Coulomb's law

Rub a plastic ruler on your hair and it will pick up tiny scraps of paper — a party trick everyone has done. Two thousand years passed between the Greeks noticing that rubbed amber attracts straw and 1785, when the French engineer Charles-Augustin de Coulomb finally put a number on the attraction. Using a delicate torsion balance sensitive enough to measure forces smaller than the weight of a mosquito, Coulomb showed that the push or pull between two charged pith balls obeys a beautifully simple rule: double the charge, double the force; double the distance, quarter the force. That rule is Coulomb's law, and it is the single equation on which the whole edifice of electrostatics — electric field, electric potential, capacitance, even Gauss's theorem — is built. It looks almost identical to Newton's law of gravitation, with one crucial twist: charge comes in two signs, so the electric force can pull things together or shove them apart, while gravity only ever pulls. This article takes Coulomb's law from statement to skill. We derive the force formula from the inverse-square structure Coulomb measured, work through two fully-numeric examples — one board-level, one competitive-exam level — that build on each other, catalogue the mark-losing traps examiners love to set, and finish with a quick-recall exam card. By the end you should be able to calculate the force between any two point charges in your sleep, and know exactly when Coulomb's law does and does not apply.

Key takeaways

Coulomb's law gives the force between two point charges: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$ , acting along the line joining them.
The constant $\dfrac{1}{4\pi\varepsilon_0} \approx 9\times10^9\ \text{N m}^2/\text{C}^2$ in vacuum or air, where $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N m}^2$ is the permittivity of free space.
Like charges repel, unlike charges attract; by Newton's third law the two forces in a pair are equal and opposite.
The force is an inverse-square law, structurally identical to gravitation, $F = Gm_1m_2/r^2$ — but electric force can be attractive or repulsive, and is vastly stronger between elementary particles.
When more than two charges are present, the net force on any one charge is the vector sum of the individual pairwise Coulomb forces — the superposition principle.
Coulomb's law strictly applies to point charges, or to spherically symmetric charge distributions treated as concentrated at the centre, with $r$ measured centre to centre.
Key idea: the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them, acts along the line joining them, and its sign — attractive or repulsive — follows straight from the signs of the two charges.

The core insight

Coulomb's law, $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ , says the force between two point charges grows with the product of the charges and shrinks with the square of the separation, is directed along the line joining them, and is repulsive for like signs, attractive for unlike signs.

Where it fits in the syllabus

Coulomb's law opens the CBSE Class 12 Physics chapter "Electric Charges and Fields" (Unit I: Electrostatics), coming right after the properties of charge — additivity, conservation, and quantisation $q = ne$ . It is the prescribed core formula of the chapter: the syllabus asks for the law itself, the value and meaning of $\varepsilon_0$ , its application to pairs and small systems of point charges via the superposition principle, and a qualitative comparison with Newton's law of gravitation. Everything that follows in the chapter — the electric field of a point charge, the field and torque of a dipole, and Gauss's theorem — is built directly on top of this one force law, so getting comfortable here pays off for the rest of the chapter.

The physical picture

Picture two small charged spheres, so small compared to the distance between them that we can treat each as a mathematical point. Coulomb's careful torsion-balance measurements established that the force between them depends on three things only: the two charge magnitudes, and the distance separating them, combined as follows.

F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}

where: $F$ is the magnitude of the electrostatic force (newtons) that each charge exerts on the other, $q_1$ and $q_2$ are the magnitudes of the two point charges (coulombs), $r$ is the distance between them (metres), and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space, a constant fixed by nature that measures how "permissive" vacuum is to electric field lines. The combination $\dfrac{1}{4\pi\varepsilon_0}$ is used so often that it is given its own symbol, $k \approx 9\times10^9\ \text{N m}^2/\text{C}^2$ , so the law is also written compactly as $F = k\,q_1q_2/r^2$ . The physical reading is direct: the force does not care about the sign of the product for its size — only the magnitudes enter the formula — but the sign of the product $q_1q_2$ tells you the nature of the force. If $q_1q_2 > 0$ (same sign) the force pushes the charges apart; if $q_1q_2 < 0$ (opposite signs) the force pulls them together. By Newton's third law, the force $q_1$ exerts on $q_2$ is always equal in magnitude and opposite in direction to the force $q_2$ exerts on $q_1$ , both acting along the straight line joining the two charges.

