In short

Coulomb's law quantifies the electrostatic force between two stationary point charges. If two charges q_1 and q_2 are separated by a distance r in vacuum, the force between them has magnitude

\boxed{\;F \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{|q_1 q_2|}{r^2} \;=\; k\,\frac{|q_1 q_2|}{r^2}\;}

where \varepsilon_0 = 8.854 \times 10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2) is the permittivity of free space and k = 1/(4\pi\varepsilon_0) = 8.988 \times 10^9 \;\text{N·m}^2/\text{C}^2 \approx 9 \times 10^9 N·m²/C² is Coulomb's constant. The direction is along the line joining the two charges — repulsive if the charges have the same sign, attractive if opposite. In vector form, the force on charge q_2 due to charge q_1 at position \vec{r}_1 is

\vec{F}_{1\to 2} \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{|\vec{r}_{12}|^2}\,\hat{r}_{12}, \qquad \vec{r}_{12} \;=\; \vec{r}_2 - \vec{r}_1,

which builds the sign logic directly into the vector algebra (the unit vector points from source to target; the product q_1 q_2 carries the sign).

Coulomb's law has the same mathematical form as Newton's law of universal gravitation: both are inverse-square, both are central, both superpose linearly. The physical differences are dramatic: charge comes in two signs (so Coulomb's force can repel), while gravitational mass is strictly positive (gravity always attracts); and the electric force between an electron and a proton is about 2.3 \times 10^{39} times stronger than the gravitational force between the same pair — an astonishing disparity that explains why the gravitational interaction of atoms is negligible while the electric interaction holds matter together.

On a dry Delhi morning in December, blow up two rubber balloons, tie them to threads of equal length, and hang them side by side from a single hook. They hang straight down, touching. Now rub each balloon vigorously on a synthetic sari or a polyester kurta for fifteen seconds and let them settle again. They do not hang straight down any more — they splay outward, the threads making a sharp V, visibly repelling each other, as if each balloon were being pushed away from its neighbour by an invisible hand. Rub a fresh balloon, hold it close to your forearm in a tanned patch of afternoon sunlight, and watch the fine hair on your arm lean toward it — visibly attracted.

Both the repulsion of the two like-charged balloons and the attraction of the charged balloon for your arm hair are the same force, obeying the same equation. It is called Coulomb's law — after the 18th-century physicist who first pinned down its form quantitatively, using a delicate apparatus called a torsion balance — and it is the mathematical foundation for every phenomenon in electrostatics: every field line, every capacitor, every atomic bond, every molecule, every chemical reaction. If you understand Coulomb's law, you understand the machinery behind almost every piece of structure in the visible universe that is not gravity and is not nuclear.

This article does three jobs. First, it derives the force law from Coulomb's observations — the torsion-balance experiment that established F \propto q_1 q_2 / r^2 in the 1780s, and the reasoning that lets you pin down the constant k. Second, it writes the law in vector form, so that the direction of the force — attract or repel — is automatic rather than a special case. Third, it compares Coulomb's law to Newton's law of universal gravitation, the only other inverse-square force law in fundamental physics, and extracts the lessons the comparison teaches about the nature of the two forces. Three worked examples (balloons on strings, a chloride ion in a salt crystal, and a proton-electron pair in a hydrogen atom) anchor the formula to physically concrete scenarios. The going-deeper section handles dielectric media, the limits of the point-charge idealisation, and the conceptual shift from Coulomb to Maxwell.

The empirical starting point — Coulomb's torsion balance

By the 1770s, it was known — from experiments like the balloon-on-sari observation above, and more systematic work with charged pith balls — that like charges repel and unlike charges attract. It was suspected, by analogy with the inverse-square law already established for gravitation, that the electric force probably fell off as 1/r^2. But "suspected" is not "established." There was a beautiful indirect argument (the interior of a hollow charged conductor must be force-free, which by a shell-theorem argument implies inverse-square), but nobody had measured the force as a function of distance and confirmed the 1/r^2 dependence directly.

Coulomb did exactly that in 1785, using a device of his own design — the torsion balance. The device is delicate, elegant, and conceptually simple:

Varying the distance r (by moving the fixed charged ball on its handle) and measuring the corresponding equilibrium twist, Coulomb found the force fell off as 1/r^2 to within the precision of the instrument: doubling the distance quartered the force; tripling it reduced the force to a ninth. Separately, by touching two identical gilded balls together (so charge redistributes equally) he could halve a charge in a controlled way, and he confirmed that the force was proportional to each charge: halving either charge halved the force.

Putting these two empirical results together:

F \;\propto\; \frac{q_1 q_2}{r^2}. \tag{1}

The proportionality constant is a matter of convention — it depends on the units you choose for charge, force, and distance.

