Coulomb's Law

Coulomb's law is the electrostatic counterpart of Newton's law of gravitation: a clean inverse-square law that gives the force between two point charges. Charles-Augustin de Coulomb measured this force with a delicate torsion balance in the 1780s, and to the precision of his apparatus he confirmed that the force was exactly proportional to the inverse square of the separation. The law has since been tested to one part in $10^{16}$1016 — it is one of the most precisely verified statements in all of physics, and it is the algebraic engine behind every electric-field, electric-potential and energy calculation that follows in Module 6.2.

Spec mapping (OCR H556 Module 6.2 — electric fields). This lesson covers the statement of Coulomb's law in the form $F = Q_1 Q_2 / (4\pi\varepsilon_0 r^2)$F=Q1​Q2​/(4πε0​r2), the meaning of the permittivity constant $\varepsilon_0$ε0​, the sign conventions for attractive versus repulsive forces, the inverse-square scaling, and the structural comparison with Newton's law of gravitation. It connects directly to the radial electric field formula derived in the preceding "electric-field-strength" lesson, and to the potential and potential-energy formulae in the next lesson.

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Statement of the Law

For two point charges $Q_1$Q1​ and $Q_2$Q2​ separated by a distance $r$r in vacuum (or, to a very good approximation, in air), the magnitude of the electrostatic force each exerts on the other is

$F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}$F=4πε0​r2Q1​Q2​​

where the permittivity of free space is

$\varepsilon_0 \approx 8.85 \times 10^{-12}\ \text{F m}^{-1} = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ε0​≈8.85×10−12 F m−1=8.85×10−12 C2N−1m−2

and the compact combination that appears in the law is

$k = \frac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9\ \text{N m}^2\,\text{C}^{-2}.$k=4πε0​1​≈8.99×109 N m2C−2.

Both forms appear in textbooks: $F = Q_1 Q_2 / (4\pi\varepsilon_0 r^2)$F=Q1​Q2​/(4πε0​r2) is the OCR data-booklet form, while $F = kQ_1 Q_2 / r^2$F=kQ1​Q2​/r2 is the more compact engineering shorthand. They are identical.

Sign convention — attractive vs repulsive

Substitute the signs of $Q_1$Q1​ and $Q_2$Q2​ literally into the formula:

• Like charges (both positive, or both negative): the product $Q_1 Q_2 > 0$Q1​Q2​>0, so $F > 0$F>0 — interpreted as repulsive, directed along the line joining the charges, outward from each.
• Unlike charges (opposite signs): the product $Q_1 Q_2 < 0$Q1​Q2​<0, so $F < 0$F<0 — interpreted as attractive, directed along the line joining the charges, inward toward each other.

Many textbooks (and the OCR data booklet) quote the law for magnitudes only and then state the direction in words; both conventions are acceptable in the exam, but you must always communicate whether the force is attractive or repulsive whenever a numerical answer is required.

By Newton's third law the two forces are equal in magnitude and opposite in direction: $\vec{F}_{1 \to 2} = -\vec{F}_{2 \to 1}$F1→2​=−F2→1​. The forces never act along anything other than the line joining the two charges.

Units check

Verify that the law really delivers newtons. With $\varepsilon_0$ε0​ in $\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$C2N−1m−2,

$[F] = \frac{[\text{C}]^2}{[\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}] \times [\text{m}^2]} = \frac{\text{C}^2}{\text{C}^2\,\text{N}^{-1}} = \text{N}.$[F]=[C2N−1m−2]×[m2][C]2​=C2N−1C2​=N.

The units work out cleanly — which they must, since $\varepsilon_0$ε0​ is defined precisely to make this so.

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Comparison with Newton's Law of Gravitation

Compare

$$\begin{aligned} F_{\text{grav}} &= \frac{Gm_{1}m_{2}}{r^{2}} \quad \text{(always attractive)} \\ F_{\text{elec}} &= \frac{Q_{1}Q_{2}}{4\pi\varepsilon_{0}r^{2}} \quad \text{(attractive or repulsive)} \end{aligned}$$Fgrav​Felec​​=r2Gm1​m2​​(always attractive)=4πε0​r2Q1​Q2​​(attractive or repulsive)​

The two laws have exactly the same mathematical form: a product of source strengths divided by 4π r² (with a dimensional constant). But they differ in crucial physical ways:

Feature	Gravitation	Electrostatics
Source	Mass m	Charge Q
Constant	G = 6.67 × 10⁻¹¹	1/(4πε₀) = 8.99 × 10⁹
Sign	Always attractive	Attractive or repulsive
Strength (per coulomb vs per kg)	Very weak	Vastly stronger
Distance dependence	1/r²	1/r²
Shielding	None	Possible (conductors)

The dramatic difference in strength is worth dwelling on. For two electrons, the ratio of the electrostatic repulsion to the gravitational attraction is about 10⁴². Electric forces utterly dominate on the atomic scale; gravity only wins on astronomical scales because matter is overwhelmingly electrically neutral — positive and negative charges cancel — whereas mass is always positive and cannot cancel itself.

