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Physics · Electric Charges and Fields

Coulomb's Law

Coulomb's law is the quantitative foundation of electrostatics: it fixes how strongly two point charges push or pull on each other. Following NCERT Section 1.5, this note develops the law in scalar and vector form, pins down the constant $k = 1/4\pi\varepsilon_0 \approx 9 \times 10^9~\text{N·m}^2/\text{C}^2$, contrasts it with Newtonian gravitation, and tracks how a dielectric medium weakens the force. Every subsequent topic in this chapter — fields, dipoles, flux, Gauss's law — rests on the relation derived here, and NEET tests it almost every year.

The statement of Coulomb's law

When the linear size of two charged bodies is much smaller than the distance separating them, the bodies may be treated as point charges and their size ignored. Charles Augustin de Coulomb, using a torsion balance in 1785, measured the force between two such charges and established a law with three intertwined claims: the force varies inversely as the square of the separation, varies directly as the product of the two charge magnitudes, and acts along the line joining the charges.

For two point charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum, the magnitude of the force is therefore

$$F = k\,\dfrac{|q_1 q_2|}{r^2}$$

The repulsion of like charges and the attraction of unlike charges both lie along the same line — the line connecting the two points. There is no transverse component for an isolated pair. The figure below shows the two signs of interaction sharing one geometry.

Figure 1 · Like vs unlike charges + + Like charges — repel (force outward) $\\hat r_{21}$ along the line joining + − Unlike charges — attract (force inward)

The constant k and permittivity of free space

In SI units the proportionality constant $k$ is about $9 \times 10^9~\text{N·m}^2/\text{C}^2$. Coulomb arrived at the law without knowing the magnitude of charge at all; in fact the choice of $k$ is what defines the unit of charge. Putting $q_1 = q_2 = 1~\text{C}$ and $r = 1~\text{m}$ gives $F = 9 \times 10^9~\text{N}$ — so one coulomb is the charge that, placed 1 m from an equal charge in vacuum, repels it with a force of nine billion newtons. The coulomb is evidently an enormous unit; practical electrostatics uses microcoulombs and nanocoulombs.

For later convenience the constant is written $k = 1/4\pi\varepsilon_0$, so that Coulomb's law takes its standard form

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

Here $\varepsilon_0$ is the permittivity of free space, with the NCERT value $\varepsilon_0 = 8.854 \times 10^{-12}~\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$. The two constants are mutually consistent: substituting $\varepsilon_0$ into $1/4\pi\varepsilon_0$ recovers $9 \times 10^9~\text{N·m}^2/\text{C}^2$.

QuantitySymbolValue (NCERT)
Coulomb constantk = 1/4πε₀≈ 9 × 10⁹ N·m²/C²
Permittivity of free spaceε₀8.854 × 10⁻¹² C²·N⁻¹·m⁻²
Force between 1 C, 1 C at 1 mF9 × 10⁹ N (repulsive)
Electronic chargee1.60 × 10⁻¹⁹ C
NEET Trap

Do not confuse k with ε₀

The constant in the numerator is $k = 9 \times 10^9$; the constant buried inside it is $\varepsilon_0 = 8.854 \times 10^{-12}$. Candidates frequently quote $9 \times 10^9$ as the value of $\varepsilon_0$ or forget that $k = 1/4\pi\varepsilon_0$. They are reciprocally related through a factor of $4\pi$, not equal.

$k = \dfrac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9~\text{N·m}^2/\text{C}^2$, with $\varepsilon_0 = 8.854 \times 10^{-12}~\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$.

Vector form and direction along the line joining

Force is a vector, so Coulomb's law is best written in vector notation. Let the position vectors of $q_1$ and $q_2$ be $\mathbf{r}_1$ and $\mathbf{r}_2$. Denote the vector from charge 1 to charge 2 by $\mathbf{r}_{21} = \mathbf{r}_2 - \mathbf{r}_1$, and the unit vector along it by $\hat{\mathbf{r}}_{21}$. The force on $q_2$ due to $q_1$ is then

$$\mathbf{F}_{21} = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q_1 q_2}{r^2}\,\hat{\mathbf{r}}_{21}$$

This single equation handles both signs correctly. If $q_1$ and $q_2$ carry the same sign, their product is positive and $\mathbf{F}_{21}$ points along $\hat{\mathbf{r}}_{21}$ — away from charge 1, i.e. repulsion. If the signs are opposite, the product is negative and $\mathbf{F}_{21}$ points along $-\hat{\mathbf{r}}_{21}$ — toward charge 1, i.e. attraction. No separate formula for like and unlike charges is needed.

Interchanging the labels gives $\mathbf{F}_{12} = -\mathbf{F}_{21}$: the force on $q_1$ due to $q_2$ is equal in magnitude and opposite in direction. Coulomb's law thus obeys Newton's third law automatically.

NEET Trap

The force is always along the line joining

For an isolated pair of point charges the Coulomb force has no sideways component — it lies entirely along $\hat{\mathbf{r}}_{21}$, the line connecting the two charges. The sign of the product $q_1 q_2$ decides only whether the arrow points outward or inward, never its line of action.