Schematic of Coulomb's torsion balance A vertical silver wire suspension hangs from a fixed support at the top. Attached to the bottom of the wire is a light horizontal rod, with a small pith ball at one end and a counterweight at the other. A second pith ball on an insulating handle is positioned at a distance r from the first. When both balls are charged, the electrostatic force causes the rod to rotate, twisting the wire. A protractor scale at the top measures the angle of twist. fixed support twist angle silver wire counterwt q_1 q_2 insulating handle r (varied) F on q_1
Coulomb's torsion balance: a light rod carrying a charged ball $q_1$ is suspended by a fine silver wire. A second charged ball $q_2$ on an insulating handle is brought to distance $r$. The electrostatic force twists the wire through an angle that is proportional to the torque, and Coulomb measured that angle while varying $r$. The $1/r^2$ law was read off from the data.

From proportionality to the force law

In SI units (coulombs for charge, newtons for force, metres for distance), the constant in equation (1) is given the name k, and equation (1) is written

F \;=\; k\,\frac{|q_1 q_2|}{r^2}. \tag{2}

The magnitude of k is fixed empirically by experiment: measuring the force between known charges at a known distance pins it down. The modern accepted value is

k \;=\; 8.9875 \times 10^9\;\text{N·m}^2/\text{C}^2 \;\approx\; 9 \times 10^9\;\text{N·m}^2/\text{C}^2.

Why the weird-looking number "nine billion"? Because the coulomb is an enormous unit. One coulomb is the charge of roughly 6.24 \times 10^{18} electrons — an ordinary balloon rubbed on a sari acquires maybe 10^{-9} to 10^{-7} C. To get a force of 1 N between two 1 C charges, you need them 30 km apart: the coulomb is simply too big a unit to be practical at the balloon scale. The factor k \approx 9 \times 10^9 is the price paid for this inconvenient choice of unit — and it makes every electrostatics calculation involving ordinary charges yield forces in the micronewton to newton range, as you would expect.

The constant k and the permittivity \varepsilon_0

Historically, k is written in terms of another constant — the permittivity of free space \varepsilon_0:

k \;=\; \frac{1}{4\pi\varepsilon_0}, \qquad \varepsilon_0 \;=\; 8.8542 \times 10^{-12}\;\text{C}^2/(\text{N·m}^2). \tag{3}

At first sight this is an ugly substitution — why replace a single number 9 \times 10^9 with a reciprocal of 4\pi times an even uglier number? The reason is that when you later derive the field of a point charge (E = q/(4\pi\varepsilon_0 r^2)), and then apply Gauss's law (which requires integrating over a sphere of area 4\pi r^2), the factor of 4\pi cancels and the formula becomes clean. Written in terms of \varepsilon_0, Coulomb's law "remembers" the spherical geometry of the field; written in terms of k, it does not.

The modern SI conventions (as revised in 2019) fix e (the elementary charge) by definition, which lets \varepsilon_0 be computed theoretically rather than measured. The numerical value given above is now a derived rather than measured constant. For every practical purpose — kitchen-scale problems, JEE Advanced problems, high-energy physics — use k \approx 9 \times 10^9\;\text{N·m}^2/\text{C}^2 and be done.

Sanity check — charge and force for a typical balloon

A party balloon rubbed on a synthetic sari picks up a charge of about q \approx 10^{-8} C (a rough estimate from typical tribocharging experiments — you can look this up or measure it with an electroscope). Two such balloons held 10 cm apart would experience a mutual force of

F \;=\; k\,\frac{q^2}{r^2} \;=\; 9 \times 10^9 \times \frac{(10^{-8})^2}{(0.1)^2} \;=\; 9 \times 10^9 \times \frac{10^{-16}}{10^{-2}} \;=\; 9 \times 10^{-5}\,\text{N}.

Ninety microneutons. That is about the weight of a single grain of sand. Is that realistic? Each balloon has a mass of maybe 2 g = 2 \times 10^{-3} kg, weight about 2 \times 10^{-2} N. A force of 9 \times 10^{-5} N on each, compared to a weight of 2 \times 10^{-2} N, should cause the threads to swing out by an angle \tan\theta \approx F/W \approx 9 \times 10^{-5}/2\times 10^{-2} \approx 4.5 \times 10^{-3}, i.e., about 0.25°. The actual splay you see with party balloons is closer to 10-30°, which tells you the real charge on a well-rubbed balloon is more like 10^{-7} to 10^{-6} C — somewhat larger than our rough estimate. Worked example 1 below does this calculation properly backwards: given the observed splay angle, compute the charge.