Exam tip. OCR often asks you to "compare gravitational and electric fields". The standard marks are: same $1/r^2$1/r2 dependence; gravity always attractive while electric can be either; electric much stronger per unit source; electric can be shielded, gravitational cannot. Memorise these five.

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Worked Example 1 — Two Protons

Calculate (a) the electrostatic force and (b) the gravitational force between two protons separated by $1.0 \times 10^{-10}$1.0×10−10 m (about one ångström, the scale of an atom). Find the ratio of the two.

Data: $e = 1.60 \times 10^{-19}$e=1.60×10−19 C, $m_p = 1.67 \times 10^{-27}$mp​=1.67×10−27 kg, $G = 6.67 \times 10^{-11}$G=6.67×10−11 N m$^2$2 kg$^{-2}$−2.

(a) Electric force.

$F_e = \frac{(1.60 \times 10^{-19})^2 (8.99 \times 10^9)}{(1.0 \times 10^{-10})^2} = \frac{(2.56 \times 10^{-38})(8.99 \times 10^9)}{1.0 \times 10^{-20}} \approx 2.30 \times 10^{-8}\ \text{N}.$Fe​=(1.0×10−10)2(1.60×10−19)2(8.99×109)​=1.0×10−20(2.56×10−38)(8.99×109)​≈2.30×10−8 N.

(b) Gravitational force.

$F_g = \frac{(6.67 \times 10^{-11})(1.67 \times 10^{-27})^2}{(1.0 \times 10^{-10})^2} \approx 1.86 \times 10^{-44}\ \text{N}.$Fg​=(1.0×10−10)2(6.67×10−11)(1.67×10−27)2​≈1.86×10−44 N.

(c) Ratio.

$\frac{F_e}{F_g} \approx \frac{2.30 \times 10^{-8}}{1.86 \times 10^{-44}} \approx 1.2 \times 10^{36}.$Fg​Fe​​≈1.86×10−442.30×10−8​≈1.2×1036.

Electrostatic repulsion between two protons is some $10^{36}$1036 times stronger than their gravitational attraction. This is why gravity plays no role in atomic-scale physics, and why chemistry can be done without ever mentioning Newton's law of gravitation.

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Inverse-Square Scaling

A favourite OCR exam manoeuvre: "How does the force change if the distance is doubled?" Reason directly from $F \propto 1/r^2$F∝1/r2. The table below summarises the multipliers you should be able to write down without a calculator.

New separation	New force as a multiple of $F$F
$2r$2r	$F/4$F/4
$3r$3r	$F/9$F/9
$r/2$r/2	$4F$4F
$r/3$r/3	$9F$9F
$0.1r$0.1r	$100F$100F

The fractional statement: if $r$r is changed by a factor $\alpha$α, then $F$F is changed by a factor $1/\alpha^2$1/α2. This shows up in exam questions on the field of a charged sphere, the force between ions in a crystal lattice, and the spacing dependence of nuclear scattering experiments.

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Worked Example 2 — Charged Spheres in Equilibrium

Two identical metal spheres, each of mass $2.0$2.0 g, hang from insulating threads of length $30$30 cm from the same point. Each sphere carries the same positive charge. The threads make an angle of $15^{\circ}$15∘ with the vertical. Calculate the charge on each sphere.

Geometry. By symmetry each sphere sits at the same height. The horizontal distance between them is

$r = 2\ell \sin\theta = 2 \times 0.30 \times \sin 15^{\circ} = 2 \times 0.30 \times 0.2588 \approx 0.1553\ \text{m}.$r=2ℓsinθ=2×0.30×sin15∘=2×0.30×0.2588≈0.1553 m.

Force balance. Three forces act on each sphere: weight $mg$mg downward, thread tension $T$T along the thread, and the electric repulsion $F_e$Fe​ horizontal. Resolving:

$T\cos 15^{\circ} = mg, \qquad T\sin 15^{\circ} = F_e.$Tcos15∘=mg,Tsin15∘=Fe​.

Dividing gives the standard pendulum-with-charge result:

$F_e = mg\tan 15^{\circ} = (2.0 \times 10^{-3})(9.81)(0.2679) \approx 5.26 \times 10^{-3}\ \text{N}.$Fe​=mgtan15∘=(2.0×10−3)(9.81)(0.2679)≈5.26×10−3 N.

Apply Coulomb's law. With $Q_1 = Q_2 = Q$Q1​=Q2​=Q,

$F_e = \frac{Q^2}{4\pi\varepsilon_0 r^2} \quad \Rightarrow \quad Q^2 = F_e \times 4\pi\varepsilon_0 \times r^2.$Fe​=4πε0​r2Q2​⇒Q2=Fe​×4πε0​×r2.

Numerically, with $4\pi\varepsilon_0 = 1/(8.99 \times 10^9) \approx 1.113 \times 10^{-10}$4πε0​=1/(8.99×109)≈1.113×10−10 F m$^{-1}$−1:

$Q^2 = (5.26 \times 10^{-3})(1.113 \times 10^{-10})(0.1553)^2 \approx 1.41 \times 10^{-14}\ \text{C}^2$Q2=(5.26×10−3)(1.113×10−10)(0.1553)2≈1.41×10−14 C2

$Q \approx 1.19 \times 10^{-7}\ \text{C} \approx 0.12\ \mu\text{C}.$Q≈1.19×10−7 C≈0.12 μC.

