Coulomb's Law കൂളോം നിയമം
How strong is the push or pull between two charges, and how does it change with distance? In 1785 Charles Augustin de Coulomb answered this with a twisting wire. His law is the most used formula of this chapter – every field, every superposition problem and half the board numericals start from it. On this page you will state it, write it in vector form, play with it in a lab and solve the textbook's Examples 1.4 and 1.5.
- Point charges and q = ne (page 6); μC = 10−6 C, cm → m (page 2).
- Vectors, unit vectors and position vectors, r21 = r2 − r1 (page 1); Newton's third law and F = ma; gravitation F = Gm1m2/r² (page 3).
1. The law
Coulomb's law (കൂളോം നിയമം) is a quantitative statement about the force between two point charges (ബിന്ദു ചാർജ്) – charged bodies whose sizes are much smaller than the distance between them. Coulomb found that the force between two point charges
- is directly proportional to the product of the magnitudes of the charges, q1q2;
- varies inversely as the square of the distance r between them – an inverse square law (വിപരീത വർഗ നിയമം);
- acts along the line joining the two charges.
How did Coulomb find it? The torsion balance
Coulomb used a torsion balance – a very sensitive force meter: a light horizontal bar hangs from a thin fibre; a small force on a sphere at the end of the bar twists the fibre, and the angle of twist measures the force. (Cavendish later used one to measure the feeble gravitational force.) But Coulomb did not know the charges on his spheres! His trick: touch a charged metal sphere (charge q) with an identical uncharged sphere. By symmetry the charge is shared equally – each gets q/2. Repeating gives q/4, q/8, … (This silently uses additivity and conservation of charge: q/2 + q/2 = q.) He varied the distance for a fixed pair of charges, then varied the charges keeping the distance fixed – and arrived at Eq. (1.1).
The twist angle is proportional to the force. Halve the distance → the twist becomes 4 times. Halve the charge on one sphere → the twist halves. A cartoon with an exaggerated angle; Coulomb's real angles were small.
👤 Charles Augustin de Coulomb (1736–1806)
A French physicist who began his career as a military engineer in the West Indies. In 1776 he returned to Paris and retired to a small estate to do research. He invented a torsion balance to measure forces, used it to measure the attraction and repulsion between small charged spheres, and in 1785 arrived at the inverse square law. Priestley and Cavendish had anticipated the law (Cavendish never published). Coulomb also found the inverse square law for the force between magnetic poles. The SI unit of charge, the coulomb, is named after him.
2. The constant k, ε₀ and the size of one coulomb
Coulomb discovered his law without knowing the size of any charge. Today it works the other way: Coulomb's law can be used to define a unit of charge. In Eq. (1.1) k could be any positive number; the choice of k fixes the size of the unit of charge. In SI units
Put q1 = q2 = 1 C and r = 1 m: F = 9 × 109 × 1 × 1 ÷ 12 = 9 × 109 N. So 1 C is the charge which, placed 1 m from an equal charge in vacuum, repels it with a force of 9 × 109 N – roughly the weight of a million loaded trucks! One coulomb is far too big for electrostatics; we use mC and μC. For later convenience k is written as
ε0 is the permittivity of free space (ശൂന്യസ്ഥലത്തിന്റെ പെർമിറ്റിവിറ്റി): ε0 = 8.854 × 10−12 C2 N−1 m−2. Check: 1 ÷ (4π × 8.854 × 10−12) = 8.99 × 109 ≈ 9 × 109 N m2 C−2. ✔ The law has been tested from the laboratory scale down to subatomic distances (r ~ 10−10 m).
3. Coulomb lab
Arrow lengths grow with the logarithm of the force so that both tiny and huge forces fit. On the graph, the orange point is at 2r: the force there is exactly one quarter. Notice that the two forces are always equal and opposite, even when q₁ ≠ q₂ – Newton's third law.