The vector form — building attract/repel into the algebra

Equation (2) gives the magnitude of the force, with the sign of the product q_1 q_2 determining whether the force is attractive or repulsive (positive product → like charges → repulsion; negative product → unlike charges → attraction). But in a real problem with charges spread across two or three dimensions, working with magnitudes-and-sign-conventions gets clumsy. The clean way is to use vectors, where direction information is built into the formula from the start.

Let charge q_1 sit at position \vec{r}_1 and charge q_2 at position \vec{r}_2 in some chosen coordinate system. Define the displacement vector from 1 to 2:

\vec{r}_{12} \;\equiv\; \vec{r}_2 - \vec{r}_1. \tag{4}

Its magnitude |\vec{r}_{12}| = r is the distance between the charges. The unit vector pointing from q_1 toward q_2 is

\hat{r}_{12} \;\equiv\; \frac{\vec{r}_{12}}{|\vec{r}_{12}|}. \tag{5}

The force on q_2 due to q_1 is then

\boxed{\;\vec{F}_{1 \to 2} \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{|\vec{r}_{12}|^2}\,\hat{r}_{12}\;} \tag{6}

Why: the unit vector \hat{r}_{12} points from the source toward the target. If q_1 q_2 > 0 (like charges), the coefficient is positive and \vec{F}_{1\to 2} points along \hat{r}_{12} — that is, away from q_1, i.e., repulsive on q_2. If q_1 q_2 < 0 (unlike charges), the coefficient is negative and \vec{F}_{1\to 2} points along -\hat{r}_{12} — that is, toward q_1, i.e., attractive on q_2. No sign conventions by hand; the math does it automatically.

Newton's third law is automatic too. The force on q_1 due to q_2 is obtained by swapping subscripts:

\vec{F}_{2 \to 1} \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_2 q_1}{|\vec{r}_{21}|^2}\,\hat{r}_{21}.

But \vec{r}_{21} = \vec{r}_1 - \vec{r}_2 = -\vec{r}_{12}, so \hat{r}_{21} = -\hat{r}_{12} and |\vec{r}_{21}| = |\vec{r}_{12}|. Therefore

\vec{F}_{2 \to 1} \;=\; -\vec{F}_{1 \to 2}.

Why: Newton's third law — the force on q_1 from q_2 is equal in magnitude and opposite in direction to the force on q_2 from q_1 — falls straight out of the vector form, without needing to be added as a separate axiom.

Vector form of Coulomb's law for like and unlike charges Two panels side by side. Left panel: two positive charges, force arrows pointing outward from each charge along the line joining them — repulsion. Right panel: a positive and a negative charge, force arrows pointing toward each other — attraction. The unit vector r-hat from source to target is shown in both cases. Like charges: repulsion + q_1 > 0 + q_2 > 0 r̂_12 F_{1→2} F_{2→1} r q_1 q_2 > 0 ⇒ F_{1→2} along r̂_12 (outward) Unlike charges: attraction + q_1 > 0 − q_2 < 0 r̂_12 F_{1→2} F_{2→1} r q_1 q_2 < 0 ⇒ F_{1→2} along −r̂_12 (inward)
The vector form of Coulomb's law in two cases. Left: like charges ($q_1 q_2 > 0$), force on each charge points away from the other — repulsion. Right: unlike charges ($q_1 q_2 < 0$), force points toward the other — attraction. The unit vector $\hat{r}_{12}$ always points from the source $q_1$ to the target $q_2$; the sign of $q_1 q_2$ flips the direction of the force automatically.

Watch: force magnitude as r changes

The magnitude F = k q_1 q_2 / r^2 falls off very rapidly with r. Half the separation and the force quadruples; double it and the force drops to a quarter. The figure below lets you drag the separation r between two charges of 1 nC each and watch the force magnitude update live. For comparison, the curve shows how gravity would behave between the two balloons (same mass at each end, 2 g) — the difference is staggering.

Interactive: Coulomb force vs distance for two 1 nC charges A log-scaled plot of the Coulomb force between two one-nanocoulomb charges as a function of distance. A draggable red dot sets the distance from two centimetres to half a metre, and the current force magnitude is displayed. distance r (m) log₁₀(F / N) −7 −6 −5 −4 −3 0.05 0.10 0.20 0.30 0.40 0.50 (gravity between the same charges: log F ≈ −16, off-chart) drag the red point along the axis
Coulomb force between two 1 nC charges as a function of separation (log-linear axes). The red curve shows $F = 9\times 10^{-9}/r^2$ N. Dragging $r$ from 50 mm to 500 mm (a factor of 10) drops the force by a factor of 100 — the hallmark of an inverse-square law. Note that gravity between the same 2 g balloons (log F ≈ −16) is so much weaker it does not even appear on this plot.

