Coulomb’s law

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NCERT Class 12 Physics Chapter 1 Section 1.5

Quick answer

Coulomb’s law says that the force between two point charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them, and acts along the line joining them: F = (1/4πε_₀)|q_₁q_₂|/r^². In vacuum 1/4πε_₀ ≈ 9 × 10^⁹ N m^² C^⁻², with ε_₀ = 8.854 × 10^⁻¹² C^² N^⁻¹ m^⁻². Like charges repel and unlike charges attract, and the forces on the two charges are equal and opposite, F_₁₂ = −F_₂₁.

With your textbook

NCERT Physics Part I (Class 12), Chapter 1 “Electric Charges and Fields”, Section 1.5 “Coulomb’s Law”, pages 6–11.

Open the official NCERT chapter (PDF, 4.0 MB, ncert.nic.in) (opens in a new tab)

Checked against the 2026-27 reprint on 19 Oct 2026.

What is Coulomb’s law?

Coulomb’s law

Coulomb’s law is a quantitative statement about the force between two point charges.

NCERT Physics Part I (Class 12), Section 1.5, page 6

Point charges are charged bodies whose sizes are much smaller than the distance between them. Coulomb found that the force between two point charges varies inversely as the square of the distance between them, is directly proportional to the product of the magnitudes of the charges, and acts along the line joining them (NCERT pages 6–7):

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

In plain words: double either charge and the force doubles; double the distance and the force drops to a quarter. The force pushes like charges apart and pulls unlike charges together, always along the straight line joining them, and each charge feels the same size of force.

Force between two point charges

For point charges q₁ and q₂ a distance r apart in vacuum, the magnitude of the force is (NCERT Eqs. (1.1) and (1.2), pages 7–8):

Coulomb’s law (NCERT Eq. 1.2)

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

k = \frac{1}{4\pi\varepsilon_0}

k \approx 9 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}

How Coulomb did it (NCERT page 7): he measured the forces with a torsion balance. The charges on his metal spheres were unknown, so he made known fractions of them: touching a charged sphere with an identical uncharged one splits the charge equally by symmetry, giving q/2, then q/4 and so on. He varied the distance for a fixed pair of charges, then varied the charges at a fixed distance, and arrived at the law. It has been established from the macroscopic scale down to distances of about 10⁻¹⁰ m.

Choosing a unit of charge. In NCERT Eq. (1.1), k could be any positive number, and its value fixes the size of the unit (page 7). This explains a possible choice of electrical units, not the modern SI definition: today the ampere is defined by fixing the elementary charge, and 1 C = 1 A s (beyond the NCERT text: the SI since 2019; Wikipedia, “Ampere”). With k ≈ 9 × 10⁹ N m² C⁻², two charges of 1 C each, 1 m apart in vacuum, repel with a force of 9 × 10⁹ N: one coulomb is far too big a unit for electrostatics, which uses µC and mC instead (NCERT pages 7–8).

Permittivity of free space

The constant is written k = 1/4πε₀ for later convenience, where ε₀ is the permittivity of free space (NCERT page 8):

Permittivity of free space

\varepsilon_0 = 8.854 \times 10^{-12}

in C² N⁻¹ m⁻² (NCERT page 8).

With this value, 1/4πε₀ = 8.99 × 10⁹ N m² C⁻², the value NCERT uses in Example 1.3 (8.987 × 10⁹, page 10). NCERT rounds it to about 9 × 10⁹ (page 7), the value to use unless a problem gives another. The dimensions of ε₀ follow from its unit, with C = A s and N = kg m s⁻²: [M⁻¹ L⁻³ T⁴ A²].

The law above is for charges in vacuum. When matter lies between the charges, its own charged constituents complicate the picture; NCERT takes this up in the next chapter (page 9). Beyond the NCERT text: for fixed point charges in an infinite, homogeneous, linear isotropic dielectric of relative permittivity K, the force at the same distance is F/K, K times weaker than in vacuum. In this macroscopic model the medium weakens the field between the charges by the factor K (Wikipedia, “Relative permittivity”), so its permittivity ε = Kε₀ takes the place of ε₀. NCERT itself brings in K = ε/ε₀ only for capacitors, in Section 2.13 (page 70). For air K is very close to 1, so air is treated as vacuum.

Coulomb’s law in vector form

Force is a vector, so NCERT writes the law with position vectors r₁ and r₂ of the two charges (pages 8–9):

r₂₁ = r₂ − r₁ points from charge 1 to charge 2, and r̂₂₁ = r₂₁/r₂₁ is the unit vector along it.
F₂₁ is the force on q₂ due to q₁.