Direction = line joining the charges. Sign of $q_1q_2 > 0 \Rightarrow$ repulsion; $q_1q_2 < 0 \Rightarrow$ attraction.

Coulomb force versus Newtonian gravitation

Coulomb's law and Newton's law of gravitation are siblings in form: both fall off as the inverse square of distance and both act along the line joining the two bodies. The resemblance ends there. Gravitation couples to mass and is unconditionally attractive; the Coulomb force couples to charge and can be either attractive or repulsive. Crucially, the electric interaction is overwhelmingly stronger.

FeatureCoulomb forceGravitational force
Distance dependenceInverse square, ∝ 1/r²Inverse square, ∝ 1/r²
Source propertyElectric charge qMass m
Sign / natureAttractive or repulsiveAlways attractive
DirectionAlong the line joiningAlong the line joining
Constantk = 1/4πε₀ ≈ 9 × 10⁹G = 6.67 × 10⁻¹¹
Relative strength (e–p pair)F_electric / F_gravity ≈ 2.4 × 10³⁹

NCERT Example 1.3 makes the disparity concrete. For an electron–proton pair the ratio of the electric to the gravitational attraction is about $2.4 \times 10^{39}$, and for two protons it is roughly $10^{36}$. Gravity governs planetary motion only because bulk matter is electrically neutral; once the charges nearly cancel, the far weaker mass coupling is what remains to dominate at astronomical scales.

Builds on this

Coulomb's law gives the force for one pair. When several charges act at once, you add the pairwise forces as vectors — see Forces Between Multiple Charges for the superposition principle.

Coulomb force in a material medium

Coulomb's law in the form above gives the force in vacuum. When the same charges are immersed in a material medium, the situation is altered by the charged constituents of the matter, and the force is reduced. The medium is characterised by its relative permittivity $\varepsilon_r$, also called the dielectric constant — the ratio of the force in vacuum to the force in the medium for the same charges and separation:

$$\varepsilon_r = \dfrac{F_{\text{vacuum}}}{F_{\text{medium}}} = \dfrac{\varepsilon}{\varepsilon_0}$$

Since $\varepsilon_r > 1$ for every dielectric, the force in a medium is always smaller than in vacuum. Writing the permittivity of the medium as $\varepsilon = \varepsilon_0 \varepsilon_r$, the law becomes

$$F_{\text{medium}} = \dfrac{1}{4\pi\varepsilon_0\varepsilon_r}\,\dfrac{|q_1 q_2|}{r^2} = \dfrac{F_{\text{vacuum}}}{\varepsilon_r}$$

NEET Trap

A medium divides the force by εᵣ — it does not add

Inserting a dielectric weakens the force. The vacuum result is divided by $\varepsilon_r$, never multiplied. Water, with $\varepsilon_r \approx 80$, cuts the Coulomb force to about $1/80$ of its vacuum value, which is why ionic salts dissolve so readily in it.

$F_{\text{medium}} = F_{\text{vacuum}} / \varepsilon_r$, with $\varepsilon_r > 1$ always.

Worked applications

The figure makes the inverse-square fall-off visible: doubling $r$ quarters the force, tripling it cuts the force to one-ninth. This steep decay is the single most tested feature of the law.

Figure 2 · Force vs distance r F r 2r F F/4 F ∝ 1/r²
Worked Example

Two point charges $q_1 = +2~\mu\text{C}$ and $q_2 = +3~\mu\text{C}$ are placed 30 cm apart in vacuum. Find the magnitude and nature of the force between them. (Take $k = 9 \times 10^9~\text{N·m}^2/\text{C}^2$.)

Apply $F = k\,|q_1 q_2|/r^2$ with $q_1 = 2 \times 10^{-6}$ C, $q_2 = 3 \times 10^{-6}$ C, $r = 0.30$ m.

$$F = 9 \times 10^9 \times \dfrac{(2\times10^{-6})(3\times10^{-6})}{(0.30)^2} = \dfrac{9 \times 10^9 \times 6 \times 10^{-12}}{0.09} = 0.6~\text{N}$$

Both charges are positive, so the force is repulsive, directed along the line joining them. If the same pair were placed in a medium of $\varepsilon_r = 2$, the force would fall to $0.6/2 = 0.3$ N.

Quick Recap

Coulomb's law in one screen

  • For point charges in vacuum: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1 q_2|}{r^2}$, directed along the line joining.
  • $k = 1/4\pi\varepsilon_0 \approx 9 \times 10^9~\text{N·m}^2/\text{C}^2$; $\varepsilon_0 = 8.854 \times 10^{-12}~\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$.
  • Vector form $\mathbf{F}_{21} = k\,\dfrac{q_1q_2}{r^2}\hat{\mathbf{r}}_{21}$ encodes repulsion (like signs) and attraction (unlike signs); obeys Newton's third law.
  • Same form as gravitation (inverse square, along line joining) but far stronger and can be repulsive.
  • In a medium the force is reduced: $F_{\text{medium}} = F_{\text{vacuum}}/\varepsilon_r$, with the dielectric constant $\varepsilon_r > 1$.