Coulomb vs Newton — the two inverse-square laws

The law you have just met has an older cousin. Newton's law of universal gravitation, for two point masses m_1 and m_2 separated by r, gives

F_g \;=\; G\,\frac{m_1 m_2}{r^2}, \qquad G \;=\; 6.674 \times 10^{-11}\;\text{N·m}^2/\text{kg}^2.

Structurally the two laws are twins. Both are:

The physical differences are dramatic:

1. Sign. Gravitational mass is strictly positive — every known lump of matter in the universe has m \geq 0. Electric charge comes in two signs. This single difference is why gravity is always attractive, while the electric force can repel. It is also why large-scale gravity is so dominant in the universe: astronomical bodies contain approximately equal numbers of positive and negative charges, so their net electrostatic force on each other is negligible, while their gravitational attractions add up. You do not orbit the Sun because of its electric charge (there isn't a net one to speak of); you orbit because of its mass.

2. Strength. The ratio of the two forces between a proton (m_p = 1.67 \times 10^{-27} kg, q_p = +e) and an electron (m_e = 9.11 \times 10^{-31} kg, q_e = -e), separated by any distance r:

\frac{F_\text{elec}}{F_\text{grav}} \;=\; \frac{k e^2 / r^2}{G m_p m_e / r^2} \;=\; \frac{k e^2}{G m_p m_e}.

The r^2 cancels — this ratio is a constant — and substituting the numbers gives

\frac{F_\text{elec}}{F_\text{grav}} \;=\; \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.674\times 10^{-11})(1.67\times 10^{-27})(9.11 \times 10^{-31})} \;\approx\; 2.27 \times 10^{39}.

Why: plug k, e, G, m_p, m_e into the ratio. The electric force between the proton and electron of a hydrogen atom is about 10^{39} times the gravitational force between them. That is a ratio so large it has no physical precedent in daily life — more than a trillion trillion trillion. Electric forces dominate every atomic, molecular, chemical, and biological process; gravity is a sideshow at the atomic scale.

3. The medium matters for Coulomb, not (classically) for Newton. If you put two charges in a medium other than vacuum — say, immersed in water — the force between them is reduced by a factor \kappa called the dielectric constant of the medium (also called the relative permittivity \varepsilon_r). For water at room temperature \kappa \approx 80. This is why salt dissolves so readily in water: Na⁺ and Cl⁻ ions attract each other with a force 80 times weaker in water than in vacuum, so thermal motion can tear them apart. No such medium-dependence exists for classical gravity; gravity sees only mass.

4. Sources of the field. Both laws give rise to a field picture — gravitational field for Newton, electric field for Coulomb. In that language, electric fields are sourced by charge density, gravitational fields by mass density. But in the full relativistic extension, gravity (general relativity) is sourced not just by mass-energy density but also by pressure and momentum flux — it is much richer. And electromagnetism, extended to moving charges, gains a second field (magnetism) that gravity does not have. These extensions are beyond this article; for static, point-charge problems, the analogy between Coulomb and Newton is tight.

Direct numerical comparison

Quantity Gravitation Electrostatics
Law F = Gm_1 m_2/r^2 F = kq_1 q_2/r^2
Constant G = 6.674 \times 10^{-11} N·m²/kg² k = 8.988 \times 10^9 N·m²/C²
Sign of "charge" always \geq 0 \pm
Force direction always attractive attract or repel
F between proton & electron at 1 Å 2 \times 10^{-47} N 2 \times 10^{-8} N
Ratio F_E / F_G — \sim 2 \times 10^{39}

The two forces are mathematically identical in form and physically completely different in scale and richness. The fact that both obey inverse-square laws is not a coincidence — it traces to the geometry of three-dimensional space, in which any "flux" from a point source spreads over a surface of area 4\pi r^2, so the force per unit solid angle stays constant while the force per unit area falls as 1/r^2. Gauss's law will make this relationship exact later in the electrostatics curriculum.

Worked examples

Example 1: Two charged balloons on strings

Two identical rubber balloons (each of mass m = 2 g) are tied to threads of length L = 80 cm and suspended from a common point in a dry Delhi December room. Each balloon is rubbed on a synthetic kurta and acquires the same charge q. At equilibrium, each thread makes an angle \theta = 15° with the vertical. Find q.