Coulomb’s law in vector form (NCERT Eq. 1.3)

\mathbf{F}_{21} = k\,\frac{q_1 q_2}{r_{21}^{2}}\,\hat{\mathbf{r}}_{21}

k = \frac{1}{4\pi\varepsilon_0}

The charges go in with their signs. If q₁ and q₂ have the same sign, F₂₁ is along r̂₂₁, away from q₁: repulsion. If their signs differ, F₂₁ is along −r̂₂₁, towards q₁: attraction. One equation covers both cases (page 8).
Swapping 1 and 2 gives the force on q₁: F₁₂ = −F₂₁. Coulomb’s law agrees with Newton’s third law (page 9).

Coulomb’s law after NCERT Figure 1.3: position vectors r1 and r2 of two charges and r21 from 1 to 2; like charges pushed apart, unlike charges pulled together, by equal and opposite forces. — The geometry of Coulomb’s law and the forces for like and unlike charges, after NCERT Figure 1.3 (page 8). Schematic.

Comparison with the gravitational force

Both laws fall off as 1/r², but the electric force is enormously stronger. NCERT Example 1.3 (pages 9–10) compares them (worked below as Example 1):

For an electron and a proton, F_e/F_G = e²/(4πε₀Gm_pm_e), about 2.3 × 10³⁹. NCERT prints 2.4 × 10³⁹, but the masses NCERT gives, with e = 1.6 × 10⁻¹⁹ C and G = 6.67 × 10⁻¹¹ N m² kg⁻², give 2.3 × 10³⁹.
For two protons, the ratio is about 1.2 × 10³⁶. NCERT prints 1.3 × 10³⁶; the same values give 1.2 × 10³⁶.
NCERT slip (page 9): both printed ratios, 2.4 × 10³⁹ and 1.3 × 10³⁶, come out a little high; NCERT’s own data give 2.3 × 10³⁹ and 1.2 × 10³⁶ (see Example 1). In an exam, work with the constants given; an NCERT-based key may print 2.4 × 10³⁹ and 1.3 × 10³⁶.
Gravity always attracts; the electric force between two protons repels. Inside a nucleus, about 10⁻¹⁵ m apart, two protons repel with F_e ~ 230 N, while their gravitational attraction is only F_G ~ 1.9 × 10⁻³⁴ N (page 9).

For the gravitational side, see the universal law of gravitation.

Solved examples

Example 1: Electric against gravitational force (NCERT Example 1.3)

(a) Find the ratio of the electric to the gravitational force (i) for an electron and a proton, (ii) for two protons. (b) Find the accelerations of an electron and a proton due to their mutual electric attraction 1 Å = 10⁻¹⁰ m apart. Take m_p = 1.67 × 10⁻²⁷ kg, m_e = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, G = 6.67 × 10⁻¹¹ N m² kg⁻².

Answer: (a) The distance cancels, because both forces go as 1/r².

(i) Electron and proton:

\frac{F_e}{F_G} = \frac{e^2}{4\pi\varepsilon_0\,G m_p m_e}

With k = 9 × 10⁹ N m² C⁻² and e² = 2.56 × 10⁻³⁸ C², the top is ke² = 2.30 × 10⁻²⁸ N m². For the bottom, Gm_p = 6.67 × 10⁻¹¹ × 1.67 × 10⁻²⁷ = 1.11 × 10⁻³⁷, and multiplying by m_e = 9.11 × 10⁻³¹ kg gives Gm_pm_e = 1.01 × 10⁻⁶⁷ N m². So

\frac{F_e}{F_G} \approx 2.3 \times 10^{39}

(ii) Two protons: replace m_e by m_p, which gives about 1.2 × 10³⁶.

(NCERT prints 2.4 × 10³⁹ and 1.3 × 10³⁶, page 9; recomputed with these constants the ratios are 2.3 × 10³⁹ and 1.2 × 10³⁶.)

(b) The force has the same size on both particles:

F = \frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r^2}

With 1/4πε₀ = 8.987 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C and r = 10⁻¹⁰ m:

F = 2.3 \times 10^{-8}\ \text{N}

So a_e = F/m_e = (2.3 × 10⁻⁸ N)/(9.11 × 10⁻³¹ kg) = 2.5 × 10²² m s⁻², and a_p = F/m_p = 1.4 × 10¹⁹ m s⁻²: gravity is negligible beside these (NCERT page 10).

