Coulomb's Law: Force Between Charges Revision Notes | A Level AQA Physics

1. Why Two Charges Push or Pull Each Other

Picture two small charged spheres held a fixed distance apart. You already know from GCSE that like charges repel and unlike charges attract — but how strong is that force, and what does it depend on? Coulomb's law gives you the precise, quantitative answer.

The force turns out to depend on three things: how large each charge is, how far apart they are, and a property of the medium between them. Get those three ingredients right and you can calculate the force to any precision the exam demands.

2. Coulomb's Law: The Equation and What Every Symbol Means

The electrostatic force $F$ between two point charges $Q_1$ and $Q_2$ separated by a distance $r$ in a vacuum is given by Coulomb's law. The constant out front packages up the permittivity of free space $\varepsilon_0$ in a way that keeps the maths clean.

The factor $\frac{1}{4\pi\varepsilon_0}$ is sometimes written as the single constant $k \approx 8.99 \times 10^9$ N m² C⁻². You will see both forms in textbooks, but AQA mark schemes accept either — provided you use the correct numerical value.

Notice the structure of the equation: $F \propto Q_1 Q_2$ and $F \propto \frac{1}{r^2}$. Doubling either charge doubles the force; doubling the separation reduces the force to one quarter. This inverse square relationship is the same mathematical form as Newton's law of gravitation — a pattern worth remembering when you compare the two field types later in the course.

3. Air as a Vacuum: A Justified Approximation

Coulomb's law as written applies strictly to charges in a vacuum. In reality, most experiments and exam questions involve charges in air. The AQA specification explicitly states that air can be treated as a vacuum when calculating the force between charges.

The reason this works is that the relative permittivity of air is approximately 1.0005 — so close to 1 that the correction is less than 0.05%. For all practical purposes at A-level, using $\varepsilon_0$ for air introduces negligible error. You do not need to apply any correction factor; state the approximation and proceed with the standard equation.

4. Charged Spheres and the Point-Charge Approximation

Real charged objects are not mathematical points — they have physical size. For a uniformly charged sphere, however, the AQA specification states that the charge may be considered to be concentrated at the centre of the sphere. This means you measure $r$ from centre to centre, not from surface to surface.

This result is analogous to the gravitational case, where a uniform sphere behaves as if all its mass is at its centre. The condition is that the charge distribution must be spherically symmetric — which is the case for a conducting sphere in electrostatic equilibrium, where charge distributes uniformly over the surface.

When two charged spheres are far apart relative to their radii, the point-charge approximation is excellent. As they get very close, the charge distributions on each sphere can distort slightly due to mutual induction, but at A-level you assume uniform distribution and measure $r$ between centres throughout.

5. Worked Calculations Using Coulomb's Law

The equation looks straightforward, but exam questions test whether you can handle unit conversions, interpret the sign of the force, and apply the sphere approximation correctly. The worked examples below build from a direct substitution to a more involved rearrangement.

Worked example: Calculating the Force Between Two Point Charges

Two point charges, $Q_1 = +3.0 \times 10^{-6}$ C and $Q_2 = -5.0 \times 10^{-6}$ C, are placed 0.12 m apart in air. Calculate the electrostatic force between them. ($\varepsilon_0 = 8.85 \times 10^{-12}$ C² N⁻¹ m⁻²)

1. $F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$ — State the formula first — this earns the method mark [M1]. Air is treated as a vacuum, so $\varepsilon_0$ is used directly.
2. $\frac{1}{4\pi\varepsilon_0} = \frac{1}{4 \times \pi \times 8.85 \times 10^{-12}} = 8.99 \times 10^{9}$ N m² C⁻² — You can use $k = 8.99 \times 10^9$ N m² C⁻² directly. This value is given in the AQA data booklet as part of the formula.
3. $r^2 = 0.12 \times 0.12 = 0.0144$ m² — Expand the square explicitly: $0.12 \times 0.12 = 0.0144$ m². Never skip this step — squaring errors are very common.
4. $|Q_1 Q_2| = 3.0 \times 10^{-6} \times 5.0 \times 10^{-6} = 15.0 \times 10^{-12} = 1.50 \times 10^{-11}$ C² — Use magnitudes for the force calculation. The opposite signs tell you the force is attractive — state this separately rather than letting a negative sign appear in the magnitude calculation.
5. $F = 8.99 \times 10^{9} \times \frac{1.50 \times 10^{-11}}{0.0144}$ $F = 8.99 \times 10^{9} \times 1.0417 \times 10^{-9}$ $F = 9.36$ N — Carry the division through carefully: $\frac{1.50 \times 10^{-11}}{0.0144} = 1.0417 \times 10^{-9}$. Then multiply by $8.99 \times 10^9$. [A1] for correct answer with units.
6. $F \approx 9.4$ N (attractive, to 2 s.f.) — Because $Q_1$ is positive and $Q_2$ is negative, the force is attractive. Always state the direction or nature of the force — examiners award a separate mark for this in some questions.