Two charged balloons on threads at equilibrium angle Two balloons hang from a common point by threads of length L, each making angle theta with the vertical. Forces on the right balloon: tension T along the thread, weight mg downward, Coulomb repulsion F horizontal pointing right. The free body triangle gives tan theta equals F over mg. θ q, 2 g q, 2 g F mg T r = 2 L sin θ
Each balloon experiences three forces: the thread tension $T$ (along the thread, up-and-inward), gravity $mg$ (straight down), and the Coulomb repulsion $F$ from the other balloon (horizontal, outward). At equilibrium the three forces sum to zero.

Step 1. Find the distance r between the two balloons at equilibrium.

From the geometry, each balloon hangs at a horizontal distance L\sin\theta from the vertical below the support. The two balloons are on opposite sides, so

r \;=\; 2 L \sin\theta \;=\; 2 \times 0.80 \times \sin 15°.
\sin 15° \;\approx\; 0.2588
r \;=\; 2 \times 0.80 \times 0.2588 \;=\; 0.4141\,\text{m}.

Why: the geometry is a symmetric V. Each balloon is displaced L\sin\theta horizontally from the vertical, so the two balloons are 2L\sin\theta apart along the horizontal.

Step 2. Write the equilibrium equations for one balloon.

Consider the right balloon. Three forces act: tension T along the thread (directed up and to the left, toward the support), weight mg (straight down), and Coulomb force F (horizontal, outward, to the right). At equilibrium, horizontal and vertical components each sum to zero.

Horizontal: T \sin\theta \;=\; F

Vertical: T \cos\theta \;=\; mg

Why: the tension has a horizontal component (inward, leftward) T\sin\theta that must balance the outward Coulomb force F. Its vertical component (upward) T\cos\theta must balance the weight. Two equations, two unknowns (T and F).

Step 3. Eliminate T by dividing.

\frac{T\sin\theta}{T\cos\theta} \;=\; \tan\theta \;=\; \frac{F}{mg}.
F \;=\; mg \tan\theta \;=\; (2 \times 10^{-3})(9.8)\tan 15° \;=\; 1.96 \times 10^{-2} \times 0.2679 \;=\; 5.25 \times 10^{-3}\,\text{N}.

Why: dividing the two equilibrium equations eliminates the tension and isolates the ratio F/mg. The result F = mg\tan\theta is a workhorse formula for pendulum/balloon problems where an unknown horizontal force tilts a weight.

Step 4. Apply Coulomb's law to get the charge.

F \;=\; k\,\frac{q^2}{r^2} \;\Rightarrow\; q^2 \;=\; \frac{F r^2}{k} \;=\; \frac{(5.25\times 10^{-3})(0.4141)^2}{9\times 10^9}.
q^2 \;=\; \frac{5.25\times 10^{-3} \times 0.1715}{9\times 10^9} \;=\; \frac{9.00 \times 10^{-4}}{9\times 10^9} \;=\; 1.00 \times 10^{-13}\,\text{C}^2.
q \;=\; \sqrt{1.00 \times 10^{-13}} \;=\; 3.16 \times 10^{-7}\,\text{C} \;=\; 316\,\text{nC}.

Why: same Coulomb's law as before, solved for q^2. Taking the square root gives the charge magnitude on each balloon (both carry the same charge q because the setup is symmetric).

Result: Each balloon carries about q \approx 3.2 \times 10^{-7} C \approx 320 nC. That is roughly 2 \times 10^{12} electrons' worth of charge — a huge absolute number, but a tiny fraction (about 10^{-14}) of the total number of electrons in the 2 g balloon itself.

What this shows: The static electricity you can feel on a rubbed balloon corresponds to hundreds of nanocoulombs, generating forces of millinewtons at separations of tens of centimetres. This is enough to visibly tilt a balloon on a thread, to make hair stand up on your arm, and to give you a small shock when you touch a metal tap. It is the same mechanism, scaled up by eight orders of magnitude in charge, that makes lightning.

Example 2: Force on a chloride ion in a salt crystal

Common salt (NaCl) crystallises in a simple cubic lattice in which Na⁺ and Cl⁻ ions alternate. The ions are approximately point charges of magnitude e = 1.6 \times 10^{-19} C, and the nearest-neighbour distance is a = 2.82 \times 10^{-10} m (2.82 Å). Consider a chloride ion at the origin. One of its six nearest neighbours is a sodium ion directly above it at distance a. What is the force on the Cl⁻ ion due to that single Na⁺ ion, and what is the electrostatic potential energy of the pair?

Pair of ions in a salt crystal A sodium ion with charge plus e above a chloride ion with charge minus e, separated vertically by lattice spacing a equals 2.82 angstroms. The attractive Coulomb force pulls them toward each other. +e Na⁺ −e Cl⁻ a = 2.82 Å F on Na⁺ (downward, attractive) F on Cl⁻ (upward, attractive)
Two ions in a sodium-chloride crystal, separated by the lattice spacing 2.82 Å. The unlike charges attract each other with the same Coulomb force, in opposite directions on each (Newton's third law).