Figure

Two point charges on a horizontal line separated by distance r, with outward arrows on both charges when they carry the same sign showing a repulsive pair, and inward arrows in a second row when the charges carry opposite signs showing an attractive pair

Superposition principle. Real problems rarely involve just two charges. When three or more point charges are present, the net electrostatic force on any one of them is found by computing the Coulomb force due to each other charge as if the rest were absent, and then adding all these individual forces as vectors — not as plain numbers. This superposition principle is what turns Coulomb's law for a pair into a working tool for any collection of charges, and it is the method used in the very next topic of this chapter, forces between multiple charges.

Coulomb's law versus Newton's law of gravitation. The two laws share the identical inverse-square skeleton — $F \propto 1/r^2$ — and both forces act along the line joining the two bodies. But gravitation, $F = Gm_1m_2/r^2$ , is always attractive because mass has only one sign, whereas the electrostatic force can be either attractive or repulsive because charge has two signs. The two constants also differ enormously in scale: for two electrons, the electrostatic repulsion outweighs their mutual gravitational attraction by a factor of roughly $10^{42}$ , which is why gravity is completely negligible in atomic and everyday electrostatic problems and Coulomb's law dominates whenever charge is involved.

Derivation ABuilding Coulomb's law from Coulomb's measured proportionalities

The goal is to assemble the single quantitative statement $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ out of the two separate experimental proportionalities that Coulomb established with his torsion balance — that force grows with each charge and shrinks with the square of the distance — and to fix the constant of proportionality using the SI value of $\varepsilon_0$ , so that a student sees not just the formula but where each piece of it comes from and why the constant looks the way it does.

Step 1. State the two experimental proportionalities separately. Coulomb varied one charge at a time, holding the other and the distance fixed, and found the force directly proportional to each charge magnitude: $F \propto q_1$ and, independently, $F \propto q_2$ . Taken together these combine into $F \propto q_1 q_2$ — the force grows with the product of the two charges, not their sum.

Step 2. State the distance proportionality. Holding both charges fixed and varying only the separation $r$ , Coulomb found the force falls off as the inverse square of the distance: $F \propto \dfrac{1}{r^2}$ . Doubling $r$ cuts the force to a quarter; tripling $r$ cuts it to a ninth.

Step 3. Combine the two proportionalities into a single relation with one constant of proportionality, call it $k$ : $F = k\,\dfrac{q_1q_2}{r^2}$ . At this stage $k$ is just an unknown number waiting to be fixed by experiment and by the choice of units for charge.

Step 4. Fix the constant using the SI system. In SI units, charge is measured in coulombs and the constant is written in the rationalised form $k = \dfrac{1}{4\pi\varepsilon_0}$ , where $\varepsilon_0$ is the permittivity of free space, measured independently in electromagnetism experiments to be $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ . Substituting, $k = \dfrac{1}{4\pi \times 8.85\times10^{-12}} \approx 8.99\times10^9\ \text{N m}^2/\text{C}^2$ , which is rounded to $9\times10^9\ \text{N m}^2/\text{C}^2$ for numerical work.

Step 5. Write the final law and attach direction. $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ gives the magnitude of the force each point charge exerts on the other, and this force acts along the straight line joining the two charges — repulsive if $q_1$ and $q_2$ have the same sign, attractive if they have opposite signs, with the two forces of the pair equal and opposite by Newton's third law.

Worked examples

Board level - force between two point charges

Question: Two point charges $q_1 = +4\ \mu\text{C}$ and $q_2 = +9\ \mu\text{C}$ are placed $0.30\ \text{m}$ apart in air. Find the magnitude and nature of the force between them.

