0, so the force is divided by εr. Same shape as Newton’s law of gravitation — but charge has two signs, and the force is 1036 times stronger.
One thing should be nagging at you. Coulomb’s law needs two charges — but a single charge sitting alone in space is clearly doing something, waiting for a partner to arrive. That something is a field. Instead of asking “what force do these two exert on each other”, we ask “what would this charge do to a test charge placed here?” Divide the force by the test charge and the second charge disappears from the equation entirely. Next page: Electric Field Strength.
Inverse squares and permittivity not clicking?
Book a free meeting and we’ll drill the unit conversions, the ratio shortcuts and the εr trap until they’re second nature.