Example 2: Halving the charges and the distance (NCERT Example 1.4)

Charged metal spheres A and B, 10 cm apart, repel with a force F. Identical uncharged spheres C and D touch A and B, and are removed; then B is brought to 5.0 cm from A. What is the new force?

Answer: touching splits each charge equally, so A and B now carry q/2 and q′/2. The new force is

F' = \frac{1}{4\pi\varepsilon_0}\,\frac{(q/2)(q'/2)}{(r/2)^2}

F' = \frac{1}{4\pi\varepsilon_0}\,\frac{q q'}{r^2} = F

The force is unchanged: each charge was halved, but the distance was halved too, and the 1/r² factor quadruples the force (NCERT pages 10–11).

Example 3: Two charges in vacuum

Charges of +2 µC and −3 µC are 30 cm apart in vacuum. Find the force between them.

Answer: here |q₁q₂| = 2 × 10⁻⁶ × 3 × 10⁻⁶ = 6 × 10⁻¹² C² and r² = (0.30 m)² = 0.09 m². Then k|q₁q₂| = 9 × 10⁹ × 6 × 10⁻¹² = 5.4 × 10⁻² N m², so

F = \frac{5.4 \times 10^{-2}}{0.09}\ \text{N}

F = 0.60\ \text{N}

The charges have opposite signs, so the force is attractive: each charge is pulled towards the other with 0.60 N (NCERT Eq. (1.2), page 8).

Example 4: Using the vector form

A charge q₁ = +1.0 µC is at the origin and q₂ = +2.0 µC is at the point (3 m, 4 m). Find the force F₂₁ on q₂ and the force F₁₂ on q₁.

Answer: r₂₁ = r₂ − r₁ = (3î + 4ĵ) m, so r₂₁ = 5 m and r̂₂₁ = 0.6î + 0.8ĵ. With kq₁q₂ = 9 × 10⁹ × 2.0 × 10⁻¹² = 1.8 × 10⁻² N m², the magnitude is

F = \frac{1.8 \times 10^{-2}\ \text{N m}^2}{25\ \text{m}^2}

F = 7.2 \times 10^{-4}\ \text{N}

Both charges are positive, so F₂₁ is along r̂₂₁: F₂₁ = (4.32î + 5.76ĵ) × 10⁻⁴ N, pushing q₂ away from the origin. By the third law, F₁₂ = −F₂₁ (NCERT Eq. (1.3), pages 8–9).

Common mistakes to avoid

Common mistake

Dropping the square on r

The force falls as 1/r^²: doubling the distance cuts the force to a quarter, not a half (NCERT page 7).

Common mistake

Mixing up F_₁₂ and F_₂₁

NCERT’s F_₂₁ is the force on q_₂ due to q_₁, along r̂_₂₁ from charge 1 to charge 2 for like charges. The two forces are equal and opposite (NCERT pages 8–9).

Common mistake

Applying the law to large, close bodies

Coulomb’s law is for point charges: bodies much smaller than the distance between them. For big spheres close together the charges are not at single points (NCERT page 6).

Quick revision summary

Coulomb’s law: the force between two point charges is proportional to the product of their magnitudes, inversely proportional to the square of the distance, and acts along the line joining them.
F = (1/4πε_₀)|q_₁q_₂|/r^², with 1/4πε_₀ ≈ 9 × 10^⁹ N m^² C^⁻² and ε_₀ = 8.854 × 10^⁻¹² C^² N^⁻¹ m^⁻² (in vacuum).
Vector form: F_₂₁ = (1/4πε_₀)(q_₁q_₂/r_₂₁^²) r̂_₂₁, with the signs of the charges; F_₁₂ = −F_₂₁, as Newton’s third law requires.
Coulomb found the law with a torsion balance, halving charges by touching identical spheres.
The electric force between an electron and a proton is about 10^³⁹ times their gravitational attraction.

Formula sheet

F = (1/4πε_₀) |q_₁q_₂|/r^²
1/4πε_₀ ≈ 9 × 10^⁹ N m^² C^⁻²
ε_₀ = 8.854 × 10^⁻¹² C^² N^⁻¹ m^⁻²
F_₂₁ = (1/4πε_₀)(q_₁q_₂/r_₂₁^²) r̂_₂₁, r_₂₁ = r_₂ − r_₁
F_₁₂ = −F_₂₁

Questions students ask

What are the limitations of Coulomb’s law?