A small charge — about $7 \times 10^{11}$7×1011 electronic charges, less than a millionth of a coulomb — is enough to hold two 2 g spheres apart against gravity at a 15° angle. The lesson: at A-Level scale, charges of even nanocoulombs to microcoulombs produce easily observable forces.

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Worked Example 3 — Electron in a Hydrogen Atom (Bohr-style)

A hydrogen atom in its ground state can be modelled, crudely, as an electron in a circular orbit of radius $a_0 = 5.29 \times 10^{-11}$a0​=5.29×10−11 m around a proton. Treating the electron as a classical particle, find its orbital speed.

Approach. The electrostatic attraction provides the centripetal force:

$\frac{e^2}{4\pi\varepsilon_0 a_0^2} = \frac{m_e v^2}{a_0}.$4πε0​a02​e2​=a0​me​v2​.

Rearranging,

$v = \sqrt{\frac{e^2}{4\pi\varepsilon_0\, m_e\, a_0}}.$v=4πε0​me​a0​e2​​.

Plugging in $e = 1.60 \times 10^{-19}$e=1.60×10−19 C, $m_e = 9.11 \times 10^{-31}$me​=9.11×10−31 kg, $a_0 = 5.29 \times 10^{-11}$a0​=5.29×10−11 m:

$v = \sqrt{\frac{(8.99 \times 10^9)(1.60 \times 10^{-19})^2}{(9.11 \times 10^{-31})(5.29 \times 10^{-11})}} \approx 2.19 \times 10^6\ \text{m s}^{-1}.$v=(9.11×10−31)(5.29×10−11)(8.99×109)(1.60×10−19)2​​≈2.19×106 m s−1.

About 0.7 % of the speed of light. This is the Bohr velocity $v_B = \alpha c$vB​=αc, where $\alpha \approx 1/137$α≈1/137 is the fine-structure constant — a deep number whose appearance here connects A-Level Coulomb's law directly to undergraduate quantum mechanics.

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Link to Electric Field

Divide Coulomb's law by the second charge $Q_2$Q2​ (the "test charge") and you get the field strength due to $Q_1$Q1​ alone at distance $r$r:

$E = \frac{F}{Q_2} = \frac{Q_1}{4\pi\varepsilon_0 r^2}.$E=Q2​F​=4πε0​r2Q1​​.

This is exactly the radial-field formula derived in the previous lesson. Coulomb's law and the radial-field formula are two views of the same physics: Coulomb's law speaks of the force between two charges, while $E = Q/(4\pi\varepsilon_0 r^2)$E=Q/(4πε0​r2) speaks of the field set up by one charge, ready to act on any test charge brought into it. The field formula has one charge; Coulomb's law has two.

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Synoptic Links

• Module 5.3 (Newton's law of gravitation). Coulomb's law shares its inverse-square form. The exam table comparing the two is the most reliable two-mark gain on every Module 6.2 paper.
• Module 5.2 (circular motion). Coulomb attraction supplies the centripetal force in the Bohr-style hydrogen atom (Worked Example 3); the same recipe (Coulomb $=$= centripetal) reappears in Rutherford scattering and in the geometry of ion traps.
• Module 4 (work, energy and power). The work done assembling charges against Coulomb repulsion is the route to electric potential energy in the next lesson.
• Module 6.2 (electric field strength). $E = Q/(4\pi\varepsilon_0 r^2)$E=Q/(4πε0​r2) is Coulomb's law divided by a test charge.
• Module 6.4 (nuclear physics). Alpha-particle scattering by a gold nucleus is Coulomb's law applied at femtometre range; the closest-approach calculation uses $\tfrac{1}{2}mv^2 = Q_\alpha Q_{\text{Au}}/(4\pi\varepsilon_0 r_{\min})$21​mv2=Qα​QAu​/(4πε0​rmin​).
• Chemistry — ionic bonding. Lattice energies of ionic solids are summed Coulomb interactions; Madelung constants arise from the geometry of the lattice.

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Specimen question modelled on the OCR H556 paper format

Question (8 marks): Two small spheres each carry charge $Q = +3.0$Q=+3.0 nC and are separated in air by $r = 4.0$r=4.0 cm.

(a) State Coulomb's law in symbols, defining each symbol used. [2]

(b) Calculate the magnitude of the electrostatic force on one sphere and state its direction relative to the other sphere. [3]

(c) The separation is now changed to $1.0$1.0 cm with the charges unchanged. Without further numerical work, state the new force as a multiple of your answer to (b) and justify your factor. [2]

(d) One sphere is now replaced by a sphere of charge $-3.0$−3.0 nC, with the separation restored to $4.0$4.0 cm. State the effect on the magnitude and direction of the force. [1]

AO breakdown