Step 1. Compute the magnitude of the force.

F \;=\; k\,\frac{|q_1||q_2|}{r^2} \;=\; \frac{(9\times 10^9)(1.6\times 10^{-19})^2}{(2.82\times 10^{-10})^2}.
F \;=\; \frac{9\times 10^9 \times 2.56 \times 10^{-38}}{7.95 \times 10^{-20}} \;=\; \frac{2.304 \times 10^{-28}}{7.95\times 10^{-20}} \;=\; 2.90 \times 10^{-9}\,\text{N}.

Why: plug the numbers into Coulomb's law directly. The magnitudes are both e (elementary charge), the separation is the lattice spacing a. The force is nanonewton-scale — small in absolute terms, but enormous relative to the weight of the ion (about 10^{-25} N for an ion of mass 10^{-26} kg), which is what lets ionic crystals hold together rigidly at room temperature.

Step 2. Determine the direction.

The charges are opposite (+e on Na⁺, -e on Cl⁻), so the product q_\text{Na}\, q_\text{Cl} = -e^2 < 0. The force on the Cl⁻ ion points toward the Na⁺ ion — that is, upward in the diagram. The force on Na⁺ points downward, toward the Cl⁻.

Step 3. Compute the potential energy of the pair.

The electrostatic potential energy of two point charges separated by r is

U \;=\; k\,\frac{q_1 q_2}{r}.

For this Na⁺–Cl⁻ pair:

U \;=\; \frac{(9\times 10^9)(+e)(-e)}{a} \;=\; -\frac{ke^2}{a} \;=\; -\frac{(9\times 10^9)(2.56\times 10^{-38})}{2.82 \times 10^{-10}}.
U \;=\; -\frac{2.304 \times 10^{-28}}{2.82 \times 10^{-10}} \;=\; -8.17 \times 10^{-19}\,\text{J} \;=\; -5.10\,\text{eV}.

Why: potential energy between two charges is kq_1q_2/r, not kq_1q_2/r^2 — the potential is the integral of the force from infinity. The negative sign reflects the attraction: work must be done against an external agent to separate the two ions to infinity. 5.1 eV is the binding energy of this single bond, and summing over all pairs gives the lattice energy of NaCl (~ 7.9 eV per ion pair, after including attractions and repulsions from all neighbours, not just the six nearest).

Result: Force = 2.90 \times 10^{-9} N, attractive; potential energy = -5.10 eV, negative (bound).

What this shows: At the scale of a single bond in an ordinary salt crystal, Coulomb's law gives the force directly. The nanonewton-scale force, on the femtogram-scale mass of an ion, sets the oscillation frequencies (optical phonons) of the crystal and its melting point (NaCl melts at 801 °C — high, because the binding is strong). Every lattice vibration, every elastic modulus, every thermal property of salt is ultimately a sum over Coulomb's law applied pair-by-pair.

Example 3: Electron in a hydrogen atom

A hydrogen atom consists of a proton (charge +e) and an electron (charge -e). In the Bohr model, the electron orbits at a radius of r_0 = 0.529 \times 10^{-10} m (the Bohr radius). Treating them as point charges: (a) What Coulomb force binds the electron? (b) What orbital speed does this imply, using F = m_e v^2 / r_0 for circular motion? (c) Compare F_\text{elec} with F_\text{grav} between the same two particles.

Hydrogen atom with proton and electron in circular orbit A proton at the centre and an electron orbiting in a circle of radius r naught equals 0.529 angstrom. The Coulomb attraction provides the centripetal force. The electron's velocity is tangent to the orbit. +e p⁺ e⁻ r₀ = 0.529 Å F (centripetal) v (tangent)
The proton-electron pair in the Bohr model: the electron moves in a circle of radius $r_0$ around the (fixed) proton, with Coulomb attraction playing the role of centripetal force.

Step 1. Compute the Coulomb force at the Bohr radius.

F \;=\; k\,\frac{e^2}{r_0^2} \;=\; \frac{(9\times 10^9)(1.6\times 10^{-19})^2}{(0.529 \times 10^{-10})^2}.
F \;=\; \frac{9\times 10^9 \times 2.56\times 10^{-38}}{2.80\times 10^{-21}} \;=\; \frac{2.304\times 10^{-28}}{2.80\times 10^{-21}} \;=\; 8.24 \times 10^{-8}\,\text{N}.