Coulomb’s law, F = k|q₁q₂|/r^², is a law for point charges at rest in vacuum: charged bodies much smaller than their separation (NCERT pages 6–7 and 9). It also holds for two non-overlapping spherically symmetric charge distributions whose charge profiles remain fixed, with r measured between centres, since outside such a sphere the field is as if all its charge were at the centre (NCERT page 40; see field of a shell). But two charged metal bodies brought close push each other’s free charges around, so the formula no longer gives the force. In an ordinary linear dielectric filling the space, the macroscopic force is reduced by its relative permittivity, and moving charges also exert magnetic forces, the subject of NCERT Chapter 4.

Why does Coulomb’s law follow an inverse square law?

Coulomb’s law is an inverse square law because that is what experiment shows: Coulomb measured the force at different separations with a torsion balance and found that it varied as 1/r^² (NCERT pages 6–7). A picture makes it natural: field lines from a point charge spread evenly over spheres of area 4πr^², so their density, like the field, falls as 1/r^² (see density of field lines). Gauss’s law rests on this exact power (NCERT page 31); beyond the NCERT text, Williams, Faller and Hill found in 1971 that the power of r differs from 2 by at most a few parts in 10^¹⁶ (Physical Review Letters, volume 26).

Why is there a 4π in Coulomb’s law?

The 4π in Coulomb’s law comes from writing the constant as k = 1/4πε₀, which NCERT does “for later convenience” (NCERT page 8). Since 4πr^² is the area of a sphere, this choice makes the flux of a point charge through a sphere around it (q/4πε₀r^²) × 4πr^² = q/ε₀, with no 4π left (NCERT pages 29–30; see Gauss’s law). So the 4π is moved into Coulomb’s law to keep Gauss’s law, and the fields worked out from it, simpler; the value of the constant is unchanged, 1/4πε₀ ≈ 9 × 10^⁹ N m^² C^⁻².

What are the unit and dimensional formula of k in Coulomb’s law?

The constant k = 1/4πε₀ in Coulomb’s law has the SI unit N m^² C^⁻² and the dimensional formula [M L^³ T^⁻⁴ A^⁻²]: rearranging the law gives k = Fr^²/(q₁q₂), with force [M L T^⁻²], r^² [L^²] and charge [A T] (see dimensional formulae). In vacuum k ≈ 9 × 10^⁹ N m^² C^⁻² (NCERT page 7). The permittivity ε₀ has the reciprocal unit, C^² N^⁻¹ m^⁻², and the reciprocal dimensions.

Why is the force between two charges weaker in a medium than in vacuum?

The force between two charges is weaker in a medium because the medium is polarised: each charge’s field slightly separates, or lines up, the charges in nearby molecules, so each charge is surrounded by induced charge of the opposite sign that partly cancels its field (NCERT Chapter 2, Section 2.10, pages 65–67; see polarisation). Beyond the NCERT text, for a homogeneous linear isotropic dielectric filling all the space around the charges the force falls to F/K, where K = ε/ε₀ > 1 is the dielectric constant, the factor by which, NCERT shows, a dielectric cuts the field in a capacitor (page 70).

Practice questions

Try each question first, then open the answer to check your working.

Q1. The distance between two point charges is tripled. By what factor does the force change?

Show answerHide answer

It becomes 1/9 of its value, because F ∝ 1/r^² (NCERT page 7).
Q2. Two equal charges 1 m apart in vacuum repel with a force of 9 N. Find each charge.

Show answerHide answer

q^² = Fr^²/k = 9 × 1/(9 × 10⁹) = 10⁻⁹ C^², so |q| ≈ 3.2 × 10⁻⁵ C, about 32 µC. Both charges may be positive or both negative; the given repulsion does not determine their common sign (NCERT Eq. (1.2), page 8).
Q3. In the vector form of Coulomb’s law, what does a negative product q_₁q_₂ tell you about the direction of F_₂₁?

Show answerHide answer

That F_₂₁ points along −r̂_₂₁, towards q_₁: the force is attractive (NCERT page 8).

Coulomb’s law

What is Coulomb’s law?

Force between two point charges

Permittivity of free space

Coulomb’s law in vector form

Comparison with the gravitational force

Solved examples

Example 1: Electric against gravitational force (NCERT Example 1.3)

Example 2: Halving the charges and the distance (NCERT Example 1.4)

Example 3: Two charges in vacuum

Example 4: Using the vector form

Common mistakes to avoid

Quick revision summary

Questions students ask

What are the limitations of Coulomb’s law?

Why does Coulomb’s law follow an inverse square law?

Why is there a 4π in Coulomb’s law?

What are the unit and dimensional formula of k in Coulomb’s law?

Why is the force between two charges weaker in a medium than in vacuum?

Practice questions

Sources and review

Topics in Electric Charges and Fields