4. Coulomb's law in vector form (Fig 1.6)
Force is a vector (സദിശം), so we write the law with vectors. Let the position vectors of q1 and q2 be r1 and r2. F12 is the force on q1 due to q2; F21 is the force on q2 due to q1. The vector from 1 to 2 is r21 = r2 − r1, and from 2 to 1, r12 = r1 − r2 = −r21; their magnitudes are equal (r12 = r21). Directions are given by the unit vectors (യൂണിറ്റ് സദിശം) r̂21 = r21/r21 and r̂12 = r12/r12 = −r̂21.
Drag the two dots (mouse or finger). Grey: position vectors from the origin O. Blue: r21 = r2 − r1 and its unit vector. Red: the forces – they always lie along the line joining the charges and are equal and opposite.
- One equation for both cases. If q1q2 > 0 (like charges), F21 is along r̂21 – repulsion. If q1q2 < 0 (unlike), F21 is along −r̂21 (= r̂12) – attraction. No separate equations are needed.
- Newton's third law. Interchanging 1 and 2: F12 = 14πε0 q1q2r122 r̂12 = −F21.
- Vacuum only. The law gives the force between charges in vacuum. If matter is present, its charged constituents complicate the situation (Chapter 2).
5. Example 1.4 – electric force versus gravity
Q. Coulomb's law for the electrostatic force between two point charges and Newton's law for the gravitational force between two stationary point masses both have inverse-square dependence on distance. (a) Compare the strength of these forces by finding the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of the electron and the proton due to their mutual electric attraction when they are 1 Å (= 10−10 m) apart. (mp = 1.67 × 10−27 kg, me = 9.11 × 10−31 kg)
Write Fe = ke²/r² and FG = Gm1m2/r². In the ratio, r² cancels – so the answer does not depend on distance. Use k = 9 × 109, G = 6.67 × 10−11 N m² kg−2, e = 1.6 × 10−19 C. For (b), the same force acts on both particles (third law): a = F/m.
(b) F = 2.3 × 10−8 N; ae = 2.5 × 1022 m s−2; ap = 1.4 × 1019 m s−2. Compared with g ≈ 9.8 m s−2, gravity is negligible for the motion of the electron.
Note: the textbook prints 2.4 × 1039 and 1.3 × 1036. With the data given, the ratios work out to 2.27 × 1039 and 1.24 × 1036; either is accepted – what matters is the order of magnitude, 1039 and 1036.
6. Example 1.5 – halve the charges, halve the distance (Fig 1.7)
Q. A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B, held by an insulating handle, is brought close to A so that the distance between their centres is 10 cm [Fig 1.7(a)]. The repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively [Fig 1.7(b)]. C and D are removed and B is brought closer to A, to a distance of 5.0 cm between their centres [Fig 1.7(c)]. What is the expected repulsion of A on the basis of Coulomb's law? A and C, and B and D, are identical. Ignore the sizes of A and B compared with their separation.
Touching an identical uncharged sphere halves the charge. Write F′ with q/2, q′/2 and r/2, and compare with F = kqq′/r².
- 🧪 Coulomb's Law (PhET simulation) PhET, University of Colorado Boulder
Change the charges and the distance and read the force. - 📄 Coulomb's Law The Physics Classroom
Worked examples of the inverse-square law. - 📄 Coulomb's law Wikipedia
The law, its vector form and the superposition principle. - 📄 Charles-Augustin de Coulomb Wikipedia
The French physicist and his torsion balance.
7. Check yourself
- F = (1/4πε0) q1q2/r², along the line joining two point charges in vacuum; 1/4πε0 ≈ 9 × 109 N m² C−2, ε0 = 8.854 × 10−12 C² N−1 m−2.
- 1 C: two 1 C charges 1 m apart repel with 9 × 109 N.
- Vector form: F21 = (1/4πε0)(q1q2/r21²) r̂21; F12 = −F21.
- Electric force between e and p ≈ 1039 times their gravitational attraction.
- Double the distance → ¼ of the force; halve both charges and the distance → same force.