Answer: $F \approx 9.4$ N, attractive

Now try a rearrangement. Coulomb's law can be rearranged to find the separation $r$ given the force and the charges. The key step is isolating $r^2$ before taking the square root — a two-stage process that students often compress incorrectly.

Worked example: Finding the Separation Given the Force

Two point charges $Q_1 = +2.0 \times 10^{-6}$ C and $Q_2 = +2.0 \times 10^{-6}$ C repel each other with a force of 0.36 N in air. Calculate their separation. ($\varepsilon_0 = 8.85 \times 10^{-12}$ C² N⁻¹ m⁻²)

1. $F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$ $r^2 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{F}$ — Multiply both sides by $r^2$ and divide both sides by $F$ to isolate $r^2$. [M1] for correct rearrangement.
2. $r^2 = 8.99 \times 10^{9} \times \frac{(2.0 \times 10^{-6}) \times (2.0 \times 10^{-6})}{0.36}$ — Use $k = 8.99 \times 10^9$ N m² C⁻². Both charges are positive, confirming a repulsive force — consistent with the question.
3. $Q_1 Q_2 = 2.0 \times 10^{-6} \times 2.0 \times 10^{-6} = 4.0 \times 10^{-12}$ C² — $2.0 \times 2.0 = 4.0$ and $10^{-6} \times 10^{-6} = 10^{-12}$.
4. $r^2 = 8.99 \times 10^{9} \times \frac{4.0 \times 10^{-12}}{0.36}$ $= 8.99 \times 10^{9} \times 1.111 \times 10^{-11}$ $= 0.09988 \approx 0.100$ m² — $\frac{4.0 \times 10^{-12}}{0.36} = 1.111 \times 10^{-11}$. Then $8.99 \times 10^9 \times 1.111 \times 10^{-11} = 0.09988$ m².
5. $r = \sqrt{0.100} = 0.316$ m — $\sqrt{0.100} = 0.316$ m (to 3 s.f.). [A1] for correct answer with units. A sanity check: two microcoulomb charges 0.32 m apart producing a fraction-of-a-newton force is physically reasonable.

Answer: $r \approx 0.32$ m (to 2 s.f.)

6. Comparing Coulomb's Law with Newton's Law of Gravitation

At A-level you are expected to recognise the structural similarity between Coulomb's law and Newton's law of gravitation. Both are inverse square laws; both involve a product of source quantities divided by $r^2$; and both use a constant that depends on the medium or space. The table below makes the comparison explicit.

One critical difference: gravitational force is always attractive, whereas the electrostatic force can be either attractive or repulsive depending on the signs of the charges. This distinction is frequently tested in 'compare' questions on the exam.

7. Exam Technique and Common Errors

Coulomb's law questions appear in multiple formats: direct calculation, rearrangement for $r$ or $Q$, ratio questions where you compare forces under changed conditions, and 'explain' questions about the direction of force. Each format has its own pitfalls.

For 'explain' or 'describe' questions about Coulomb's law, examiners reward precise language. Saying the force 'decreases' when separation increases is insufficient — you must say it 'decreases in proportion to $1/r^2$' or 'decreases by a factor of four when $r$ doubles' to access the higher marks.

Coulomb's law is the quantitative foundation of the entire electric fields topic: every result that follows — electric field strength, potential, and the motion of charges in fields — builds directly on the inverse square force law you have studied here.