Why: Bohr radius is 0.529 Å — smaller than the NaCl spacing of 2.82 Å by a factor of 5.3 — so the force is larger by 5.3^2 \approx 28 compared to the salt-crystal example. At atomic dimensions, Coulomb forces are of order 10^{-7} N; this looks small but compared to an electron mass of 9.1 \times 10^{-31} kg it gives enormous accelerations.

Step 2. Use circular motion to get the speed.

F \;=\; \frac{m_e v^2}{r_0} \;\Rightarrow\; v \;=\; \sqrt{\frac{F r_0}{m_e}} \;=\; \sqrt{\frac{(8.24 \times 10^{-8})(0.529\times 10^{-10})}{9.11 \times 10^{-31}}}.
v \;=\; \sqrt{\frac{4.36 \times 10^{-18}}{9.11\times 10^{-31}}} \;=\; \sqrt{4.79 \times 10^{12}} \;=\; 2.19 \times 10^6\,\text{m/s}.

Why: centripetal force equals mass times velocity squared over radius. Rearranging gives v. The result is about 2.2\times 10^6 m/s — astonishingly, this is about 1/137 of the speed of light. (The ratio v/c = \alpha \approx 1/137 is the fine-structure constant, one of the most celebrated dimensionless numbers in physics.)

Step 3. Compare with gravity.

F_\text{grav} \;=\; G\,\frac{m_p m_e}{r_0^2} \;=\; \frac{(6.674 \times 10^{-11})(1.67\times 10^{-27})(9.11\times 10^{-31})}{(0.529\times 10^{-10})^2}.
F_\text{grav} \;=\; \frac{1.015 \times 10^{-67}}{2.80 \times 10^{-21}} \;=\; 3.63 \times 10^{-47}\,\text{N}.

Ratio:

\frac{F_\text{elec}}{F_\text{grav}} \;=\; \frac{8.24\times 10^{-8}}{3.63\times 10^{-47}} \;=\; 2.27 \times 10^{39}.

Why: plug in G, m_p, m_e, r_0 for gravity. The ratio F_\text{elec}/F_\text{grav} is 2.3 \times 10^{39} — this is the number derived in general terms above (ratio of k e^2 to G m_p m_e), confirmed here by direct numerical substitution.

Result: Coulomb force: 8.2 \times 10^{-8} N, giving orbital speed 2.2 \times 10^6 m/s. Gravitational force: 3.6 \times 10^{-47} N — smaller by a factor of 2.3 \times 10^{39}.

What this shows: The electron is held to the proton almost entirely by the Coulomb force; gravity is irrelevant by forty orders of magnitude. This is why chemistry is electromagnetic, not gravitational — and why every atom, every molecule, every biological macromolecule, every crystal, every piece of ordinary matter, owes its structure to Coulomb's law and nothing else. Gravity rules the cosmos; electromagnetism rules everything else you will ever touch.

Common confusions

If you came here to understand Coulomb's law, use the formula, and set up simple electrostatics problems, you have what you need. What follows is for readers curious about how the law extends to continuous charge distributions, how dielectric media modify it, and the conceptual promotion from Coulomb's law to field theory.

From two charges to many — superposition and integration

Coulomb's law is stated for two point charges. For a collection of N point charges q_1, q_2, \ldots, q_N, the net force on a test charge Q at position \vec{r} is the vector sum of the individual Coulomb forces:

\vec{F}_\text{on Q} \;=\; \sum_{i=1}^{N} \frac{1}{4\pi\varepsilon_0}\,\frac{q_i Q}{|\vec{r} - \vec{r}_i|^2}\,\hat{r}_{iQ}.

This is the principle of superposition, discussed at greater length in superposition of electrostatic forces. It is not a theorem; it is an experimental fact (equivalent to saying the electromagnetic field equations are linear), but it is the fact that makes calculation feasible.

For a continuous charge distribution (a charged rod, a ring, a disk, a sphere), the sum becomes an integral:

\vec{F}_\text{on Q} \;=\; \int \frac{1}{4\pi\varepsilon_0}\,\frac{Q\,dq}{|\vec{r} - \vec{r}'|^2}\,\hat{r}',

where dq is an infinitesimal element of charge at position \vec{r}', and the integral runs over the distribution. This kind of integral — handled case-by-case in the electrostatics curriculum — is what you use to derive the field of a ring, a disk, an infinite plane, and so on.

Coulomb's law in a dielectric medium

If you put two charges not in vacuum but in a medium (water, glass, air, oil), the force between them is modified. The medium's molecules polarise in response to the charges — positive nuclei shift slightly toward the negative charge, negative electron clouds shift slightly toward the positive charge — and these induced dipoles create a field that opposes the original one. The net field is weakened by a factor called the dielectric constant \kappa (or relative permittivity \varepsilon_r) of the medium:

F_\text{in medium} \;=\; \frac{1}{4\pi\kappa\varepsilon_0}\,\frac{|q_1 q_2|}{r^2}.