NEET PYQ Snapshot — Coulomb's Law

Coulomb's law is one of the most reliably tested ideas in this chapter — the contact-charge ratio and inverse-square dependence recur often.

NEET 2025

Two identical charged conducting spheres A and B (centres a fixed distance apart) each carry charge $q$ and repel with force $F$. A third identical uncharged sphere is touched to A first, then to B, then removed. The new force of repulsion between A and B (treating them as point charges) is:

  1. 3F/8
  2. 3F/5
  3. 2F/3
  4. F/2
Answer: (1) 3F/8

Initially $F = kq^2/r^2$. The uncharged sphere touches A: charge shares equally, A becomes $q/2$ and the third sphere carries $q/2$. It then touches B (which has $q$): total $q/2 + q = 3q/2$ shared equally, so each becomes $3q/4$; B is left with $3q/4$. Now A has $q/2$, B has $3q/4$, giving $F' = k(q/2)(3q/4)/r^2 = \tfrac{3}{8}\,kq^2/r^2 = 3F/8$. This is exactly the contact-and-share trick Coulomb himself used to vary charges.

NEET 2016

Two identical charged spheres hang from a common point by strings of length $l$, initially a distance $d$ apart ($d \ll l$) due to mutual repulsion. Charge leaks off both spheres at a constant rate, so they approach with velocity $v$. Then $v$ varies with the separation $x$ as:

  1. $v \propto x^{-1}$
  2. $v \propto x^{-1/2}$
  3. $v \propto x$
  4. $v \propto x^{1/2}$
Answer: (2) v ∝ x^(−1/2)

For small angles, $T\sin\theta \approx kq^2/x^2$ and $T\cos\theta \approx mg$, giving $\tan\theta \approx (x/2)/l = kq^2/(x^2 mg)$, so $x^3 \propto q^2$, i.e. $q \propto x^{3/2}$. Differentiating with $dq/dt$ constant: $\tfrac{dq}{dt} \propto x^{1/2}\tfrac{dx}{dt} = x^{1/2}v = $ constant, hence $v \propto x^{-1/2}$.

FAQs — Coulomb's Law

The recurring conceptual points NEET aspirants ask about Coulomb's law.

What is Coulomb's law?
Coulomb's law states that the electrostatic force between two stationary point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them, acting along the line joining the two charges. In vacuum, F = (1/4πε₀)·q₁q₂/r², where 1/4πε₀ = k ≈ 9 × 10⁹ N·m²/C².
What is the value of the Coulomb constant k and of ε₀?
The Coulomb constant k = 1/4πε₀ ≈ 9 × 10⁹ N·m²/C² in SI units. The permittivity of free space ε₀ = 8.854 × 10⁻¹² C²·N⁻¹·m⁻². With these values, two charges of 1 C placed 1 m apart in vacuum repel with a force of 9 × 10⁹ N, which is why the coulomb is a very large unit.
How does the vector form of Coulomb's law handle attraction and repulsion?
The vector form is F₂₁ = (1/4πε₀)·q₁q₂/r²·r̂₂₁, where r̂₂₁ points from charge 1 to charge 2. The single equation handles both cases: if q₁ and q₂ have the same sign the product is positive, so F₂₁ is along r̂₂₁ (repulsion); if they have opposite signs the product is negative, so F₂₁ points along −r̂₂₁ (attraction). No separate formula is needed.
How does Coulomb's law compare with Newton's law of gravitation?
Both are inverse-square laws acting along the line joining the two bodies. The differences: gravitation depends on mass and is always attractive, whereas the Coulomb force depends on charge and can be attractive or repulsive. The electric force is vastly stronger — for an electron–proton pair the electric force is about 2.4 × 10³⁹ times the gravitational force.
How does a medium change the Coulomb force?
When the charges are placed in a material medium instead of vacuum, the force is reduced by the relative permittivity (dielectric constant) εᵣ of the medium. The force in the medium is F = F_vacuum/εᵣ, equivalently F = (1/4πε₀εᵣ)·q₁q₂/r². Since εᵣ > 1 for every dielectric, the force always decreases compared to vacuum.
Why must charges be point charges for Coulomb's law to apply directly?
Coulomb's law in the form F = (1/4πε₀)q₁q₂/r² is defined for point charges, where the size of the bodies is negligible compared with the separation r. For extended bodies the charge distribution and the distance to each element vary, so the force must be obtained by treating the body as a collection of point charges and applying superposition.
Related Topics
Electric Charges and Fields Return to the full chapter overview and subtopic map. Properties of Electric Charge Quantisation, additivity and conservation of charge. Forces Between Multiple Charges Superposition of pairwise Coulomb forces. Electric Field and Field Lines From force per charge to the field concept. Electric Dipole A pair of equal, opposite charges and its field. Gauss's Law and Applications Flux and symmetry as a shortcut to fields.