Step 1. Write down Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ , using $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2/\text{C}^2$ for air (essentially the same as vacuum).

Step 2. Convert every quantity to SI units before substituting: $q_1 = 4\times10^{-6}\ \text{C}$ , $q_2 = 9\times10^{-6}\ \text{C}$ , $r = 0.30\ \text{m}$ , so $r^2 = 0.09\ \text{m}^2$ .

Step 3. Substitute into the formula: $F = 9\times10^9 \times \dfrac{(4\times10^{-6})(9\times10^{-6})}{0.09}$ .

Step 4. Work the numerator first: $9\times10^9 \times 36\times10^{-12} = 324\times10^{-3} = 0.324$ .

Step 5. Divide by $r^2$ : $F = \dfrac{0.324}{0.09} = 3.6\ \text{N}$ .

Step 6. Decide the nature of the force from the signs: both charges are positive, so $q_1q_2 > 0$ and the force is repulsive.

Answer: $F = 3.6\ \text{N}$ , repulsive, directed along the line joining the two charges, each charge pushing the other away.

Competitive level - identical charges and a doubled distance

Question: Two identical point charges of equal magnitude $q$ exert a force of $0.90\ \text{N}$ on each other when separated by $0.10\ \text{m}$ in air. Find (a) the magnitude of each charge, and (b) the force between them if the distance is doubled to $0.20\ \text{m}$ while both charges are simultaneously reduced to a third of their original value.

Step 1. Since the charges are identical, $q_1 = q_2 = q$ , so Coulomb's law reduces to $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$ .

Step 2. Rearrange for $q^2$ : $q^2 = \dfrac{F r^2}{1/4\pi\varepsilon_0} = \dfrac{F r^2}{9\times10^9}$ .

Step 3. Substitute $F = 0.90\ \text{N}$ and $r = 0.10\ \text{m}$ , so $r^2 = 0.01\ \text{m}^2$ : $q^2 = \dfrac{0.90 \times 0.01}{9\times10^9} = \dfrac{9\times10^{-3}}{9\times10^9} = 1\times10^{-12}$ .

Step 4. Take the square root: $q = \sqrt{1\times10^{-12}} = 1\times10^{-6}\ \text{C} = 1\ \mu\text{C}$ . This answers part (a).

Step 5. For part (b), track both changes at once using the proportionality directly, without recomputing from scratch: $F \propto \dfrac{q_1q_2}{r^2}$ . Reducing each charge to a third multiplies the numerator $q_1q_2$ by $\dfrac{1}{3}\times\dfrac{1}{3} = \dfrac{1}{9}$ ; doubling $r$ multiplies $r^2$ by $4$ , so the denominator effect is a further factor of $\dfrac{1}{4}$ .

Step 6. Combine the two factors on the original force: $F_{\text{new}} = F \times \dfrac{1}{9} \times \dfrac{1}{4} = 0.90 \times \dfrac{1}{36} = 0.025\ \text{N}$ .

Answer: (a) Each charge has magnitude $1\ \mu\text{C}$ ; (b) after the changes, the force falls to $0.025\ \text{N}$ , still repulsive since both charges keep the same sign. Timing note: In an exam, resist the urge to recompute both new charge values and the new distance from scratch and resubstitute into the full formula — that costs an extra thirty to forty seconds of arithmetic. Spotting that $F$ obeys a clean proportionality and scaling the original answer by the two ratio factors is the faster, safer route, and it is exactly the kind of shortcut competitive exams reward under time pressure.