Equivalently, define the permittivity of the medium \varepsilon = \kappa \varepsilon_0 and write F = q_1 q_2/(4\pi\varepsilon r^2). Some values of \kappa (near room temperature):

Medium \kappa
Vacuum 1 (by definition)
Dry air 1.0006
Mumbai humid air (50 % RH) 1.0008
Glass ~5
Ethanol ~24
Water ~80
Barium titanate ceramic ~1200

The fact that water has a very large dielectric constant is why it is a universal solvent for ionic compounds. The Coulomb force holding a Na⁺ ion to a Cl⁻ ion in a NaCl crystal is reduced by a factor of 80 in water, which means the thermal agitation of the water molecules (kT at room temperature) is enough to pull the two apart — the salt dissolves. In oil or air the force is essentially unchanged, and salt does not dissolve.

In the Indian-monsoon context: the dielectric constant of air rises slightly with humidity. In Mumbai on a humid afternoon, this tiny increase in \kappa translates into slightly weakened electrostatic forces — so your rubbed comb picks up less dust, and your polyester kurta is less "sticky" than it is on a dry Delhi December day. Static electricity is fundamentally a dry-weather phenomenon; the Indian subcontinent demonstrates this geographically, with winter Delhi producing hair-raising static and summer Chennai producing almost none.

Why 1/r^2, really — the geometric reason

The inverse-square form of Coulomb's law is not an accident. If the electric field lines from a point charge spread out uniformly in all directions, they pass through a sphere of radius r with surface area 4\pi r^2 — and since the total flux must be conserved (by Gauss's law), the density of field lines (and hence the field strength) at distance r must fall as 1/r^2. This is the deep geometric reason both Coulomb's law and Newton's gravity law are inverse-square: three-dimensional space.

If space had four dimensions, Coulomb's law would be F \propto 1/r^3 (sphere area would go as r^3). If space were two-dimensional, it would be F \propto 1/r (circle circumference goes as r). The dimensionality of space is written into the shape of the force law.

From Coulomb to Maxwell — the conceptual promotion

Coulomb's law, taken literally, says: "charge q_1 exerts an instantaneous force on charge q_2 across all of space." This is "action at a distance" — the force appears without any mediating physical process. Newton was profoundly uncomfortable with this idea for gravity, and physicists of the 19th century were no happier about it for electromagnetism.

The resolution came in two stages.

Stage 1: Faraday's field concept. Instead of saying "q_1 acts on q_2 across the gap," say "q_1 creates an electric field \vec{E} everywhere in space, and q_2 experiences a force \vec{F} = q_2 \vec{E} at its own location." The force is now a local phenomenon — the field at q_2's position is what matters, not the distant q_1. Coulomb's law becomes the statement \vec{E}(\vec{r}) = k q_1 \hat{r}/r^2 for the field due to a point charge.

Stage 2: Maxwell's equations. The electric field (together with a companion magnetic field \vec{B}) is governed by four differential equations. These equations predict electromagnetic waves — disturbances in \vec{E} and \vec{B} that propagate through vacuum at the finite speed c = 1/\sqrt{\mu_0 \varepsilon_0} = 3 \times 10^8 m/s, the speed of light. Coulomb's law, as stated above, is the static limit — valid when charges do not move, or move slowly compared to c. For rapidly varying charges, Coulomb's law is replaced by a richer theory in which fields propagate at the speed of light, and information cannot travel faster.

The promotion is not cosmetic — it changes the ontology of the theory. In Coulomb's picture, there are charges, and forces between them. In Maxwell's picture, there are charges, and there is an electromagnetic field filling all of space, with a life of its own (it can detach from its source and propagate as light). This picture is the foundation of modern physics, from radio broadcasting to laser interferometry to the Large Hadron Collider. Coulomb's law is the opening chapter; Maxwell's equations are the whole book.

The magnetic force — what Coulomb misses

Two moving charges exert not only an electric force on each other (Coulomb) but also a magnetic force (which depends on their velocities). For charges moving at speeds \ll c, the magnetic force is smaller than the Coulomb force by a factor of roughly (v/c)^2 and can usually be neglected. For high-speed charges — electrons in an accelerator, currents in a wire — the magnetic force is central. The full story is given by the Biot-Savart law and the Lorentz force, which you will meet later in the electromagnetism curriculum. Coulomb's law is the subset of that story in which nothing moves.

Where this leads next