Common traps

Forgetting that Coulomb's law strictly applies to point charges, or to spherically symmetric charge distributions treated as if concentrated at the centre with $r$ measured centre to centre; it cannot be applied blindly to arbitrarily shaped extended bodies, especially at close range.
Mixing up units — charge must be in coulombs and distance in metres before substituting into $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ ; forgetting to convert $\mu\text{C}$ or $\text{nC}$ to coulombs (a factor of $10^{-6}$ or $10^{-9}$ ) is one of the most common numerical slips in board exams.
Treating the force as always attractive, the way gravity is — the sign of the charges must be checked explicitly to decide whether the Coulomb force is attractive or repulsive; a wrong sign call loses marks even when the magnitude is correct.
Ignoring the vector nature of force when more than two charges are involved — Coulomb's law by itself gives magnitude and direction for a single pair; combining several pairwise forces on one charge requires vector addition via the superposition principle, not simple algebraic addition of magnitudes.
Squaring only the distance and forgetting to square it correctly when the distance itself is doubled or halved — many marks are lost by writing $F \propto 1/r$ instead of $F \propto 1/r^2$ under time pressure.

You should be able to…

State Coulomb's law precisely and identify the physical meaning and SI value of $\varepsilon_0$ and $k = 1/4\pi\varepsilon_0$ .
Calculate the magnitude of the force between two point charges given their charge magnitudes and separation, converting units correctly first.
Determine whether a Coulomb force is attractive or repulsive purely from the signs of the two charges, and state this as a separate conclusion alongside the magnitude.
Use proportional reasoning to find how the force changes when the charges and/or the distance are scaled, without recomputing from scratch.
Solve for an unknown charge or an unknown distance given the force and the remaining quantities.
Compare the inverse-square form of Coulomb's law with Newton's law of gravitation, explaining both the structural similarity and the key physical differences.

Ask the tutor: want to practise a few more Coulomb's law numericals, or see how the superposition principle extends this force law to three or more charges at once?

Why does Coulomb's law only work for point charges — what if the charged object is a big sphere?

Coulomb's law was derived and measured for charges small enough, compared to the separation, that their shape does not matter — that is what "point charge" means. For an extended object, different parts of it sit at different distances from the other charge, and in general you would need calculus to add up the contributions from every bit of charge. The one convenient exception the syllabus gives you is a uniformly charged sphere: because of its perfect symmetry, it behaves — for any point outside it — exactly as if all its charge were concentrated at its centre, so Coulomb's law can be applied with $r$ measured from centre to centre.

If Coulomb's law and gravity have the same 1/r squared shape, why does gravity feel so much weaker in everyday life?

It is not that gravity is weaker as a law — it is that the constant $G$ in Newton's law of gravitation is astronomically smaller than the constant $k = 1/4\pi\varepsilon_0$ in Coulomb's law, and everyday objects carry only tiny net charges (most matter is electrically neutral overall) while carrying large masses. Compare two electrons: their electric repulsion is about $10^{42}$ times stronger than their gravitational attraction. Gravity only dominates our daily experience because planet-sized amounts of mass build up a large pull, while ordinary objects almost always have their positive and negative charges nearly perfectly balanced.

Does Coulomb's law change if the two charges are placed in water instead of air?

Yes, though it is not tested in detail in this chapter beyond the idea. The formula $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ is written for vacuum, and air is close enough to vacuum that the same constant $9\times10^9\ \text{N m}^2/\text{C}^2$ is used without correction. Inside a medium such as water, the permittivity is larger than $\varepsilon_0$ , and the measured force between the same two charges at the same separation comes out smaller — this "weakening" of the force inside a dielectric medium is picked up properly in the next chapter, Electrostatic Potential and Capacitance, when dielectrics are introduced.

In numericals, how do I quickly tell if I should get attraction or repulsion without drawing anything?

Just multiply the signs of the two charges mentally before you touch the magnitude formula. If both charges are positive, or both are negative, the product $q_1q_2$ is positive and the force is repulsive — same-sign charges push each other away. If one is positive and the other negative, the product is negative and the force is attractive. Write this one-word conclusion (attractive or repulsive) right next to your numeric answer for $F$ ; examiners specifically award a mark for stating the nature of the force, separate from the mark for the magnitude.